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Chapter 4
(Equilibrium)
Dr.VikramPanchal
Institute Of Chemistry
Worksheet
Sem–2
Section – B
(1). What is meant by reversible reaction?
In reversible reactions, forward and reverse reactions continuously occur.
(2). What is called chemical equilibrium?
The equilibrium established in chemical reactions is called chemical equilibrium.
(3). How can be said that equilibrium is dynamic in nature?
The equilibrium is dynamic and not steady as the forward and the reverse reactions occur with the same
velocity at the equilibrium time in equilibrium reactions in closed vessel.
(4). How many types of equilibrium are there? Which are they?
There are two types of equilibrium.
Physical
(ii)
Chemical
(5). What is called reverse reaction?
The change of product to reactant is called the reverse reaction.
(6). What is meant homogeneous equilibrium? Give its example.
When all reactants and products are in same physical state in a reversible reaction, it is called
homogeneous equilibrium. E.g. Haber process
(7). What is meant by heterogeneous equilibrium? Give its example.
When all reactants and products are not in same physical state in a reversible reaction, it is called
heterogeneous equilibrium. E.g. decomposition of CaCO3
(8). Mention requirements of chemical equilibrium.
The reverse reaction should occur in a closed system with constant temperature and pressure.
(9). Mention the law of active masses.
‘The driving force of chemical reaction is directly proportional to the active masses of reactants.’
(10). Which scientists gave the law of mass action?
Norvegian scientist Guldberg and Waage in 1864 proposed the law of mass action.
(11). Interpret Ke = Kc.
Ke = [conc. Of products] / [conc. Of reactants]
Then, Keq = Kc
(12). Interpret Keq = Kp.
Ke = partial pressure of products / partial pressure of reactants
Keq = Kp
(13). Write formula of Kp for NH4COONH2(s)
2NH3(g) + CO2(g)
3
Kp = 4/27 p
(14). Give formula of equilibrium constant for decomposition of NH4HS(s).
NH4HS(s)
NH3(g) + H2S(g)
Kp = p2/4
(15). Write formula showing relation between Kc and Kp.
Kp = Kc (RT)∆n
∆n= np-nr
(16). Give the law of chemical equilibrium.
(17) Mention the formula for equilibrium constant for the reaction
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(l)
(18). Under which conditions Kp and Kc will be equal?
When ∆n = np – nr = 0, then Kp = Kc.
(19). When Kp > Kc?
When ∆ng > 0, then Kp > Kc.
(20). Write le- Chatelier’s principle.
If from the factors determining the equilibrium state, any one factor is changed, there will be such a
change in the system that the effect will be nullified or made negligible so that the value of equilibrium
constant at that temperature will remain constant.
(21). What idea can be obtained by Le- Chatelier’s principle?
We can study in detail the effect of change in external factors like concentration, pressure and
temperature at equilibrium.
(22). Why concentrations of components remain constant at equilibrium?
Equilibrium state is the end of reaction and at the end conc. Of each component is constant.
(23). Which are the factors affecting equilibrium?
The external factors like, concentration, pressure and temperature can affect equilibrium.
(24). Give equation showing relation between ∆G and Qc
∆G = ∆G0 + RTlnQc
(25). Give unit for the equilibrium constant of the reaction N2(g) + 3H2(g)
2NH3(g)
M2-4 = M-2
Kp = p2NH3/(pN2 x p3H2) (atm)-2
(26). If Qc < Kc and Qc > Kc then in which directions the reaction will occur?
If Qc < Kc, the reaction will occur in forward direction.
If Qc > Kc, then the reaction will occur in reverse direction.
(27). Give formula of Kc for the reaction Ag2O(s) + 2HNO3(aq)
H2O(l)
2AgNO3(aq) +
Because [Ag2O], [H2O] are constants.
(28). Write relation between ∆G0 and equilibrium constant K.
∆G0 = -RT ln K or K = e-∆G0/RT
(29). What will be the effect of catalyst on equilibrium?
The catalyst reduces time required to complete the reaction but does not affect equilibrium state.
(30). In the reaction A2 + B2
2AB + Q cal., what should be decreased to
increase the proportion of AB?
On decreasing temperature, the rate of endothermic i.e., forward reaction increases.
(31). Mention function of catalyst.
The catalyst decreases the energy of activation.
(32). Complete the reaction
Fe3+(aq) + SCN-(aq)
[Fe(SCN)]2+(aq)
Light yellow colour colourless
colour like blood
(33).In how many sections, Michael Faraday classified substances?
(i) electrolyte
(ii) Non- electrolyte.
(34). What is called an electrolyte? Give example.
When a substance dissolves in water and ionized so its aqueous solution can conduct electricity, it is
known as electrolyte. Eg. : NaCl
(35). What is called non electrolyte? Give example.
When a substance dissolves in water or remain insoluble in water and cannot conduct electricity is known
as non – electrolyte. Eg.: C6H12O6
(36). What is called strong electrolyte? Give example.
An electrolyte whose solubility is more than 0.01 M or completely ionized in aqueous solution, known as
strong electrolyte.
(37). What is called dissociation?
In the dissociation, the positive ion and the negative ion present in the original substance are separated
showing dissociation, viz.,
Na+Cl-(s) + H2O(l)
Na+(aq) + Cl-(aq)
(38). What is called ionization?
In ionization original substance is changed into ion form in aqueous solution , viz.
CH3COOH(l) + H2O(l)
CH3COO-(aq) + H3O+(aq)
(39). Approximately how many litres of HCl is secreted in the stomach of human being?
In the stomach of a human being about 1.2 to 1.5 litres HCl is secreted.
(40). Which acid is included in vinegar?
CH3COOH is included in vinegar.
(41). What can be called acid and base according to Arrhenius theory?
Substance which dissociate in water and give H+ are called acids and substance which dissociate in water
and give out ions are called bases.
(42). Mention two drawbacks of Arrhenius theory.
(i) Proton is highly unstable.
(ii) It cannot exist independently.
(43). Give definitions of acid-base according to Lowry – Bronsted theory.
The substance which gives proton is called the acid and substance which receives proton is called base.
(44). Give definitions of Lewis acid – base.
The substance which accept electron pair is acid and the substance which donate electron pair is base.
(45). Which substances do not obey the definition of Lowry – Bronsted acid – base?
BF3, SiO2, Ag+, Cu+2 etc do not obey definition of Lowry – Bronsted theory.
(46). What is used to measure strength of different acid – base according to Lowry – Bronsted theory.
According to Lowry – Bronsted theory, the substance has more capacity to donate proton is acid and more
capacity to accept proton is base.
(47). Write conjugate acids of water and dimethyl amine.
H2O – H3O+
(conjugate acid of H2O)
(CH3)2NH – (CH3)2N+H2
(conjugate acid of dimethyl amine)
(48). What will be the conjugate acid of (C6H5)2NH ?
(C6H5)2NH → (C6H5)2NH2+ is a conjugate acid.
(49). Write conjugate acid – base of H2PO4-.
H2PO4-1 - H3PO4 (conjugate acid)
H2PO4-1 - HPO42- (conjugate base)
(50). Give definition of conjugate base.
When Lowry – Bronsted acid lose proton it gives conjugate acid.
(51). Give definition of conjugate acid.
When Lowry – Bronsted base accept proton, it gives conjugate acid.
(52). What indicates the values of Ka and Kb ?
Ka = ionization constant of weak acid.
Kb = ionization constant of weak base.
(53). CH3NH2 is stronger base in comparison to NH3. Why ?
KbCH3NH2 > KbNH3
CH3NH2 has +I group with nitrogen which increases electron density on nitrogen and also increases basic
strength.
(54). Which substances act as Lewis acid – base?
The substance which can accept electron pair is acid and the substance which can donate electron pair is
base.
(55). Classify the given substances in Lewis acid – base:
AlCl3, BF3, CH3CH3OH, NH3, NH2-, Ag+
Lewis acid- AlCl3, BF3, Ag+, CH3CH2OH
Lewis base- NH3, NH2-, CH3CH2OH
(56). What is meant by self ionization of water?
H2O + H2O
H3O+(aq) + OH-1(aq)
(57). What is meant by ionic product of water?
Kw = [H3O+] = [OH-] = 1 x 10-14 at 298 K.
A product of [H3O+] and [OH-] is known as ionic product of water.
(58). What is concentration of pure water at 298 K temperature ?
[H2O] = 55.5 M
(59). Give definition of pH.
A negative logarithm to the base 10 of H3O+ ion concentration is known as pH.
(60). Give definition of pOH.
A negative logarithm to the base 10 of OH- ion concentration is known as pOH.
(61). What will be the sum of values of pH and pOH at 298 K temperature ?
At 298 K, pH + pOH = 14.
(62). On the basis of which values of pH, solutions can be said to be acidic, basic or neutral?
pH = 7
neutral solution
pH < 7
acidic solution
pH > 7
basic solution
(63). What is the pH of CH3COONa solution ?
The pH of CH3COONa is greater than 7.
(64). What is the pH of FeCl3 solution ?
The pH of FeCl3 solution is less than 7.
(65). What will be the effect of aqueous CaCl2 solution on litmus paper ?
The aqueous solution of CaCl2 is acidic therefore it turns blue litmus paper to red.
(66). What is meant by hydrolysis of salt ?
When salt is added to water, ions of salt react with water and produce acid and base is known as
hydrolysis.
(67). Which instrument is used to determine accurate pH of solution?
pH meter is used to measure pH.
(68). What is meant by hydrolysis constant ?
Hydrolysis reaction is an equilibrium reaction and so its corresponding equilibrium constant can be
calculated which is known as hydrolysis constant.
(69). What is called buffer solution ?
The solution which resists the change in pH carried out by addition of acid or base in small proportion to
them or are being diluted, and the values of their pH remain constant are called buffer solution.
(70). Give one example each of acidic, basic and neutral buffer solutions.
Acidic buffer – CH3COOH + CH3COONa
Basic buffer – NH4OH + NH4Cl
Neutral solution – CH3COOH + NH4OH
(71). Give examples of sparingly soluble salts.
AgCl, CuS, Al(OH)3, BaSO4 etc. are sparingly soluble salts.
(72). Give definition of solubility product (Ksp).
The product of concentration of ions of sparingly soluble salt in its aqueous solution is known as
solubility product.
(73). In which solvent, non polar substances like naphthalene dissolves ?
Naphthalene dissolves in non- polar solvents like benzene, toluene etc.
(74). Give formula for solubility product (Ksp) of sparingly soluble salt Sb2S3.
Sb2S3(aq)
2Sb3+(aq) + 3S2-(aq)
2s
3s
Ksp = [Sb3+]2[S2-]3 = [2S]2[3S]3 = 10855
(75). Mention the conditions between Ip and Ksp.
When Ip > Ksp, salt will precipitate.
Ip < Ksp, no precipitation.
(76). Give four examples of sparingly soluble salts.
AgCl, CuS, CdS, BaSO4.
(77). What is indicated by the value of Ksp of sparingly soluble salt ?
The value of Ksp indicates the solubility of a given salt.
(78). What is not included in the equation of Ksp equilibrium constant of a sparingly soluble salt?
Concentration of sparingly soluble salt and concentration of H2O because both are constant.
(79). What is called common ion effect ?
When concentration of one of the ions of sparingly soluble salt is increased, so that solubility of sparingly
soluble salt decreases and the rate of reverse reaction increases.
(80). The ions of which group are not precipitated in presence of NH4OH and NH4Cl in qualitative
analysis?
The ions of III B, IV, IV A groups are not precipitated in presence of NH4Cl and NH4OH.
(81). The ions of which group are precipitated in presence of HCl and H2S in qualitative analysis?
The ions of II groups are precipitated in presence of HCl and H2S.
(82). What is meant by ionic product Ip?
The product of cations and anions of sparingly soluble salt obtained from two different soluble salts is
known as Ip.
(83).How can acidic buffer solution be prepared? Give example.
Acidic buffer solution can be prepared by mixture of weak acid and its salt with strong base.
Eg: CH3COOH + CH3COONa
(84). How can basic buffer solution be prepared? Give example.
Basic buffer solution can be prepared by mixture of weak base and its salt with strong acid.
Eg: NH4OH + NH4Cl
(85). Mention the importance of buffer solutions.
The pH of solution can be maintained by adding buffer solution . eg: pH of blood is 7.35.
(86). How can neutral buffer solution be prepared?
Neutral buffer solution can be prepared by neutralization of weak acid and weak base.
Section-C
1. Mention the operational and conceptual definitions of acid and base.
Ans. Acid means such a substance which is (1) Sour in taste (2) Turns wet Blue litmus paper
red. (3) forms salt and water by reacting with base. (4) In certain circumstances produces
hydrogen gas by reacting with metals. Similarly, base means such a substance which is (1) bitter
in taste. (2) turns wet red litmus blue. (3) form salt and water by reacting with acid.
These are operational definitions. The modern definitions used at present are called conceptual
definitions in which Arrhenius, Bronsted-Lowry and Lewis definitions of acid base are included.
Arrhenius concept about Acid and Base:
According to Arrhenius concept, substances which dissociate in water and give hydrogen ion
(H+) are called acids and substances which dissociate in water and give hydroxyl ions (OH-) are
called bases.
Concept of Bronsted – Lowry for acid and base:
The substance which gives a proton or donates a proton is called the acid and the substance
which receives the proton or accepts a proton is called the base.
Lewis Concept of Acid and Base :
Lewis mentioned that acid means a substance which can accept a pair of electrons and base
means a substance which can donate a pair of electrons.
2. Derive the formula for dissociation constant of a weak acid.
Ans. Ionization Constant of Weak Acid (Ka) :
In aqueous solution of weak monobasic acid HA, there is a partial ionization and so the
equilibrium is obtained as below :
HA(aq) + H2O(l)
H3+O(aq) + A-(aq).
Suppose the initial concentration of weak acid (HA) is C mollit-1 and degree of ionisation
is α, then following can be written :
Reaction :
HA(aq) + H2O(l)
H3+O(aq) + A-(aq).
Initial
Concentration(M)
C
0
0
Degree of ionisation(α)
1-α
α
α
Concentration at equilibrium(M) (1 - α)C
αC
αC
[H3O+][A-]
Equilibrium constant Ke= [HA][H2O]
=
(αC)(αC)
….1
(1 - α)C[H2O]
But [H2O] is accepted as constant and so
Ke[H2O] = (αC)(αC)
(1 - α)C
= α2C
= Ka
…2
(1 - α)
Where Ka is the ionization constant or dissociation constant of the Weak acid HA.
3. Derive the formula for dissociation constant of a weak base.
Ans. ionisation constant (Kb) of weak base:
The ionization of monoacidic weak base MOH will take place in aqueous solution as follows:
H2O
MOH(aq)
M+(aq) + OH-(aq)
As base is weak, incomplete ionization will occur and so equilibrium will be obtained and it can
be expressed as below:
[M+] [OH-] …………1 and
Kc = [MOH] [H2O]
[M+] [OH-]
=Ke ………. 2
Kex[H2O]=
[MOH]
Where Kb is the ionization or dissociation constant of monoacidic weak base. If we know the
initial concentration of weak base and its degree of ionization.
4. Obtain the relation between equilibrium constants Kc and Kp.
Ans. Relation between Kp and Kc
As seen earlier the equilibrium constant of a gaseous reaction can be written as
…. 1
But we know that according to simple gas equation PV = nRT. Hence, it can be written as
 P = (n/V)RT = CRT
(where n/V = C = concentration in mol lit-1)
Substituting the values of p in the above equation 1, it can be written as
….2
….3
= Kc X (RT)ng …. 4
Where ∆ng = (c+d) – (a+b)
Means number of total moles of gaseous products minus number of total moles of gaseous
reactants.
Hence, it can be written as
Kp = Kc X (RT)ng…. 5
5. Derive the formula for solubility product of sparingly soluble salt CaF2.
Ans. CaF2
Ca2+ + 2FS
2S
2+
- 2
Ksp = [Ca ][F ] = [S][2S]2 = 4S3.
6. Mention the use of effect of common ion in qualitative analysis.
Ans. The use of effect of common ion can be made to separate one ion from the other in presence
of other ion in qualitative analysis. It can also be used for decrease in solubility of the
components in the mixture. In qualitative analysis, the solubility products of sulphides of metal
ions of second group are less in comparison to solubility products of sulphides of metal of III B
group ions, therefore, HCl is added before adding H2S water to test the second group ions.
H2S(aq)
2H+(aq) + S2-(aq)
HCl(aq)
H+(aq) + Cl-(aq)
The common ion available from HCl creates common ion effect on the equilibrium and
decreases concentration of S2- ions. Hence, the sulphides of the ions included in second group
can only be precipitated because their solubility products are less. In the same way, for
precipitation of ions of group III A ammonium chloride (in excess) is added along with NH4OH.
The concentration of OH- available from ionization of NH4OH gets decreased due to common
ion effect due to NH+4 available from NH4Cl. Hence, the hydroxides of the ions of III A group
only will be precipitated because the values of solubility products of the hydroxides of III A
group are low. Above this, if HCl gas is passed through saturated solution of NaCl, NaCl
becomes insoluble and separates as precipitates due to the common ino effect of Cl- ions.
It is necessary to note that under certain situations the solubility increases instead of decreasing.
The solubility of salt like phosphates increase when acid is added to their solutions or pH of the
solution decreases. The reason for this is that, phosphate ion combines with H+ available from
acid and converts into acid by accepting proton. Hence, solubility of phosphate salt increases.
7. Give definition of buffer, explain acidic, basic and neutral buffer, giving suitable
examples.
Ans. “ The solution which resists the change in pH carried out by addition of acid or base in
small proportion to them or are being diluted, and the values of their pH remain constant are
called buffer solution.”
(i) Acidic buffer solution: Acidic buffer solution can be prepared by mixture of weak acid and its
salt with strong base.
(ii) Basic buffer solution: Basic buffer solution can be prepared by mixture of weak base and its
salt with strong acid.
(iii) Neutral buffer solution: Neutral buffer solution can be prepared by neutralization of weak
acid and weak base. These type of buffer solutions are shown below:
Type
Substances
Value of pH
Acidic
CH3COOH + CH3COONa
<7
Basic
NH4OH + NH4Cl
>7
Neutral CH3COOH + NH4OH
=7
8. Bronsted-Lowry acid can be a Lewis acid but all Lewis acids cannot be Bronsted-Lowry
acid. Explain giving suitable example.
Ans. As far as bases are concerned, there is not much difference between Lowry-Bronsted and
Lewis concepts as a base provides a lone-pair in both the cases. Further, the compounds which
are classified as acids by Lowry-Bronsted concepts also acts as a lewis acids. However, Lewis
concept fro an acid includes many substances that do not have a proton. Some such compounds
are given below
Lewis acid + Lewis Base
BF3
+ :NH3
F3B
NH3
BF3
+ :F
BF4
Ag+(aq) + 2:CN-(aq)
Ag(CN)-2(aq)
+
Ag (aq) + 2:NH3(aq)
Ag(NH3)+2(aq)
AlCl3 + :Cl
AlCl4In the above reactions, :NH3, :F , :CN-, :Cl- all contain a lone-pair of electrons (depicted as : )
Utilizing which they react with species like BF3, Ag+ and AlCl3. Here BF3, Ag+ AlCl3 accept a
lone-pair of electrons so they are lewis acid.
While species like : NH3, :F-, :CN-, :Cl- donate electron pair and hence they are lewis bases. Thus
a molecule acting as a lewis acid can be a positive ion as a molecule having a deficit of an
electron pair. Similarly, a molecule acting as a Lewis acid can be an anion or a molecule having
a lone-pair of electrons.
9. Derive the formula of ionic product of water
Ans. Ionic Product of Water
Water is an amphoteric oxide when acid is added to it; it accepts the proton and act as a base and
when base is added to it, it donates a proton and acts as acid. When reaction between two
molecules of water takes place, one molecule donates proton and other molecule receives proton
and shows conjugate acid-base reaction.
H2O(l) + H2O(l)
H3O+(aq)
+
OH-(aq)
Acid-1
Base-2
Conjugate acid-2
Conjugate base-2
If we express the equilibrium constant of above reaction them.
[H3O+][OH-]
Ka =
[H2O][H2O]
Where Ka is the dissociation constant of acid. There is no significant change in concentration of
water (55.5M) because H2O is a weak acid (possesses about 10-7M H+). Thus, if H2O is
considered constant.
Ka x [H2O]2 = [H3+O] [OH-] = Kw.
Where Kw is ionic product of water. Water is neutral and so [H3+O] and [OH-] in it are 10-7 M.
Hence, Kw = [H3+O] [OH-] = (10-7)(10-7) = 10-14 which is constant and equilibrium constant
remains constant at constant temperature; so the value of Kw will be constant at constant
temperature, viz., the value of Kw is 1 x 10-14 at 298 K.
10. Explain calculation of pH of a solution from dissociation constant
Ans. In aqueous solution of weak monobasic acid HA, there is a partial ionization and so the
equilibrium is obtained as below :
HA(aq) + H2O(l)
H3+O(aq) + A-(aq).
Suppose the initial concentration of weak acid (HA) is C mollit-1 and degree of ionisation
is α, then following can be written :
Reaction :
HA(aq) + H2O(l)
H3+O(aq) + A-(aq).
Initial
Concentration(M)
C
0
0
Degree of ionisation(α)
1-α
α
α
Concentration at equilibrium(M) (1 - α)C
αC
αC
+
[H3O ][A ]
Equilibrium constant Ke= [HA][H2O]
= (αC)(αC)
…1
(1 - α)C[H2O]
But [H2O] is accepted as constant and so
Ke[H2O] = (αC)(αC)
(1 - α)C
= α2C
= Ka
…2
(1 - α)
Where Ka is the ionization constant or dissociation constant of the s\acid HA. (At this stage we
consider both the terms same) Hence, for any weak monobasic acid, following can be written
[H3O+][A-]
Ke=
[HA]
…3
-1
The unit of Ka will be mollit .
Similarly Kb = [OH-][M+]
[MOH]
As the values of Ka depend on [H3+O] the values of ka will be different for different.
Lesser the value of Ka, weaker will be the acid. The value of Ka will be constant at definite
temperature.
According to the relations seen earlier,
pH = -log10[H3+O]
pOH = -log10[OH-]
Similarly, pKa = -log10[Ka]
PA- = -log10[A-] can be written. Where [A-] is the concentration of negative ion.
11. Discuss the dynamic nature of equilibrium.
Ans. Dynamic Nature of Equilibrium
The most important matter in the case of equilibrium is that there is a continuous transformation
of reactant to product and product to reactant. This state appears to be steady but it is not so. This
type of reaction which takes place in both the directions is called reversible reaction and it is
expresses by the symbol of two half arrows(↔). This symbol indicates that such reaction occurs
simultaneously in both (forward and reverse) directions. Generally, the change of reactant to
product is called forward reaction and the change of product to reactant is called reverse reaction.
Thus, in reversible reactions, forward and reverse reactions continuously occur and we find it as
equilibrium state. The mixture of reactants and products obtained at equilibrium time is called
equilibrium mixture. The decomposition reaction of solid calcium carbonate in a cold vessel, at
high temperature can be shown as below.
∆
CaCO3(s) ↔CaO(s) + CO2(g)
The equilibrium is dynamic and not steady or static as the forward and the reverse
reactions occur with the same velocity at the equilibrium time in equilibrium reactions in closed
vessels. In the above reaction obtaining CaO and CO2 by decomposing of CaCO3and obtaining
CaCO3 by combination of CaO and CO2 continuously take place. Suppose, if we deposit, some
amount in our bank account and withdraw the same amount, then balance in the account appears
steady or static. But this can be considered operative or dynamic and not closed or static. It is
very difficult to determine the dynamic nature of equilibrium, even then with the help of
radioactive isotope, it can be proved, viz. 14CO2 gas containing radioactive isotope, 14C and
CaCO3 are taken in two different flasks and CO2 obtained by decomposition is connected with
vessel containing 14CO2 gas, after sometime, Ca14CO3 will be formed in the vessel of CaCO3 and
CO2 will be obtained in the vessel containing 14CO2. Thus if the equilibrium would have been
steady, there must not be exchange of 12C and 14C. With the help of suitable counter, the
radioactivity can be measured and the proof for the dynamic nature of equilibrium can be
obtained through the proportions of concentrations of reactants and products remain constant.
The reaction can be fast or slow depending upon the nature of the reactant and the experimental
conditions.
Equilibrium reactions can be divided into following three categories:
(i) Reactions which are almost at the extent of completion and concentration of
reactants may be negligible. It is not possible to detect this experimentally.
(ii) Reactions in which the products are formed in very less proportions and most part
of the reactant remains unchanged at the equilibrium.
(iii) Reactions in which the concentrations of reactants and products are in comparable
proportions at equilibrium.
12. Discuss the dynamic nature of chemical equilibrium
Ans. Dynamic Nature of Chemical Equilibrium
The dynamic nature of chemical equilibrium can be demonstrated by taking example of the
production of ammonia. By keeping known quantities of dinitrogen and dihydrogen at high
temperature and pressure in a closed vessel, the amount of ammonia formed can be determined at
constant intervals by a series of experiments. The quantities of unreacted dinitrogen and
dihydrogen also can be determined. From this it is concluded that even if the reactants and
products are in different proportions, their concentrations are same at equilibrium. This
constancy in composition indicates dynamic nature of equilibrium. For this, in synthesis of
ammonia, deuterium (D2) instead of dihydrogen, (H2) is used and ammonia gas is produced by
Haber process and it is studied. The results obtained are similar to those obtained above. In the
mixture proportions of N2, D and ND3 instead of N2, H2, NH3 can be determined and equilibrium
can be obtained. If D2 is added after the formation of ammonia by reaction of N2 and H2, the
reaction may not occur but H in NH3 is displaced by D and ND3 can be determined by mass
spectrometer. Thus, it is proved that in the reaction
That the reactions from reactants to products and products to reactants that is forward and
reverse reactions continuously occur with the same rates and so ND3 instead of NH3is obtained.
By the use of radioactive isotope, the dynamic nature of equilibrium can be proved viz. For the
reaction, H2(g) + I2(g)
2HI(g)
131
Radioactive isotope I of iodine can be used to study the dynamic nature of chemical
equilibrium. As the equilibrium is dynamic, certain properties or factors are found similar. E.g.
Intensity of colour, constant pressure, constant concentration etc.
13. Explain effect of catalyst on equilibrium
Ans. Effect of catalyst:
The use of suitable catalyst always helps to increase the rate of reaction, viz. iron powder is used
as the catalyst in the production of ammonia by Haber’s process. Use of catalyst is associated
with chemical kinetics because it affects the rate of reaction. The function of catalyst is to
decrease the energy of activation. Hence, the reaction easily moves forward towards product.
During this reaction, the energy of activation decreases but has no effect on equilibrium constant,
that is, more proportion of product cannot be obtained. Let us examine the case of reaction of
ammonia gas obtained by combination of dinitrogen gas and dihydrogen gas with the help of
Haber’s process.
N2(g) + 3H2(g)
2NH3(g)
∆H = -92.38 KJ molIn the above reaction, total 2 moles product are obtained from total 4 moles of reactants.
Hence, according to Le Chatelier’s principle, increase in pressure is advantageous to get more
ammonia but for the reactions occurring in closed vessel and so the pressure has to be kept
limited. In addition, this reaction is exothermic and so according to Le Chatelier’s principle, the
decrease in temperature is advantageous but decrease in temperature affects the rate of reaction.
Hence, more time is required for completion of the reaction, which is not economically
advantageous in industry. Hence, temperature is also to be restricted. By making compromise
with these two and using catalyst whereby the energy of activation is decreased and increasing
rate of reaction, more possible product is obtained in less time. Hence, Haber used iron powder
as catalyst in the production reaction of ammonia and satisfactory results were obtained. In the
production of ammonia, the values of optimum pressure and temperature respectively and iron
powder is used as catalyst. If the value of equilibrium constant of the reaction Kc is very low then
the use of catalyst is not fruitful or helpful.
14. Explain effect of concentration of equilibrium
Ans. Effect of change in concentration:
If we add or remove the reactant or the product from the reaction in equilibrium, its effect on
equilibrium according to Le Chatelier’s principle will be as follows:
(a)If the concentration of reactant or product is increased by addition of reactant or product, the
reaction will occur in such a way that the increase in concentration will be taken for use, i.e. the
increase in concentration of reactant, the concentration of product will increase and if
concentration of product is increased the reaction will result in the direction of increase in
concentration of the reactant.
(b)If the concentration of reactant or product is decreased by removing reactant or product, the
reaction will occur in such a way that product or reactant will be established again.
Hence, if any change in concentration of reactant of product is carried out then the equilibrium
will try to make this effect minimum and equilibrium will be established accordingly. If we take
this as an example, in the reaction
If concentration of reactants dinitrogen or dihydrogen is increased, the reaction will proceed
towards right hand side and the product ammonia obtained will be more. If the concentration of
nitrogen or hydrogen is decreased, the reaction will proceed towards left hand side and reactants
of nitrogen or hydrogen will be obtained back i.e. production of ammonia will decrease. Here, it
is necessary to remember that 1 mole dinitrogen combines with 3 moles of dihydrogen in this
reaction and forms 2 moles of ammonia. Hence, increase in concentration of dihydrogen rather
than dinitrogen, will give more product. We take another example of heterogeneous equilibrium.
If solid CaCO3(s) is heated in closed vessel, the following decomposition reaction will occur.
CaCO3(s)
CaO(s) + CO2(g)
Hence, if more CaO(s) is to be obtained then, the CO2(g) formed in the reaction should be removed
because CO2(g), gas can combine with solid CaO(s) and carry out reverse reaction then proportion
of product will decrease. Hence, by removing CO2(g) from the reaction vessel, more CaO(s) can
be obtained.
15. Do the concentration of reactant and product remain constant at the time of
equilibrium? Why?
Ans. In the beginning of the reaction, the reactants which are in closed vessels, slowly change
into products and the concentration of reactant decreases as the time proceeds. Along with this
phenomenon, there is increase in concentration of product as the time proceeds. A certain point
of time comes when there is no change in the concentration of reactant and product even if the
time proceeds. This situation is called equilibrium state.
16. Deduce the formula of decomposition constant of solid NH4HS in a closed vessel.
Ans. NH4HS(s)
NH3(g) + H2S(g)
x
x
= total 2x moles
Kp = (PNH3) x (PH2S) = ( ½)P x (½)P = ( ¼)P2 atm2
17. Explain the effect of temperature on equilibrium by taking suitable example.
Ans. Effect of temperature:
The value of equilibrium constant is associated with temperature, i.e. the value of equilibrium is
constant at constant temperature. If temperature changes its value also changes. Reactions can be
of two types (1) Exothermic and (2) Endothermic, with the change in temperature, there is a
change in absorbed or evolved heat. In exothermic reaction when heat is absorbed it works as a
reactant. In exothermic reaction when heat is evolved, it works as a product. Hence, it can be
said that the equilibrium constant increases with increase in temperature in endothermic reaction;
equilibrium constant decreases with increase in temperature in exothermic reaction. Especially,
let us make it clear that the increase or decrease in temperature affects the rate of reaction. Let us
think about production of NH3(g).
∆H means change in enthalpy which is the difference in total enthalpies of products and
reactants, its positive value indicates endothermic reaction and negative value indicates
exothermic reaction. Hence, the above reaction is exothermic. As seen earlier that increase in
temperature is not favourable for exothermic reaction because reaction will go in reverse
direction and will decrease the product. Hence, decrease in temperature is advantageous to obtain
more products but by decreasing the temperature the rate of reaction decreases. Hence, more
time is required for completion of the reaction.
Hence, as a compromise, at the lowest possible temperature and at the highest possible
pressure, catalyst is used. Hence, more possible product is obtained in less possible time, this
type of state is called optimization state.
18. Give Le-Chatelier’s principle and mention its importance.
Ans. “ If from the factors determining the equilibrium state, any one factor is changed, there will
be such a change in the system that the effect will be nullified or made negligible so that the
value of equilibrium constant at that temperature will remain constant.” This principle can be
applied to both physical and chemical equilibrium.
19. Explain the effect of change in pressure on equilibrium by taking suitable example.
Ans. Effect of change of pressure
The change in pressure can be carried out by increasing or decreasing the concentration of the
gas or respectively decreasing or increasing the volume of the vessel. By carrying out this type
of change, there will be change in proportions of gaseous reactants or products or total products.
Le Chatelier’s principle can also be applied to such reactions. In heterogeneous equilibrium if we
do not take into consideration the effect of pressure on solid or liquid substances in equilibrium,
it can work because their volumes and concentration are independent of change of pressure. Let
us take the following example:
In this reaction 1 mole of reactant CO(g) reacts with 3 moles of reactant H2(g) and forms 1 mole
product CH4(g) and 1 mole product H2O(g). thus 2 moles product is obtained from 4 moles of
reactants. Hence, there is decrease in number of moles during reaction.
Suppose, the pressure on the closed vessel in which the reaction is carried out at constant
temperature and the volume of equilibrium mixture is made half, then what will happen? Total
pressure will be double because PV = constant. The concentrations or pressures of reactants or
products are increased. Hence, according to Le Chatelier’s principle, the equilibrium will try to
attain the original state. As pressure is doubled ad 2 moles of products are obtained and 4 moles
of reactants, there is decrease in number of moles and the reaction will go in forward direction,
i.e. more product will be obtained.
As a reverse of this, if in reaction CaCO3(s)
CaO(s) + CO2(g)
The number of moles of products increases (from 0 to 1). If pressure is increased by addition of
CO2(g), then, the reaction will become reverse reaction and will decrease the product.
20. Explain the effect of increase in temperature on the system 2SO2(g) + O2(g)
2SO3(g)
+ heat
Ans. The above reaction is an exothermic reaction. According to Le-Chatelier’s principle,
increase in temperature shifts equilibrium to the left and decrease in concentration of products
21. What will be the effect on product in the system N2(g) + 3H2(g)
2NH3(g) by increasing
the pressure? Explain on the basis of Le-Chatelier’s principle.
Ans. N2(g) + 3H2(g)
2NH3(g)
1 mole 3 mole
2 mole
In this reaction total 2 moles of products are obtained from 4 moles of reactants. According to
Le-Chatelier’s principle, as pressure is increased and 2 moles of products are obtained and 4
moles of reactants, there is decrease in number of moles and the reaction will go in forward
direction i.e more product will be obtained
22. Discuss homogeneous and heterogeneous equilibrium taking suitable examples
Ans. (i) Homogeneous Equilibrium: In homogeneous equilibrium all the reactants and products
will be in one similar phase. viz. The product ammonia produced by the reaction of reactants
dinitrogen and dihydrogen, all are in gaseous phase. N2(g) + 3H2(g)
2NH3(g)
Similarly, the hydrolysis of methyl acetate is also homogeneous equilibrium because, the
reactants and the products in it are also in the same phase (liquid phase)
CH3COOCH3(l) + H2O(l)
CH3COOH(l) + CH3OH(l)
In addition to this, the reaction between aqueous solution containing Fe3+ ion and aqueous
solution containing SCN- ions is also an example of homogeneous equilibrium. But the only
change in it is that this equilibrium is ionic equilibrium.
Fe3+(aq) + SCN-(aq)
[Fe(SCN)]2+(aq)
All the equilibrium reactions and their equilibrium constants that we discussed earlier are the
examples of homogeneous equilibrium and also derived from the relations between Kp, Kc and
Kx related to them. For calculation of values of Kp pressures must be expressed in unit of bar
because, bar is the unit in standard condition but in SI unit it is Pascal. The relation between
them is as follows:
1 Pascal Pa = 1 Nm-2(Newton meter-2) and
1 bar = 105 Pa
(ii) Heterogeneous Equilibrium :
If reactants and products, posses more than one phase, the equilibrium is called heterogeneous
equilibrium,
The equilibrium between water (liquid) and water vapour(steam) in a closed vessel is the
example of heterogeneous equilibrium.
H2O(l)
H2O(g)
Similarly, Ca(OH)2(s)
Ca2+(aq) + 2OH-(aq)
Is also the example of heterogeneous equilibrium.(Equilibrium between solid and liquid). It is
necessary to note here that this is an example for ionic equilibrium.
Generally, in heterogeneous equilibrium, pure solid or liquid are associated, the concentrations
of reactants and products can be separated, viz. The concentration of pure solid or liquid is its
density and density is constant at constant temperature. Hence, concentration can also be taken as
constant or the pure solid or liquid which are present will be independent of concentration.
Suppose, some substance X is involved in this, then concentrations X(s) and X(l) will be taken as
constant in whatever proportion they are present while concentrations of X(g) and X(aq) will
change and will vary with volume.
23. What will be the effect on equilibrium system H2O(g) + CO(g)
H2(g) + CO2(g) if water
vapour is introduced ?Explain.
Ans. Increase in vapour will lead to increase in reactants because H2O(g) is in vapour state.
Hence, acc. to law, equilibrium is shifted in forward direction and hence, product increases and
Kc also increases.
24. What will happen if Cl2 is removed from the equilibrium system PCl5(g)
PCl3(g) +
Cl2(g)? Explain.
Ans. If Cl2(g) is removed (i.e product is removed) from the system, conc. Of product will
decrease and acc. to law, equilibrium shifts in forward direction resulting in increased conc. Of
products [i.e Cl2] and hence increases Kc.
25. Mention general characteristics of equilibrium involved in physical process.
Ans. The following are the general characteristics of systems in equilibrium for the physical
systems
(i) At constant temperature, equilibrium is possible only in closed system.
(ii) Both the processes (reactions) opposite to each other, that is, the forward and the reverse
reactions occur at the same rates and the equilibrium involved in it remains dynamic but
condition remains steady (static).
(iii)All the properties f the system which can be measured remains constant.
(iv)For physical processes, when equilibrium is obtained then at constant temperature, the value
of one of the factors remains constant. The list of these properties is shown in Table 4.1
(v)At any stage magnitude of such quantities show to what extent the physical process has
advanced before reaching to equilibrium.
26. Predict the direction of forward or reverse reaction
Ans. R
P
If concentration of [R] increases, reaction is in forward direction.
If concentration of [R] decreases, reaction is in reverse direction.
If concentration of [P] increases, reaction is in reverse direction.
If concentration of [P] decreases, reaction is in forward direction.
27. Give conjugate acid-base theory giving suitable example.
Ans. Danish chemist Bronsted, an English chemist Lowry presented the concept of acid and
base. They made H+ (proton) as a base. According, to their concept , the substance which gives a
proton or donates a proton is called the acid and the substance which receives the proton or
accepts a proton is called the base. Thus, acid is a proton donor and base is proton acceptor. Let
us take the dissociation reaction of hydrogen chloride in water.
Base
Acid
H3+O (aq) + Cl-(aq)
HCl(g) + H2O(l)
Acid
Similarly,
Base
Acid
NH3(g) + H2O(l)
Base
NH4+(aq)+ OH-(aq)
Base
Acid
We shall understand in detail the first from the above reactions :
HCl
H+
+ ClAcid-1 proton conjugate base-1
As it gives proton, HCl is an acid
H2O + H+
H3O+
Base-2
Proton
Conjugate acids-2
As it accepts a proton, H2O is a base.
Total reaction:
HCl + H2O
H3O+(aq) + Cl-(aq).
Acid-1
Base-2
Acid-2
Base-1.
In the above reaction giving-taking of proton is not shown. Hence, it can be said that only
transfer of proton takes place, it is not obtained free. Every acid will lose proton and so its
conjugate base will be formed and every base will accept a proton and so its conjugate acid will
be formed. Hence, this concept is known as proton transfer or conjugate acid-base concept.
28. Mention the relation between solubility product and ionic product
Ans. Ksp > Ip then precipitation will take place
Ksp < Ip then precipitation will not take place.
Ksp = Ip then reaction will be in equilibrium.
29. Water is amphoteric. Explain it on the basis of Lowry-Bronsted theory
Ans. Danish scientist Bronsted and English chemist Lowry presented the concept of acid and
base. They made H+(proton) as a base. According to their concept, the substance which gives a
proton or donates a proton is called the acid and the substance which receives a proton or accepts
a proton is called the base. Water also donates H+ and also receives a proton hence water is
amphoteric.
30. How can it be said that pure water is neutral?
Ans. Ionic product of water is 10-14 and moreover [H3O+] = [OH-] = 10-7 which is constant and
equilibrium constant remains constant at constant temperature. Hence, water is neutral.
31. What is the value of ionic product of water at 298 K temperature? What is the effect of
temperature in its value? why?
Ans. The value of ionic product of water at 298K is 1 x 10-14. As the process of self ionization of
water is endothermic, with increase in temperature, reaction goes in forward direction resulting
in increase concentration of [H+] and [OH-] ions. Hence, with increase in temperature, Kw
increases.
32. Aqueous solution of FeCl3 is acidic. Explain
Ans. FeCl3 is a salt of strong acid and weak base.
FeCl3
Fe3+(aq) + 3Cl-(aq)
HCl
H+ + Cl- (acid is strong, so it gets fully dissociated)
Fe(OH)3
Fe3+ + 3OH- (weak base so partially dissociates)
Moreover, hydrolysis constant for strong acid and weak base is Kh = Kw/Kb = [OH-]/Co. Hence
pH of solution is < 7. Hence it is acidic.
33. Aqueous solution of CH3COONa is basic. Explain
Ans. CH3COONa is a salt of strong base and weak base.
CH3COONa
CH3COO- + Na+
CH3COOH
CH3COO- + H+ (partial dissociates)
+
NaOH
Na + OH- (completely dissociates)
Moreover Kh of weak acid and strong base is Kh = Kw/Ka = [H3O]2/Co. Hence pH of solution is
>7. Hence it is basic.
34. Deduce the formula of Ksp of sparingly soluble salt bismuth sulphide – Bi2S3.
Ans. Bi2S3
2Bi3+ + 3S23+ 2 2- 3
Ksp = [Bi ] [S ] = (2S)2(3S)3 = 108S5
35. Deduce the formula of Ksp of sparingly soluble salt Zinc Phosphate- Zn3(PO4)2.
Ans Zn3(PO4)2
3Zn2+ + 2PO433S
2S
2+ 3
3- 2
3
Ksp = [Zn ] [PO4 ] = [3S] [2S]2 = 27S3 x 4S2 = 108S5
36. Deduce the formula of Ksp of sparingly soluble salt aluminium hydroxide Al(OH)3
Ans. Al(OH)3
Al3+ + 3OHS
3S
3+
- 3
3
 Ksp = [Al ][OH ] = [S][3S] = 27S4
37. The solubility of AgCl decreases in presence of NaCl. Explain giving reason
Ans. A saturated aqueous solution of silver chloride has following equilibrium :
AgCl(s)
Ag+(aq) + Cl-(aq)
If, to this solution, a small amount of NaCl is added, the concentration of Cl - will be increased.
Therefore, some cl- ions will combine with Ag+ ions to form insoluble AgCl. Thus the solubility
of AgCl decreases because of common ion effect
38. Derive formula of Ksp for sparingly soluble salt- Ca3(PO4)2.
Ans. Ca3(PO4)2
3Ca2+ + 2PO433S
2S
2+ 3
3- 2
3
Ksp = [Ca ] [PO4 ] = [3S] [2S]2 = 27S3 x 4S2 = 108S5
39. The solubility of AgCl decreases by addition of AgNO3 to a saturated solution of AgCl.
Explain
Ans. AgCl(s)
Ag+(aq) + Cl-(aq)
If, to this solution, a small amount of AgNO3 is added, the concentration of Ag+ will be
increased. Therefore, some Ag+ ions will combine with Cl- ions to form insoluble AgCl. Thus the
solubility of AgCl decreases because of common ion effect
40. White precipitates are obtained when HCl gas is passed through a saturated solution of
NaCl. Give reason
Ans. In the solution of some electrolytes the concentration of undissociated electrolysis increases
under the effect of common ion effect. For e.g, on passing HCl gas through a saturated solution
of NaCl, the concentration of Cl- will increase. As a result of this NaCl is precipitated. Common
ion effect is utilized for the purification of impure NaCl obtained from sea water.
41.After addition of small amount of HCl in a mixture of Cu2+ and Zn2+, if H2S gas is
passed from the solution, CuS is precipitated but ZnS is not precipitated. Give reason
Ans. Use is made of common ion effect in qualitative analysis. If H2S gas is passed through an
acidified solution (acidified by HCl) containing Cu2+ and Zn2+ ions, only copper is precipitated as
CuS. Zns is not precipitated. In the presence of HCl, the ionization of H2S is decreased.
Consequently, the concentration of S2- ions being much low, only CuS is precipitated.
42. The value of Kp for the reaction NH2COONH4(s)
atm3. What will be the value of Kc for this reaction?
Ans. Data: Kp = 3.2 x 102 atm3
T = 298K
R = 0.082 atm lit mol- k-.
∆n(g) = np- nr (in gaseous state)
= 3- 0 = 3.
Kp = Kc x (RT)∆n(g) So, Kc = Kp
(RT)∆n(g)
So, Kc = 3.2 x 102
(0.082 x 298)3
Kc = 3.2 x 102
(24.436)2
2NH3(g) + CO2(g) is 3.2 x 102
So, Kc = 2.193 x 10-2 (mol/litre)3.
43. The value of Kp = 600 atm3 is found for the reaction when solid NH4COONH2(s) is
heated at 400 K temperature; What will be the value of total pressure of the system at
equilibrium state?
Ans. NH4COONH2(s)
2NH3(g) + CO2(g)
From, the equation, use have at equilibrium, that PNH3 will be twice than that of Pco2.
At equilibrium, let Pco2 be x, then PNH3 will be 2x
KP = [NH3]2 x [CO2]
= (2x)2 x X
= 4x3
Kp = 600 atm3 (from data)
= 4x3
 x3 = 150 atm3
X = 5.172 atm (from log)
Now, total pressure
Ptotal = PNH3 + Pco2
= 2x + x
= 3x
= 3 (5.172)
Ptotal = 15.516 atm.
44. The value of Kp = 0.05 atm for the reaction A(g) + 2B(g)
3C(g) + D(g). Write the
value Kc terms of R at 1000 K temperature.
Ans. Data: Kp = 0.05 atm
T = 1000 K
∆n(g) = np – nR ( In gaseous state)
=4–3
=1
Kp = Kc x (RT)∆n(g)
Kc = Kp
(RT)∆n(g)
= 0.05
(R x 1000)-1
= 5 x 10-2
(R x 103)1
Kc = 5 x 10-5 atm K-1
R
45. What will be the value of Kp if the pressure at equilibrium is 3 atm for the reaction.
NH3COONH2(s)
2NH3(g) + CO2(g) ?
Ans. NH3COONH2(s)
2NH3(g) + CO2(g)
From the equation, use have, at equilibrium PNH3 will be there than that of PCO2.
At equilibrium, at PCO2 = x  PNH3 = 2x
Ptotal = PNH3 + PCO2 = 2x + x = 3x
Nowwe are given, Ptotal = 3atm
3x = 3 atm
 x = 1 atm
Now, Kp = (PNH3)2 x PCO2 = (2x)2 x X
= 4x3
Kp = 4x3 = 4 x (1 atm)3
Kp = 4 atm3.
46. In the reaction H2(g) + I2(g)
2HI(g), 0.4 moles H2 and I2 are taken. At equilibrium
0.5 mole HI is formed. Find equilibrium constant Kp.
Ans. Data:
Reaction:
H2(g) + I2(g)
2HI(g)
Conc. at
Equilibrium 0.15 0.15
0.15
[ Total no. of moles of reactant initially of H2 and I2 are 0.4 moles each, hence total no. of moles
= 0.8 moles.
While at equilibrium, 0.5 moles of product are formed and at equilibrium [H2] = [I2]
→ 0.3 moles are unreacted at equilibrium
[H2] = [I2] = 0.15 moles (at equilibrium)]
Kc = [HI]2
[H2] [I2]
= (0.5)2
(0.15) (0.15)
= 0.25
0.0225
Kc = 11.11
Now, Kp = Kc x (RT)∆n(g)
∆n(g) = np – nR (In gaseous state)
= 2 – (H1)
=0
Kp = Kc x (RT)0
 Kp = Kc
Kp = 11.11 [Kp has no unit because ∆n(g) = 0]
47. Calculate the pH of 0.5M HF(aq). Ka = 2 x 10-1.
Ans. Data:
Ka = 2 x 10-4
H2O
HF
H3O+ + F0.5M
Ionisation constant, Ka = [H3O+] [F-]
[HF]
[As, ratio of H3O+ and F- ion is 1:1, then moles would be equal
Ka = [H3O+]2
HF
-4
2 x 10 = [H3O2]
 [H3O+]2 = 1 x 10-4
0.5
+ 2
[H3O ] = 1 x 10-2 M
PH = -log [H3O+] or –log [H+]
= -log [1 x 10-2]
= -log1 + -log(10-2)
= 0 + 2 (log 10)
PH = 2
 [H3O+] = [F-]]
log (axb) = log a + log b
log ab = b log a
log10 = 1
log1 = 0
48. KSP of Mg(OH)2 = 1 x 10-12 At what pH 0.01M Mg(OH)2 will be precipitated?
Ans. Mg(OH)2(s)
Mg2(aq) + 2OH-(aq)
Let the solubility be S mol/litre
 [Mg2+] = S mol/litre
[OH-] = 2S mol/litre
KSP = [Mg2+][OH-]2
= [S] [2S]2
= 4S3
KSP = 1 x 10-12 given
KSP = 4S3 = 1 x 10-12
→ S3 = ¼ x 10-12 M3
S = (¼)1/3x 10-4 M
[
(¼)1/3= log 1/3 (1-4)
= 1/3 (log1 – log4)
= 1/3 (0 – 0.6021)
= - 0.2007
= anti log (- 0.2007)
= 1 + antilog (0.7993)
= 6.299 x 10-1
S = 6.299 x 10- M ]
 [OH-] = 2 x 6.299 x 10-5 M
= 12.598 x 10-5 M
= 1.2598 x 10-4 M
POH = -log [OH-]
= -log (1.2598 x 10-4)
= 4 log 10 – log (1.2598)
= 4 – 0.01
= 3.99
PH + POH = 14
→ PH = 14 - POH
= 14 – 3.99
PH = 10.01
49. What will be the concentration of H+ when 0.1M HCN is mixed with 0.2M NaCN ?
Ka = 6.2 x 10-10
[Hint : [H+] = Ka x Concentration of Acid solution
Concentration of Salt solution
Ans. NaCN
Na+ + CN- (Salt solution) 0.2M
HCN
H+ + CN- (Acid solution) 0.1M
+
[H ] = Ka x Concentration of Acid solution
Concentration of Salt solution
= 6.2 x 10-10 x 0.1
0.2
[H ] = 3.1 x 10 M.
50. Find the change in pH of 1.0ml 0.10M HCL when its volume is made 50 ml by adding
water?
Ans. HCl
H+ + Cl- (As ratio of 1:1)
0.1M
0.1M 0.1M
H
P 1 = -log [H+]
= -log [1 x 10-1]
 pH1 = 1
→ (1)
Now, as volume 50 mol is added to the solution, udarity of solution nill change.
M1 V1 = N2 V2
Now, M1 = 0.10M
V1 = 1ml
V2 = 50ml
N2 = ?
(0.10) x (1) = N2 x (50)
 N2 = 0.1
50
-3
M2 = 2 x 10 M
PH2 = - log10 [H+] 9after adding 50 ml to solution)
= - log [2 x 10-3]
= - [log 2 + -3 (log10) ]
= - (0.3010 -3)
[ log 10 = 1]
PH2 = 2.699
→ (2)
H
H
H
N our change in p = p 2 - p 1
= 2.699 – 1
[ from equation (1) and (2) ]
= 1.699
Change in pH = 1.699
51. The dissociation constant of HCOOH is 1.8 x 10-4. What will be the diassociation
constant of conjugate base-formate ion?
Ans. Dissociation Constant, Ka = 1.8 x 10-4
HCOOH
HCOO-(aq) + H2(aq)
+
-10
Weak Acid
Conjugate Base
HCOO- + H2O
HCOOH(aq) + OH-(aq)
Hydrolysin constant, Kn
Now, Kn = Kw
= Ionic product of water
Ka
Dissouolion constant of HCOOH
-14
Kn = 1 x 10
1.8 x 10-4
Kn = 5.55 x 10-9
52. A saturated solution can be prepared by dissolving 0.08gram. CaF2 in 2.901 litre at 298
K tempareture. Find its KSP. (Molecular mass = 78 grammol-1).
Ans. W = 0.08 gm CaF2
Molecula mass = 78 gm mol-1
Moles of CaF2 = 0.08
78
= 1.0256 x 10-3 moles
Conc. of CaF2 = Moles
Litre
= 1.0256 x 10-3
2.901
= 3.53547 x 10-4 mol/litre
→ (1)
CaF2
Ca2+
+
2FS mol/litre
2S mol/litre
Let Dolubility Product of CaF2 be S mol/litre
Then [Ca2+] =
S mol/litre
[ F-]
=
2S mol/litre
KSP = [Ca2+] [ F-]2
= [S] [2S]2
= 4S3
= 4 x (3.5347 x 10-4)3 M3
= 4 x 44.1629 x 10-12 M3
= 1.7665 x 10-10 (mol/litre)3
KSP = 1.76 x 10-10 M3
53. The value of Ka of a weak acid is 109 times more than the value of Kw. Calculate pH and
pOH of 0.1 m aqueous solution of this acid.
Ans. For weak acid,
HA(1) + H2O
H3O+(aq) + A-(aq)
Ka = [H3O+] [A-] { Concentration of positive ion [H3O+] = concentration of
[HA] negetive ion [A-]}
Now, Weak acid will ionize to much less ectent and so concentration of its unionized part will
remain nearly same as its original concentration i.e. O.1M
[HA] = 0.1M
Ka = 109 Kw (as given)
But, Kw = 1 x 10-14 (Ionic produce of water)
Ka = 109 x (1 x 10-14)
Ka = 10.5
→ (1)
Now, for Ionisatin of weak acid,
Ka = [H3O+]2
[Ha]
1.5 = [H3O+]2
1 x 10-1
+ 2
[H3O ] = 10.6
[H3O+]2 = 1 x 10-3 M
Now,
PH = -log10 [H3O+]
= -log [1 x 10-3]
= -[log 1 + -3(log10)]
= - (-3)
( log 1 = 0)
H
P =3
( log 10 = 1)
Now, pH + pOH = 14
pOH = 14 – pH
= 14 – 3
pOH = 11
54 .In aqueous solution of 0.05 M C6H5OH the ionisation constant Ka = 1.28 x 10-10
calculate concentration of H3O+ ion in the solution.
Ans. C6H5OH(1) + H2O
H3O+(aq) + C6H5O-(aq)
Phenol
(phenoxy Ion)
Now, Ka=[H3O+]2
[C6H5OH]
[Concentration of H3O+ ion and C6H5O- ion will remain nearly equal.
Moreover, weak acid will ionize to much less extant and hence, concentration of unionised part
will remain nearly same as original concentration]
Given, Ka = 1.28 x 10-10
[C6H5OH] = 5 x 10-2M
Ka=[H3O+]2
[C6H5OH]
[H3O+] = Ka. [C6H5OH]
+
-10
-2 1/2
[H3O ] = (1.28 x 10 x 5 x 10 )
= (6.4 x 10-12)1/2
+
[H3O ] = 2.529 x 10-6M
= 2.53 x 10-6M
55. Calculate pH of 0.002M HNO3 solution.
Ans. HNO3
H+(aq) + NO3-(aq)
0.002M
0.002M
[As ratio is 1:1]
+
-3
[H ] = 2 x 10 M
PH = -log [H+]
= -log [2 x 10-3]
= -[log 2 + log (10-3)]
[log(axb) = log a + log b]
= -[0.3010 + -3(log 10)]
[logab = b loga]
= -[0.3010 – 3]
[ log 10 = 1]
H
p = 2.6990
56. Calculate values of pH and POH of 0.06 M H2SO4 solution.
Ans. H2SO4
2H+ + SO420.06M
0.12M 0.06M
[ As ratio of H+ ion production is 2:1, hence, molar concentration will be doubled
H+ = 2 x 0.06M = 0.12M]
Now, pH = -log [H+]
= -log [0.12]
= -log [1.2 x 10-1]
= -log 1.2 – log (10-1)
= -0.0792 + 1
[ log 10 =1]
= 0.9208
pH = 0.9208
Now, pH + pOH = 14
0.9208 + pOH = 14
OH
p = 14 – 0.9208
= 13.0792
OH
p = 13.0792
57. Whatwill be [OH-] in 100 ml. NaOH solution having pH = 10.0?
Ans. Data: pH = 10.0 [OH-] = ?
pH = -log [H+] = 10
= -log [H+] = 10
= -log [H+] = -10
[H+] = 1 x 10-10M
Now, H2O(1)
H+(aq) + OH-(aq)
→ (1)
[ Self – lonization of water ]
Weak electrolyte
Kw = [H+] [OH-] = 1 x 10-14 ( found at 250 C)
Now, from equation (1), [H+] = 1 x 10-10
1 x 10-10 x [OH-] = 1 x 10-14
[OH-] = 1 x 10-4M
58. [H3O+] in aqueous solution of CH3COOH is 1 x 10-3M. What will be its intial
concentration if Ka = 1.25 x 10-5M ?
Ans. [H3O+] = 1 x 10-3M
Ka = 1.75 x 10-5M
For weak acid,
CH3COOH(1) + H2O(1)
H3O+(aq) + CH3COO-(aq)
+
Ka = [H3O ] [CH3COO ]
[CH3COOH]
Now, the concentration of positive ion [H3O+] =concentration of positive ion [CH3COO-]
→ Moreover, weak acid (CH3COOH) will ionize to much less extent and so concentration of its
unionized part will remain nearly same as its original concentration
[CH3COOH] = Co
Ka = [H3O+]2
Ka = [H3O+]2
Co
Co = (1 x 10-3)2
1.75 x 10-5
Co = 5.714 x 10-2M
59. KSP of Mg(OH)2 = 1.2 x 10-11. Find solubility in pure water. (S = 1.442 x 10-4M)
Ans. Suppose, the solubility of Mg(OH2) is S mol/litre.
Then, the concentration of [Mg2+] and [OH-] ion in saturated solution will be as follows.
Mg2+(aq) + 2OH-(aq)
S mol/litre 2S mol/litre
Now, for Mg (OH)2,
KSP = [Mg2+] [OH-]2
= (S) (2S)
= 4S3
→ (1)
KSP = 4S3
KSP = 1.2 x 10-11 (from data)
From eqn(1) KSP = 1.2 x 10-11 = 4S3
S3 = 3 x 10-12
S = 1.442 x 10-4M
[ log (S3) = log (3)
3 log S = 0.4771
log S = 0.4771
3
Mg(OH)2
= 0.1590
 S = 1.442]
Section –D
1. Explain dynamic nature of chemical and physical equilibrium.
Ans. Dynamic Nature of Equilibrium
The most important matter in the case of equilibrium is that there is a continuous transformation
of reactant to product and product to reactant. This state appears to be steady but it is not so. This
type of reaction which takes place in both the directions is called reversible reaction and it is
expresses by the symbol of two half arrows(↔). This symbol indicates that such reaction occurs
simultaneously in both (forward and reverse) directions. Generally, the change of reactant to
product is called forward reaction and the change of product to reactant is called reverse reaction.
Thus, in reversible reactions, forward and reverse reactions continuously occur and we find it as
equilibrium state. The misture of reactants and products obtained at equilibrium time is called
equilibrium misture. The decomposition reaction of solid calcium carbonate in a cold vessel, at
high temperature can be shown as below.
∆
CaCO3(s) ↔CaO(s) + CO2(g)
The equilibrium is dynamic and not steady or static as the forward and the reverse
reactions occur with the same velocity at the equilibrium time in equilibrium reactions in closed
vessels. In the above reaction obtaining CaO and CO2 by decomposing of CaCO3and obtaining
CaCO3 by combination of CaO and CO2 continuously take place. Suppose, if we deposit, some
amount in our bank account and withdraw the same amount, then balance in the account appears
steady or static. But this can be considered operative or dynamic and not closed or static. It is
very difficult to determine the dynamic nature of equilibrium, even then with the help of
radioactive isotope, it can be proved, viz. 14CO2 gas containing radioactive isotope, 14C and
CaCO3 are taken in two different flasks and CO2 obtained by decomposition is connected with
vessel containing 14CO2 gas, after sometime, Ca14CO3 will be formed in the vessel of CaCO3 and
CO2 will be obtained in the vessel containing 14CO2. Thus if the equilibrium would have been
steady, there must not be exchange of 12C and 14C. With the help of suitable counter, the
radioactivity can be measured and the proof for the dynamic nature of equilibrium can be
obtained through the proportions of concentrations of reactants and products remain constant.
The reaction can be fast or slow depending upon the nature of the reactant and the experimental
conditions.
Equilibrium reactions can be divided into following three categories:
(i) Reactions which are almost at the extent of completion and concentration of
reactants may be negligible. It is not possible to detect this experimentally.
(ii) Reactions in which the products are formed in very less proportions and most part
of the reactant remains unchanged at the equilibrium.
(iii) Reactions in which the concentrations of reactants and products are in comparable
proportions at equilibrium.
Dynamic Nature of Chemical Equilibrium
The dynamic nature of chemical equilibrium can be demonstrated by taking example of the
production of ammonia. By keeping known quantities of dinitrogen and dihydrogen at high
temperature and pressure in a closed vessel, the amount of ammonia formed can be determined at
constant intervals by a series of experiments. The quantities of unreacted dinitrogen and
dihydrogen also can be determined. From this it is concluded that even if the reactants and
products are in different proportions, their concentrations are same at equilibrium. This
constancy in composition indicates dynamic nature of equilibrium. For this, in synthesis of
ammonia, deuterium (D2) instead of dihydrogen, (H2) is used and ammonia gas is produced by
Haber process and it is studied. The results obtained are similar to those obtained above. In the
mixture proportions of N2, Dand ND3 instead of N2, H2, NH3 can be determined and equilibrium
can be obtained. If D2 is added after the formation of ammonia by reaction of N2 and H2, the
reaction may not occur but H in NH3 is displaced by D and ND3 can be determined by mass
spectrometer. Thus, it is proved that in the reaction
That the reactions from reactants to products and products to reactants that is forward and
reverse reactions continuously occur with the same rates and so ND3 instead of NH3is obtained.
By the use of radioactive isotope, the dynamic nature of equilibrium can be proved viz. For the
reaction, H2(g) + I2(g)
2HI(g)
Radioactive isotope 131I of iodine can be used to study the dynamic nature of chemical
equilibrium. As the equilibrium is dynamic, certain properties or factors are found similar. E.g.
Intensity of colour, constant pressure, constant concentration etc.
2. Explain equilibrium solid in solution and gas in solution.
Ans. Equilibrium Involving Dissolution of Solid in Solution
For the study of this type of equilibrium, the example of sugar and its solution in water can
be taken. At constant temperature, take some water. Add some sugar into it and stir. In the
beginning sugar easily dissolves, but as more and more amount of sugar is added, it dissolves
according to its solubility and then some amount of sugar remains in solid form without
dissolution. We know this state as saturated solution but in this system the equilibrium is also
establishes sugar(solid) and liquid(solution of sugar). This can be expressed as below:
Sugar(s)
Sugar solution(aq)
Solid
liquid
As studies earlier, equilibrium is dynamic because the forward and the reverse reactions
continuously occur in each system. In this system, the amount of sugar that dissolves in water,
the same amount of sugar separates from solution of sugar. Hence, the number of molecules of
sugar and number of aqueous molecules of sugar in the solution remain constant in this system.
Equilibrium involving Gas and Solution
At constant temperature and pressure, carbon dioxide can be dissolved in water in a closed
vessel(system), so that a system containing gas and a solution of carbon dioxide can be formed.
As temperature and pressure are constant, carbon dioxide dissolves according to its pressure and
temperature and forms solution of carbon dioxide and the excess carbon dioxide gas remains in
equilibrium with it. Equilibrium is dynamic and so molecules have carbon dioxide gas that
dissolve in water is the same as the number of molecules of the gas that release from the solution
in the system. Thus, in this closed system, the total number of molecules(amount) of carbon
dioxide in gaseous form and those that have dissolved in water remain constant. This equilibrium
can be shown as below:
CO2(g)
CO2(soln)
All the reactions studies above are processes in which only physical change takes place and so
they all are examples of physical process equilibrium.
3. Mention the characteristics of chemical equilibrium
Ans. Characteristics of Chemical Equilibrium
(i)
During chemical equilibrium, the properties like colour, concentration, pressure or
temperature of the system remain constant and they are similar in the total area of the
system.
(ii)
When chemical equilibrium is attained, the rates of forward and reverse reactions
become equal.
(iii) If the factors like concentration, pressure, temperature etc. which affect the chemical
equilibrium are changed, they produce effect on equilibrium.
(iv)
Even if the initial concentrations are different, the equilibrium constant remains
constant at constant temperature.
(v)
The value of equilibrium constant changes if the temperature changes.
(vi)
For attainment of equilibrium, the reaction can be carried out from left to right
(reactants to products) or from right to left (products to reactants)
(vii) There is no effect of catalyst on the equilibrium constant and so the proportions of
products remain same but the rate of reaction to attain equilibrium increases.
4. Give chemical equilibrium (active masses) law and derive its formula
Ans. Law of Chemical Equilibrium and Equilibrium Constant
The mixture of reactants and products at equilibrium is called equilibrium mixture. We shall
study the relation between concentrations of reactants and products at equilibrium state.
Let us take a simple reversible reaction as follows:
A+B
C + D
In this reaction A and B are reactants and C and D are products. This means that in this reaction
moles of reactants and products are one each but in all reactions this may not happen. Hence, it is
necessary that their moles are expressed. Balanced reaction determines their moles. Viz.
N2(g) + 3H2(g)
2NH3(g)
From the experimental studies of many reversible reactions scientists of Norway, Guldberg and
Waage mentioned in 1864 that the concentrations of substances in equilibrium mixture can be
expressed by following equilibrium equation.
Where Kc is equilibrium constant and [ ] bracket expresses concentration of reactant or product
on mol/lit or M. The equilibrium equation is also known as law of active masses because in the
early years of chemistry, concentration was said to be ‘active mass’.
Now, we shall derive the equation for equilibrium constant of a general reaction. Suppose, if a
reaction takes places, as given below in which the reactants and products are shown in balanced
form with their proper moles (a, b, c or d)
aA + bB
cC + dD
On the basis of Guldberg and Waage’s law the rate of forward reaction
Vf [A]a [B]b or
Vf = Kf [A]a [B]b
Where Kf is the proportionality constant for forward reaction.
The rate of reverse reaction
Vr [C]c [D]d
Vr= Kr [C]c [D]d
Where Vr is the proportionality constant for reverse reaction.
At equilibrium the rates of forward and reverse reaction will be equal and so Vf = Vr
That is, Kf[A]a [B]b = Kr [C]c[D]d
Thus, when equilibrium is attained if we determine the concentration of the reactants and the
products in any reaction and their stoichiometric multiples, the equilibrium constant Kc can be
obtained.
5. What is meant by Kc and Kp? Deduce relation between Kc and Kp.
Ans. Relation between Kp and Kc
As seen earlier the equilibrium constant of a gaseous reaction can be written as
…. 1
But we know that according to simple gas equation PV = nRT. Hence, it can be written as
 P = (n/V)RT = CRT
(where n/V = C = concentration in mol lit-1)
Substituting the values of p in the above equation 4.10, it can be written as
….2
…3
= Kc X (RT)ng …. 4
Where ∆ng = (c+d) – (a+b)
Means number of total moles of gaseous products minus number of total moles of gaseous
reactants.
Hence, it can be written as
Kp = Kc X (RT)ng…. 4
6 Deduce the equilibrium constant Kp pCO2 for decomposition of solid CaCO3 in closed vessel
Ans. Kc = [CaO][CO2]
Now, Density of CaCO3 = Constant
[CaCO3]
Density of CaO
 Kc x Density of CaCO3 = [CO2]
Density of CaO
 Kp = PCO2
7. Deduce the formula Kp = (P2)/4 for decomposition of solid NH4HS in closed vessel.
Ans. NH4HS(s)
NH3(g) + H2S(g)
x
x
Total moles = 2x
Total pressure = Patm P/2
P/2
Kp = [PNH3][PH2S] = (P/2)(P/2) = (P2/4) atm2
8. Derive the formula for decomposition constant of solid ammonium carbonate and prove
that Kp = (4/27)P3.
Ans. NH4COONH2(s)
2NH3(g) + CO2(g)
Moles = 2x
x
Total moles = 3x
Kp = [PNH3]2[PCO2] = (2P/3)2(P/3)2 = (4P3/27) atm3
9. Give Le-Chatelier’s principle and explain in detail the effect of concentration on
equilibrium.
Ans. Equilibrium constant is independent of initial concentration of reactant but, if change is
carried out in concentration of pressure, there is an effect on equilibrium, the equilibrium tries to
nullify this effect. Reaction can be endothermic or exothermic. Hence, the heat absorbed or
evolved, functions like that of the reactant and its effect is on equilibrium which tries to nullify
the effect. Le Chatelier studies the effect of concentration and temperature and his presentation is
called Le Chatelier’s principle whose statement can be written as below:
“ If from the factors determining the equilibrium state, any one factor is changed, there
will be such a change in the system that the effect will be nullified or made negligible so that the
value of equilibrium constant at that temperature will remain constant.” This principle can be
applied to both physical and chemical equilibrium.
We shall study in detail the factors like (1) change in concentration (2) change in pressure
(3) addition of inert gases (4) change in temperature and (5) use of catalyst affecting the
equilibrium
(1)Effect of change in concentration:
If we add or remove the reactant or the product from the reaction in equilibrium, its effect on
equilibrium according to Le Chatelier’s principle will be as follows:
(a)If the concentration of reactant or product is increased by addition of reactant or product, the
reaction will occur in such a way that the increase in concentration will be taken for use, i.e. the
increase in concentration of reactant, the concentration of product will increase and if
concentration of product is increased the reaction will result in the direction of increase in
concentration of the reactant.
(b)If the concentration of reactant or product is decreased by removing reactant or product, the
reaction will occur in such a way that product or reactant will be established again.
Hence, if any change in concentration of reactant of product is carried out then the equilibrium
will try to make this effect minimum and equilibrium will be established accordingly. If we take
this as an example, in the reaction
If concentration of reactants dinitrogen or dihydrogen is increased, the reaction will proceed
towards right hand side and the product ammonia obtained will be more. If the concentration of
nitrogen or hydrogen is decreased, the reaction will proceed towards left hand side and reactants
of nitrogen or hydrogen will be obtained back i.e. production of ammonia will decrease. Here, it
is necessary to remember that 1 mole dinitrogen combines with 3 moles of dihydrogen in this
reaction and forms 2 moles of ammonia. Hence, increase in concentration of dihydrogen rather
than dinitrogen, will give more product. We take another example of heterogeneous equilibrium.
If solid CaCO3(s) is heated in closed vessel, the following decomposition reaction will occur.
CaCO3(s)
CaO(s) + CO2(g)
Hence, if more CaO(s) is to be obtained then, the CO2(g) formed in the reaction should be removed
because CO2(g), gas can combine with solid CaO(s) and carry out reverse reaction then proportion
of product will decrease. Hence, by removing CO2(g) from the reaction vessel, more CaO(s) can
be obtained.
10. Explain Arrhenius acid-base theory and mention it’s drawbacks
Ans. Arrhenius concept about Acid and Base:
According to Arrhenius concept, substances which dissociate in water and give hydrogen ion
(H+) are called acids and substances which dissociate in water and give hydroxyl ions (OH-) are
called bases. E.g;
(1) HCl(g)
+H2O
H+ + Cl-
Acid
+H2O
(2) NaOH(s)
Na+ + OHBase
The limitations of this concept are as follows:
(i) proton (H+) is highly unstable.
(ii) It can not exist independently.
(iii) It immediately combines with molecules of solvent water and gives H+3O ions. Its addition
in certain bases OH- is not present even then they show properties of base, viz., NH3. Similarly
compounds like BF3 do not possess H+, even then they act as acid.
(iv) this concept can only be applied to aqueous solution because salt like NH4Cl reacts as acid in
solvent like liquid NH3. Hence, it is difficult to accept Arrhenius concept as the universal one.
Because ionization is given importance. (In addition it is necessary to know that ionic radius of
H+ is about 10-15 meter and so it is very small in size9. Hence it easily gets combined with
molecules of water and forms H3+O ion which is called as hydronium ion. One estimate is that
one H+ can be combined with four molecules of water showing H+ + 4H2O  H9O4+ ion.
11. Explain with suitable example, Lowry-Bronsted acid-base theory(proton transfer
theory)
Ans. Concept of Bronsted – Lowry for acid and base:
Danish chemist Bronsted, an English chemist Lowry presented the concept of acid and base.
They made H+ (proton) as a base. According, to their concept , the substance which gives a
proton or donates a proton is called the acid and the substance which receives the proton or
accepts a proton is called the base. Thus, acid is a proton donor and base is proton acceptor. Let
us take the dissociation reaction of hydrogen chloride in water.
Base
Acid
H3+O (aq) + Cl-(aq)
HCl(g) + H2O(l)
Acid
Similarly,
Base
Acid
NH3(g) + H2O(l)
Base
NH4+(aq)+ OH-(aq)
Base
Acid
We shall understand in detail the first from the above reactions :
HCl
H+
+ ClAcid-1 proton conjugate base-1
As it gives proton, HCl is an acid
H2O + H+
H3O+
Base-2
Proton
Conjugate acids-2
As it accepts a proton, H2O is a base.
Total reaction:
HCl + H2O
H3O+(aq) + Cl-(aq).
Acid-1
Base-2
Acid-2
Base-1.
In the above reaction giving-taking of proton is not shown. Hence, it can be said that only
transfer of proton takes place, it is not obtained free. Every acid will lose proton and so its
conjugate base will be formed and every base will accept a proton and so its conjugate acid will
be formed. Hence, this concept is known as proton transfer or conjugate acid-base concept.
We have earlier seen in limitations of Arrhenius concept that OH- is not present in NH3 even
then it acts as a base but according to Bronsted-Lowry concept it can be explained.
NH3 + H2O
NH+4(aq) + OH-(aq)
Base-1 Acid -2
Acid-1
Base-2
In the reaction, base NH3 accepts a proton and forms conjugate acid NH4+ ion and acid H2O loses
proton and forms conjugate base OH-.
Thus the concept of Bronsted - Lowry is found to be more applicable and acceptable than
Arrhenius concept. Even then its limitations are also known. This difficulties are observed in the
study of reactions in organic chemistry and complex salts. BF3 has no proton even then it acts as
an acid. Hence, the third concept has come in to existence, which is known as Lewis acid- base
concept. Proton is given importance in Bronsted – Lowry concept.
12. Explain in detail Lewis acid-base theory.
Ans. Lewis Concept of Acid and Base :
Lewis, in presenting this concept, in 1923, mentioned that acid means a substance which can
accept a pair of electrons and base means a substance which can donate a pair of electrons. Thus,
instead of the concepts of proton, ionization, conjugate acid or base, he made electrons
associated with all reactions and substances and its arrangement as the base of the concept. As
seen earlier BF3 can be said acid or not and NH3 can be said a base or not can be solved by this
concept. It will be clear from the following reaction:
Thus, BF3 accepts the pair of electron and so it is acid and NH3 donates a pair of electron, so it is
acid and NH3 donates a pair of electron, so it is base. Electron deficient substances or ions like
AlCl3, Co3+, Mg+2, will act as acid and substances like H2O, NH3, OH-, F- will act as base. They
are respectively called Lewis acid and Lewis base.
13. Derive the formula of dissociation constant of Weak acid
Ans. Ionization Constant of Weak Acid (Ka) :
In aqueous solution of weak monobasic acid HA, there is a partial ionization and so the
equilibrium is obtained as below :
HA(aq) + H2O(l)
H3+O(aq) + A-(aq).
Suppose the initial concentration of weak acid (HA) is C mollit-1 and degree of ionisation
is α, then following can be written :
Reaction :
HA(aq) + H2O(l)
H3+O(aq) + A-(aq).
Initial
Concentration(M)
C
0
0
Degree of ionisation(α)
1-α
Concentration at equilibrium(M) (1 - α)C
Equilibrium constant Ke=
α
αC
α
αC
[H3O+][A-]
[HA][H2O]
= (αC)(αC)
…1
(1 - α)C[H2O]
But [H2O] is accepted as constant and so
Ke[H2O] = (αC)(αC)
(1 - α)C
= α2C = Ka
…2
(1 - α)
Where Ka is the ionization constant or dissociation constant of the s\acid HA. (At this stage we
consider both the terms same) Hence, for any weak monobasic acid, following can be written
[H3O+][A-]
Ke=
[HA]
…3
-1
The unit of Ka will be mollit .
As the values of Ka depend on [H3+O] the values of ka will be different for different.
Lesser the value of Ka, weaker will be the acid. The value of Ka will be constant at definite
temperature.
14. Derive the formula of dissociation constant of weak base
Ans. ionisation constant (Kb) of weak base:
The ionization of monoacidic weak base MOH will take place in aqueous solution as follows:
H2O
MOH(aq)
M+(aq) + OH-(aq)
As base is weak, incomplete ionization will occur and so equilibrium will be obtained and it can
be expressed as below:
[M+] [OH-] …………4 and
Kc = [MOH] [H2O]
+
[M ] [OH-]
=Ke ………. 5
Kex[H2O]=
[MOH]
Where Kb is the ionization or dissociation constant of monoacidic weak base. If we know the
initial concentration of weak base and its degree of ionization, we can calculate the value of Kb
as we have studied earlier. [M+] = [OH-].
15. Obtain relation between ionic equilibrium constant Ka, initial concentration Co and
concentration of H3O+ in aqueous solution of weak formic acid.
Ans. As formic acid is weak acid, it is partially ionized in water. The equilibrium between the
undissociated species of CH3COOH and the ions H+(aq) can be represented as
CH3COOH(l) + H2O(l)
H3O+(aq) + CHCOO-(aq)
The equilibrium constant K for this reaction is as follows :
K = [H3O+][CH3COO-]
[CH3COOH][H2O]
Here the change in concentration of water on dissolving the weak acid will be quite negligible
compared to its original concentration, so considering the value of [H2O] as constant and joining
it with equilibrium constant K we get a new constant Ka as follows.
K[H2O] = Ka = [H3O+][CH3COO-]
[CH3COOH]
Here formic acid will ionize to much less extent and so the concentration of its unionized part
will remain nearly same as its original concentration.
[CH3COOH] = Co So, we have
Ka = [H3O+][CH3COO-]
Co
Again, the concentration of the positive ions [H3O+] and that of the negative ions [CH3COO-]
will be the same i.e. [H3O+] = [CH3COO-]
[CH3COO-] can be replaced by [H3O+], Ka = [H3O+]2
Co
+ 2
+
1/2
[H3O ] = Ka.Co
[H3O ] = (KaCo)
16. Obtain relation between ionic equilibrium constant Kb, initial concentration Co and
concentration of OH- in aqueous solution of weak base aniline.
Ans. As aniline is weak acid, it is partially ionized in water. The equilibrium between the
undissociated species of C6H5NH2 and the ions H+(aq) can be represented as
NH3(g) + H2O(l)
NH4+(aq) + OH-(aq)
The equilibrium constant K for this reaction is as follows :
K = [NH4+][OH-]
[NH3][H2O]
Here the change in concentration of water on dissolving the weak base will be quite negligible
compared to its original concentration, so considering the value of [H2O] as constant and joining
it with equilibrium constant K we get a new constant Ka as follows.
K[H2O] = Kb = [NH4+][OH-]
[NH3]
Here aniline will ionize to much less extent and so the concentration of its unionized part will
remain nearly same as its original concentration.
[NH3] = Co So, we have
Ka = [NH4+][OH-]
Co
Again, the concentration of the positive ions [NH4+] and that of the negative ions [OH-] will be
the same
[OH-] can be replaced by [NH4+],
Kb = [OH-]2
Co
- 2
1/2
[OH ] = Kb.Co
[OH ] = (KbCo)
17. What is meant by ionic product of water? Deduce its equation.
Ans. Ionic Product of Water
Water is an amphoteric oxide when acid is added to it; it accepts the proton and act as a base and
when base is added to it, it donates a proton and acts as acid. When reaction between two
molecules of water takes place, one molecule donates proton and other molecule receives proton
and shows conjugate acid-base reaction.
H2O(l) + H2O(l)
H3O+(aq)
+
OH-(aq)
Acid-1
Base-2
Conjugate acid-2
Conjugate base-2
If we express the equilibrium constant of above reaction them.
[H3O+][OH-]
Ka =
[H2O][H2O]
Where Ka is the dissociation constant of acid. There is no significant change in concentration of
water (55.5M) because H2O is a weak acid (possesses about 10-7M H+). Thus, if H2O is
considered constant.
Ka x [H2O]2 = [H3+O] [OH-] = Kw.
Where Kw is ionic product of water. Water is neutral and so [H3+O] and [OH-] in it are 10-7 M.
Hence, Kw = [H3+O] [OH-] = (10-7)(10-7) = 10-14 which is constant and equilibrium constant
remains constant at constant temperature; so the value of Kw will be constant at constant
temperature, viz., the value of Kw is 1 x 10-14 at 298 K. If we find the ratio of concentrations of
dissociated and un dissociated water, it will remain on left hand side, i.e. the number of un
dissociated molecules or its concentration will be more.
18. Discuss the factors affecting strength of acid.
Ans. Factors Affecting strength of acid:
From the study of dissociation constant, pH value etc., it is found that their values are different.
The reason for this is that [H3+O] available can be more or less. What may be the reason for this?
If the acid is strong, its value of Ka will be high and the value of pH will be low. The dissociation
of acid will depend on strength of acid and the polarity of H-A bond. As the strength of H-A
bond decreases, the energy required for breaking that bond will decrease and HA will be
stronger. When difference between electro negativities of A and B will increase, apparently
ionization will occur and will be easy to break the structure of the bond. Hence, the acidity will
increase.
On comparing the elements of the same group of periodic table, the strength of H-A bond will be
more important factor than polar nature. As we go down in the group the size of A will increase
and so strength of H-A bond will decrease and hence acid strength will increase.
Increase in size
HF<< HCl <<HBr<<HI
Increase in acid strength
For this reason, H2O is stronger acid than H2S, but if we discuss the elements in the same period
of periodic table, the polarity of H-A bond will determine the strength of acid. As the
electronegativity of A increases, the strength of acid will increase.
Increase in electro negativity
CH4<< NH3<< H2O << HF
Increase in acid strength
19. Write a short note on buffer solutions.
Ans. Buffer solutions
The pH of the fluids like blood in our body and urine is almost constant. If there is change in this
pH, it affects biochemical reaction in the body. The pH of chemical and biochemical reactions in
our body are constant, biochemical reactions in our body are constant, viz. the pH of human
saliva is 6.4. In addition, hydrochloric acid is present in human stomach which helps in
digestion. The pH of cosmetics are also kept constant. Hence, the question arises that how pH in
any solution can be kept constant. Such solutions are called buffer solutions. Its definition can be
given as below:
“ The solution which resists the change in pH carried out by addition of acid or base in small
proportion to them or are being diluted, and the values of their pH remain constant are called
buffer solution.”
Buffer solutions can be acidic or basic. If pKa of weak acid and pKb of weak base are known,
buffer solutions of known pH can be prepared. Buffer solutions can be of three types as follows:
(i) Acidic buffer solution: Acidic buffer solution can be prepared by mixture of weak acid and its
salt with strong base.
(ii) Basic buffer solution:Basic buffer solution can be prepared by mixture of weak base and its
salt with strong acid.
(iii) Neutral buffer solution: Neutral buffer solution can be prepared by neutralization of weak
acid and weak base. These type of buffer solutions are shown below:
Type
Substances
Value of pH
Acidic
CH3COOH + CH3COONa
<7
Basic
NH4OH + NH4Cl
>7
Neutral CH3COOH + NH4OH
=7
Buffer solution of known pH can be prepared by using the following HendersonHaschelback equation.
For acidic solution,
[Salt]
pH = pKa + log [Acid] …………. 1
where [acid] is concentration a weak acid and its dissociation constant is Ka and [salt] is
concentration of the salt of this weak acid with strong base. For a buffer solution, it can be
written as
pH= pKa + log [CH3COONa]
[CH3COOH]
similarly, for basic buffer solution e.g., NH4OH + NH4Cl can be written that,
[NH4Cl]
pH = pKa + log [NH4OH]
Such buffer solution can be used in chemical and biochemical reactions and especially in
analytical chemistry. In human body buffer solutions containing [HCO-3] and [CO2-3] as well as
[H2PO-4] and [HPO2-4] are present.
20. Write a short note on pH scale.
Ans. pH Scale.
If we express the concentration of hydronium ion [H3+O] in molarity then values like 10-12 to 10-2
are possible. It is difficult to express these values on simple graph paper. Hence, scientist
Sorenson found a scale which is called pH scale. According to him pH = -log10[H3+O]. The
values 10-12 to 10-2 shown above can be converted to +12 to +2 if calculated on the basis of this
relation and plotting of graph can be easy. The definition of pH can be given like this, “pH of a
solution is the negative logarithm to the base 10, of the concentration of hydrogen or hydronium
ion”. According to thermodynamics, activity is more proper word instead of concentration but in
dilute solutions activity and concentration can be considered can be considered to be the same.
Now as seen earlier a solution containing 10-7M [H3+O] and [OH-] is neutral. Hence,
pH = -log10[H3+O] = -log1010-7M = 7 and for acidic solution [H3+O]> 10-7M, pH < 7
Similarly, for basic solution [H3+O]< 10-7M, Hence, pH > 7. Therefore, it can be written as :
pH< 7 Acidic solution
pH> 7 Basic solution
pH = 7 Neutral solution
AS seen above,
Kw = [H3+O] [OH-]
Putting the values, Kw= (10-7)(10-7) = 10-14and –log Kw = -log(10-14)
Therefore, pKw = 14
Therefore, pH + pOH =pKw = 14
Temperature affects the values of pH, pOH, pKw. The above discussion can be shown in the
following table :
Concentration (M)
Acidic
Neutral
Basic
+
-7
-7
[H3 O]
More than 10
10
Less than 10-7
[OH-]
Less than 10-7
10-7
More than 10-7
pH
Less than 7
7
More than 7
pOH
More than 7
7
Less than 7
pH paper, litmus paper or universal indicator can be used to test whether the solution is acidic or
basic but the exact values of pH can be determined with the help of instrument called pH meter.
21. Explain the calculation of pH of a solution from the dissociation constant of weak acid.
Ans. In aqueous solution of weak monobasic acid HA, there is a partial ionization and so the
equilibrium is obtained as below :
HA(aq) + H2O(l)
H3+O(aq) + A-(aq).
Suppose the initial concentration of weak acid (HA) is C mollit-1 and degree of ionisation
is α, then following can be written :
Reaction :
HA(aq) + H2O(l)
H3+O(aq) + A-(aq).
Initial
Concentration(M)
C
0
0
Degree of ionisation(α)
1-α
α
α
Concentration at equilibrium(M) (1 - α)C
αC
αC
[H3O+][A-]
Equilibrium constant Ke= [HA][H2O]
= (αC)(αC)
…1
(1 - α)C[H2O]
But [H2O] is accepted as constant and so
Ke[H2O] = (αC)(αC)
(1 - α)C
= α2C
= Ka
…2
(1 - α)
Where Ka is the ionization constant or dissociation constant of the s\acid HA. (At this stage we
consider both the terms same) Hence, for any weak monobasic acid, following can be written
[H3O+][A-]
Ke=
[HA]
…3
-1
The unit of Ka will be mollit .
As the values of Ka depend on [H3+O] the values of ka will be different for different.
Lesser the value of Ka, weaker will be the acid. The value of Ka will be constant at definite
temperature. IN table 4.3 the values of ionisation constants of some weak acids are given.
According to the relations seen earlier,
pH = -log10[H3+O]
pOH = -log10[OH-]
Similarly, pKa = -log10[Ka]
PA- = -log10[A-] can be written. Where [A-] is the concentration of negative ion.
It is apparent from above relations that if the values of initial concentration [H3+O] and Ka are
known, then at equilibrium, [H3+O] concentration can be determined and pH can be calculated
22. Explain effect of common ion on solubility of sparingly soluble salt.
Ans. Effect of Common Ion on Solubility of Sparingly Soluble Salt
The sparingly soluble salt that has dissolved in solution that is completely dissociated and so it is
ionic in form. Hence, it is a strong electrolyte. Earlier, in chemical equilibrium, effect of
concentration, application of Le Chatelier’s principle etc. have been studied. Solubility product is
ionic equilibrium and the effect of concentration can be studied. As it is equilibrium constant, it
will depend on temperature, but its value will be constant at constant temperature.
What will happen if we add soluble ionic substance like KCl in the solution of a sparing soluble
salt like AgCl?
AgCl(s)
Ag+(aq) + Cl-(aq)
KCl(s)
K+(aq) + Cl-(aq)
The Cl- from AgCl in equilibrium and Cl- ion obtained by complete ionization of KCl, the
concentration of Cl- will increase. Hence, according to Le Chatelier’s principle, the equilibrium
will shift towards left side so as to nullify the effect of Cl- i.e. more AgCl will be formed. In
other words, there will decrease in solubility of AgCl. Hence, it can be said that because of the
effect of common ion on sparingly soluble alt its solubility decreases and sparingly soluble salt
precipitates more.
The use of effect of common ion can be made to separate one ion from the other in presence of
other ion in qualitative analysis. It can also be used for decrease in solubility of the components
in the mixture. In qualitative analysis, the solubility products of sulphides of metal ions of
second group are less in comparison to solubility products of sulphides of metal of III B group
ions, therefore, HCl is added before adding H2S water to test the second group ions.
H2S(aq)
2H+(aq) + S2-(aq)
HCl(aq)
H+(aq) + Cl-(aq)
The common ion available from HCl creates common ion effect on the equilibrium and
decreases concentration of S2- ions. Hence, the sulphides of the ions included in second group
can only be precipitated because their solubility products are less. In the same way, for
precipitation of ions of group III A ammonium chloride (in excess) is added along with NH4OH.
The concentration of OH- available from ionization of NH4OH gets decreased due to common
ion effect due to NH+4 available from NH4Cl. Hence, the hydroxides of the ions of III A group
only will be precipitated because the values of solubility products of the hydroxides of III A
group are low. Above this, if HCl gas is passed through saturated solution of NaCl, NaCl
becomes insoluble and separates as precipitates due to the common ino effect of Cl- ions.
Its is necessary to note that under certain situations the solubility increases instead of decreasing.
The solubility of salt like phosphates increase when acid is added to their solutions or pH of the
solution decreases. The reason for this is that, phosphate ion combines with H+ available from
acid and converts into acid by accepting proton. Hence, solubility of phosphate salt increases.
23. Explain with suitable example the use of common ion effect in qualitative analysis.
Ans. The use of effect of common ion can be made to separate one ion from the other in presence
of other ion in qualitative analysis. It can also be used for decrease in solubility of the
components in the mixture. In qualitative analysis, the solubility products of sulphides of metal
ions of second group are less in comparison to solubility products of sulphides of metal of III B
group ions, therefore, HCl is added before adding H2S water to test the second group ions.
H2S(aq)
2H+(aq) + S2-(aq)
HCl(aq)
H+(aq) + Cl-(aq)
The common ion available from HCl creates common ion effect on the equilibrium and
decreases concentration of S2- ions. Hence, the sulphides of the ions included in second group
can only be precipitated because their solubility products are less. In the same way, for
precipitation of ions of group III A ammonium chloride (in excess) is added along with NH4OH.
The concentration of OH- available from ionization of NH4OH gets decreased due to common
ion effect due to NH+4 available from NH4Cl. Hence, the hydroxides of the ions of III A group
only will be precipitated because the values of solubility products of the hydroxides of III A
group are low. Above this, if HCl gas is passed through saturated solution of NaCl, NaCl
becomes insoluble and separates as precipitates due to the common ino effect of Cl- ions.
Its is necessary to note that under certain situations the solubility increases instead of decreasing.
The solubility of salt like phosphates increase when acid is added to their solutions or pH of the
solution decreases. The reason for this is that, phosphate ion combines with H+ available from
acid and converts into acid by accepting proton. Hence, solubility of phosphate salt increases.
24. What is meant by common ion effect? In qualitative analysis, NH4Cl is added before
addition of NH4OH in precopotation of ions of III A group. Explain giving reason.
Ans. Let us take the example of weak acid, acetic acid ( )
CH3COOH(aq)+ H2O(l)
H3+O(aq) + CH3COO-(aq)
OR
HAc + H2O(l)
H3+O(aq) + Ac-(aq)
Where HAc and Ac- are the short forms of CH3COOH and CH3COO- ion.
[H3+O][Ac-]
Ka = [HAc]
………………..1
Suppose, we add CH3COONa or HCl, to the solution of HAc in equilibrium, then what will
happen? As studied earlier in chemical equilibrium if HCl is added, [H3+O] will increase and if
CH3COONa is added, [Ac-] will increase. Hence, according to Le Chatelier’s principle, the
equilibrium will make negligible change and will keep the same value of equilibrium constant.
This means that the equilibrium will be shifted towards left and concentration of HAc will
increase i.e. the amount of un dissociated acid will increase and there will be decrease in [H3+O]
and hence, there will be increase in pH. By addition of HCl due to increase in [H3+O] similar
result will be obtained. This effect is known as effect of common ion effect on dissociation
constant of acid.
In the same way, in the case of ionization of weak base NH3, if we increase [NH4+] by adding
salt like NH4Cl then, according to Le Chatelier’s principle, as [NH4+] increases the equilibrium
will shift towards left and hence, un dissociated NH3 will increase, i.e. [OH-] will decrease. As a
result, pH will increase.
In qualitative analysis, to precipitate the radicals of III A viz Fe2+, Fe3+, Al3+, Cr3+ as their
hydroxides, NH4Cl is added to the solution before adding NH4OH. Consequently, the reverse
reaction is favoured in the equilibrium, resulting in decrease of OH- ion concentration. Because
of this, the hydroxides of III B, IV groups and that of Mg2+ are not precipitated alon with those
of group III A.
NH4Cl(aq)
NH4+(aq) + Cl-(aq)
NH4OH(aq)
NH4+(aq) + OH-(aq)
The hydroxides of the radicals of III A group are extremely less soluble compared to those of
succeeding groups. So, in presence of NH4Cl the hydroxides of III A group are precipitated by
NH4OH.
(25). 2 moles of PCl5 is heated in closed vessel of 4 liter a t definite temperature. At
equilibrium state PCl5 remained undecomposed. Calculate Kc of the reaction.
Solution: Reaction,
PCl5(g)
PCl3(g) +Cl2(g)
Now, 55% PCl5 remains undecomposed.
Therefore, 45 % PCl5 goes under reaction.
= 45 x 2 = 0.9 moles PCl5
100
As ratio of PCl5 and (PCl3 and Cl2 ) is 1 : 1 in number of moles, 0.9 moles of PCl5 will yield 0.9
moles of PCl3 and Cl2.
PCl5(g)
PCl3(g) +Cl2(g)
In the beginning:
2 moles
0
0
Used moles at
equilibrium:
0.9 moles
0.9
0.9
Unused moles at
equilibrium:
1.1 moles
Now, Kc = [concentration of products]
[concentration of reactants] ……..(i) Kc = equilibrium constant
Now, at equilibrium, moles of reactants = 1.1 moles
and moles of products = 0.9 moles ( in solution)
Now, this vessel consists of 4 liter solution
Therefore, concentration = 1.1 = 0.275 M/L (for reactant)
4
= 0.9 = 0.225 M/L (for products)
…………(ii)
4
From equation (i) Kc = [PCl3] [Cl2]
[PCl5]
From equation (ii) Kc = (0.225) (0.225)
= 0.050625
(0.275)
0.275
Therefore, Kc = 0.184 mol/lit
Kc = 1.84 x 10-1 mol/lit.
(26). The value of Kp obtained at 1060 kelvin temperature is 0.033 mollit-1 for the reaction
2NOCl(g)
2NO(g) + Cl2(g). What will be the value of Kc for the reaction.
Solution:
2NOCl(g)
2NO(g) + Cl2(g).
∆n(g) = np – nr (for gaseous molecules)
=(2+1)–2=1
Data:
Kp = 0.033 atm
T = 1060K
0.082 atm lit mol-1 K-1
Kp = Kc x (RT)∆n(g)
 Kc = Kp
(RT)∆n(g)
Substituting the values from data,
Kc = 0.033
= 0.033
(0.082 x 1060)
86.92
-4
= 3.796 x 10 M
Therefore, Kc = 3.796 x 10-4 M
(27). At definite temperature and 3 atm pressure, 75 % PCl5 is decomposed into PCl3 and
Cl2. Calculate Kp of this reaction. PCl5(g)
PCl3(g) +Cl2(g).
Solution: 75% PCl5 is decomposed into PCl3 and Cl2.
At equilibrium,
moles PCl5
decomposed
100
75
1
x [ suppose initial moles of PCl5 is 1.]
x= 75 x 1
= 0.75 moles
100
0.75 moles of PCl3 and Cl2 are formed and unreacted moles of PCl5 = ( 1 – 0.75) = 0.25 moles
PCl5(g)
PCl3(g) +Cl2(g).
Initial moles:
1
0
0
Used moles at
equilibrium:
0.75
0.75
0.75
Unused moles
at equilibrium:
0.25
Now, total moles at equilibrium = 0.25 + 0.75 + 0.75
= 1.75 moles
Total pressure in system = 3 qtm
Now, partial pressure at equilibrium: moles of 1 element x total pressure
Total moles
Hence, partial pressure of PCl3 ( Ppcl3) = 0.75 x 3 = 3 x 3
1.75
7
= 9 atm
7
………… (i)
Similarly , partial moles of Cl2
Pcl2 = 0.75 x 3
= 3 x3
=9
1.75
7
7
atm …………..(ii)
Partial pressure of PCl5
Ppcl5 = 0.25 x 3 = 1 x3
= 3 atm
1.75
7
7
……………….. (iii)
Now, Kp = (Ppcl3) x (Pcl2)
(Ppcl5)
From equation (i) , (ii) and (iii)
Kp = 9/7 x 9/7
3/7
Kp = 3.857 atm.
(28). 1 mole N2 and 3 moles H2 are heated at 473 K temperature and 100 atmosphere
pressure. If 0.50 mole of NH3 are formed at equilibrium time calculate equilibrium
constant Kp for this reaction. N2(g) + 3H2(g)
2NH3(g)
Solution: N2(g) + 3H2(g)
2NH3(g)
Now, the ratio of production of moles of this reaction is 1 : 3 : 2
Initial moles of N2 = 1mole
Data
Initial moles of 3H2 = 3 moles
Therefore, from the ratio, moles of NH3 formed = 2 moles
But from the data, moles of NH3 fromed = 0.50 moles.
Therefore, ¼ th of NH3 is formed [ because 2 /0.50 = 4]
Hence, 1/4th of reactants must be used.
For N2 , ¼ x 1 = ¼ moles
For H2 , ¼ x 3 = ¾ moles
For NH3 = 0.50 moles ……(i)
Above these are the moles formed in solution
Therefore, unused moles at equilibrium
For N2 , 1 – ¼ = ¾ moles ….(ii)
For H2 , 3 – ¾ = 9/4 moles ……(iii)
N2(g) + 3H2(g)
2NH3(g)
Initial moles:
1
3
2
Used moles at
equilibrium
¼
¾
0.50
Unused moles
at equilibrium
¾
9/4
Partial pressure = moles of 1 element x total pressure
Total number of moles
Total pressure = 100 atm
Total number of moles = ¾ + 9/4 + 0.50
= 3.50 moles
Partial pressure of N2
PN2 = ¾ x 100
3.50
= 21.428 atm …………..(iv)
Partial pressure of H2
PH2 = 9/4 x 100
3.50
= 64.285 atm ……….(v)
Partial pressure of NH3
PNH3 = 0.50 x 100
3.5
= 14.2857 atm …….(vi)
Now, from equation N2(g) + 3H2(g)
2NH3(g)
Kp = (PNH3)2
=
(14.2857)2
(PN2) x (PH2)3
21. 2428 x (64.285)3
Kp ≈ 3.616 x 10-5 atm -2
(29). On heating 0.5 mole solid calcium carbonate in a vessel of 500ml volume, at 400 K
temperature , Kc 0.9 mol lit-1 is obtained at equilibrium state, then calculate mole of CO2 at
equilibrium. How many percent the reaction would have been completed?
CaCO3(s)
CaO(s) + CO2(g)
Solution:
CaCO3(s)
CaO(s) + CO2(g)
Initial moles:
Used moles at
equilibrium:
0.5
-
-
x
x
x [as ratio is 1 : 1]
unused moles at
equilibrium :
(0.5 – x)
concentration
(0.5 – x)
x
x
(mole/ liter)
0.5
0.5
0.5 [ volume in solution is 500ml (from data)]
Kc = [CaO][CO2]
[CaCO3]
= (x/0.5) (x/0.5)
= (2x) (2x)
((0.5 –x)/0.5)
2(0.5 – x)
2
Kc = 4x
1 – 2x
……….(i)
From data, Kc = 0.9 mol/lit
Therefore, from equation (i),
0.9 = 4x2
=> 9
= 4x2
1-2x
10
1 – 2n
9 – 18 x=40x2
 40x2 + 18x -9 = 0
Comparing with quadrative equation, ax2 + bx + c = 0
a = 40
b = 18
………(ii)
c= -9
Solution, x = - b ±√ b2 – 4ac
2a
Substituting values of a, b ,c from equation (ii)
x = -18 ± √ (18)2 – 4(40) (-9)
2( 40)
x= -18 ± √ 324 + 1440
80
x = -18 ± √ 1764
= -18 + 42
80
80
x = -18 + 42
or
x = -18 -42
80
80
x= 24 or
x= -60
80
80
(not possible as solution of x is negative value)
Therefore
3
=> x = 0.3 moles
10
Therefore, moles of Co2 at equilibrium = x = 0.3 moles
Therefore, from data ,
0.5 ml of CaCO3  0.3 moles of CO2 and 0.3 moles of CaO
Therefore, moles of CO2 formed = 0.3 moles
CaCO3(s)
CaO(s) + CO2(g)
0.5
0.3
100
(?)
0.3 x 100 = 60
0.5
Therefore, % of reaction completed = 60%
(30). Mg(HCO3)2(s) is decomposed as follows in a closed vessel. If mole fraction of CO2 is 0.8
and Kp = 64 atm3 then calculate total pressure of mixture at equlibrium.
Mg(HCO3)2
MgCO3(s) + CO2(g) + H2O(g)
Solution:
Mg(HCO3)2
MgCO3(s) + CO2(g) + H2O(g)
Now, gaseous mole in system are CO2 and H2O
Mole fraction of CO2 = 0.8
Total moles = 1
Therefore, mole fraction of H2O = 1 -0.8
= 0.2 …….(i)
Therefore, mole fraction Xco2 = 0.8
Mole fraction XH2O = 0.2
Now, partial pressure = moles of element x
total pressure
Total moles
= mole fraction x total pressure
Therefore, partial pressure of CO2 (PCO2) = Xco2 x total pressure
= 0.8 x ……….(ii) [let total pressure be x]
Partial pressure of H2O (PH2O) = XH2O x total pressure
= 0.2 x ………………(iii)
Now, Kp = [PCO2] [PH2O]
But, from data, Kp = 64 atm2 and from equation (ii) and (iii)
64 = (0.8x) ( 0.2x)
Therefore, 64 = 0.16x2
Therefore, x2 = 64
0.16
2
x = 400
therefore, x = 20 atm
Therefore, total pressure in system = 20 atm
(PCO2 = 16 atm PH2O = 4 atm)
(31). If 3.65 x 10 -2 gram HCl is dissolved in 500 ml solution, then find its pH (pH = 2.699)
Solution: Data,
For HCl, weight (W) = 3.65 x 10-2 gram
M = 36.5 gram/ mole
V = 500 ml = 0.5L
Moles of HCl = W = 3.65 x 10-2
M
36.5
= 1 x 10-3 mole
Concentration of [H+] in HCl = 1 x 10-3
0.5
= 2 x 10-3 M/L
pH= -log[H+]
= -log[2 x 10-3]
= - [log 2 + log (10-3)] [ log(axb) = log a + log b]
= 3 – log2
= 3 – 0.3010
pH= 2.6990.
(32). Find pH of mixture of 50ml 0.03 M HCl and 60 ml 0.02M NaOH .
Solution: 0.03 ml HCl  1000ml HCl solution contains 0.03 mol HCl
Therefore, 50 ml solution contains = 0.03 x 50
1000
= 1.5 x 10-3 moles HCl ……..(i)
Similarly,
0.02 M NaOH  1000ml NaOH contains 0.02 mol NaOH
Therefore, 60 ml solution contains = 0.02 x 60
1000
= 1.2 x 10-3 moles NaOH …..(ii)
NaOH + HCl neutralisation
NaCl + H2O
Therefore, 1.2 x 10-3 moles of NaOH will neutralize 1.2 x 10-3 moles of HCl out of 1.5 x 10-3
moles of HCl
Therefore, moles of HCl that remains unneutralised are = (1.5 – 1.2) x 10-3
= 0.3 x 10-3
= 3 x 10-4 moles ………(iii)
Total volume of the mixture = 110ml
Therefore, molarity of HCl after neutralization = moles
= moles x 1000
Volume(L)
V (ml)
-4
= 3 x 10 x 1000
[from equation (iii)]
110
Therefore, molarity of HCl = 2.727 x 10-3 M
Now, HCl is strong acid and hence complete ionization takes place
Therefore, HCl(aq) + H2O(l)  Cl-(aq) + H3O+(aq)
2.727 x 10-3 M
2.727 x 10-3 M
pH= - log[H3O+]
= - log [ 2.727 x 10-3]
Therefore, pH = [ log 2.727 + ( -3) log 10]
pH = - log 2.727 + 3log10
therefore, pH = -0.4357 + 3
therefore, pH=2.5643
(33). Find weight of CH3COOH in 100ml aqueous solution of CH3COOH having pH 3.0 Ka
for CH3COOH = 1.75 x 10-5
Solution: CH3COOH + H2O
H3O+ + CH3COOAs CH3COOH is weak acid, partial ionization takes place
Therefore, Ka = 1.75 x 10-5
Therefore, for weak acid, Ka = [H3O+]2
Co
Where,Co = initial concentration of weak acid i.e. CH3COOH
 Co = [H3O+]2
Ka ………….. (i)
Now, pH = - log10 [H3O+]
= 3 [given]
Therefore, -log[H3O+] = 3
Therefore, [H3O+] = 1 x 10-3 M …………………..(ii)
Substituting the value of [H3O+] from equation (ii) in equation (i)
Therefore, Co = [H3O+]2 = ( 1 x 10-3)2
Ka
1.75 x 10-5
Co = 5.7142 x 10-2 M
Now, concentration = mole
Liter
Therefore, 1 liter solution conatin 5.7142 x 10-2 moles of acetic acid
1L
5.7142 x 10-2 moles
100ml(10-1L)
?
-2
= 5.7142 x 10 =
5.714 x 10-3 moles
10-1
Now, mole =
weight
Molecular weight
M.W. of CH3COOH = 60 gram / mole
Therefore, moles= W
60
Therefore, W = 60 x moles
= 60 x 5.7412 x 10-3
W = 3.428 x 10-1 gram
= 0.342 gram.
(34).0.02 M CH3NH2 is ionized to 15% at equilibrium .Find its ionization constant.
Solution: CH3NH2(g) + H2O(l)
CH3NH+3(aq) + OH-(aq)
Methyl amine ionizes to 15% at equilibrium
If 100 moles of CH3NH2 are taken, then 15 moles of OH- are formed
CH3NH2
OH100
:
15
0.02
:
(?)
= 15 x 0.02
= 0.3 x 10-2 M
100
Therefore, [OH ] = 3 x 10-3 M
Now, CH3NH2 is weak base
Therefore, weak base , Kb = [OH-]2
Co
Where, Kb = ionization constant of weak base
Co = initial concentration of base = 0.02 M
Therefore, Kb = ( 3 x 10-3)2
0.02
= 9 x 10-6
2x 10-2
Therefore, Kb = 4.5 x 10-4
(35). For NH3 Kb = 1.77 x 10-5 , what will be the pH of 500 ml solution containing 1.7 gram
ammonia?
Solution: NH3(g) + H2O(l)
NH+4(aq) + OH-(g)
For weak base, Kb = 1.77 x 10-5
V (ml) = 500 ml = 500 x 10-3 L
W = 1.7 gram
M.W. of ammonia = 17 gm/ mole
Concentration = moles
Liter
=
weight of NH3 (gm)
M.W. of NH3 (gm/mole) x volume (L)
=
1.7
17 x 500 x 10-3
= 0.2 M
Therefore, concentration = 0.2 M ……..(i)
For weak base,
Kb = [OH-]2
Co
Where, Co = initial concentration of NH3 = 0.2 M (from equation (i))
Therefore, [OH-]2 = Kb x Co
= 1.77 x 10-5 x 0.2
= 3.54 x 10-6
Therefore, [OH ] = Kb x Co
= 1.77 x 10-5 x 0.2
= 3.54 x 10-6
Therefore, [OH-] = √ 3.54 x 10-6
Therefore, [OH-] = 1.881 x 410-3 M
Now, pOH = -log10[OH-]
= -log10 [ 1.881 x 10-3]
Therefore, pOH = 3 – log(1.881)
= 3 – 0.2744
Therefore, pOH = 2.7256
Now, pH + pOH = 14
Therefore, pH = 14 – pOH
14 – 2.7256
Therefore, pH = 11.2744
(36). pH of 0.05M solution of a weak acid is 3.68 then find Ka of that acid.
Solution: for weak acid,
Initial concentration (Co) = 0.05M
pH = 3.68
Ka = ?
Now ,pH = - log10[H3O+] = 3.68
Therefore, -log[H3O+]=3.68
Shifting minus (-) sign to right side and taking anti –log on both sides
Therefore, [H3O+] = - antilog(3.68)
anti-log (-3.68)
= antilog(-3.68)
-1 +1
= antilog( 4 + 0.32)
antilog (-4 + 0.32)
Therefore, [H3O+] = antilog(0.32) x 10-4
Therefore, [H3O+] = 2.089 x 10-4 M
Now, for weak acid,
Ka = [H3O+]2
Co
Therefore, Ka = (2.089 x 10-4)2
0.05
= 87.278 x 10-8 = 8.727 x 10-7
Therefore, Ka = 8.7278 x 10-7
(37). Find pH of 0.025M CH3COONa solution. Ka of CH3COOH = 1.75 x 10-5.
Solution: CH3COONa  CH3COO(aq)- + Na(aq)+
0.025M
0.025M
0.025M
Now, CH3COO-(aq) + H2O(l)
CH3COOH(aq) + OH-(aq) . ……….(i)
[ A- + H2O
HA + OH-(aq)]
Weak acid
+
H2O(l)
H(aq) + OH-(aq)
Now, hydrolysis constant, Kn = Kw
Ka
[CH3COONa is salt of strong base (NaOH) and weak acid (CH3COOH) and hence, Kn = Kw
Ka ]
Therefore, Kn = Kw = [CH3COOH][OH ] and from reaction (i)
Ka
[CH3COO-]
As concentration of [OH ] produced by the self- ionization of water is negligible in comparison
with the concentration of OH- produced by hydrolysis reaction.
Therefore, [CH3COOH] = [OH-] (At equilibrium)
Therefore, Kn = Kw =
[OH-]2
Ka
[CH3COO-]
Kw = ionic product of water = 1x 10-14
Ka = 1.75 x 10-5
[CH3COO-] = initial concentration of CH3COONa
= Co = 0.025 M
Substituting , these values in above equation
Kw = [OH-]2
=> 1 x 10-14 = [OH-]2
Ka
[CH3COO ]
1.75 x 10-5
0.025
Therefore, [OH-]2 = 1.4285 x 10-11
= 14.285 x 10-12
Therefore, [OH-] = 3.7796 x 10-6 M
≈ 3.78 x 10-6 M
Now, pOH = -log10 [OH-]
= -log[3.78 x 10-6]
= -log 3.78 + 6log10
= 6 –log(3.78)
= 6- 0.5775
Therefore, pOH= 5.4225
Now, pH + pOH = 14
 pH= 14 –pOH
= 14 – 5.4225
= 8.5775
Therefore, pH = 8.5775
(38). Kb of NH3 is 1.8 x 10-5.Calculate pH of 0.20M solution of NH4Cl
Solution: NH4Cl  NH4+ + ClNow, NH4Cl is salt of strong acid (HCl) and weak base (NH3)
 Kn = Kw
Kb
Where, Kn = hydrolysis constant
Kw = ionic product = 1 x 10-14
Kb = ionization constant of weak base = 1.8 x 10-5.
Now,
NH+4 + H2O
NH3 + H3O+
Now,
Kn = Kw = [NH3] [H3O] = [H3O+]2
Kb
[NH+4]
[NH+4]
+
[ because at equilibrium, [NH3] = [H3O ] ]
Therefore, Kw = 1x 10-14 = [H3O+]2
Kb
1.8 x 10-5
0.20
+ 2
Therefore ,[H3O ] = 1 x 0.20 x 10-9
1.8
= 0.111 x 10-9
= 1.111 x 10-10
+
Therefore, [H3O ] = 1.054 x 10-5 M
Now, pH = -log10 [H3O+]
= -log10[1.054 x 10-5]
= -log 1.054 + 5 log10
= 5 – log(1.054)
(log10 = 1)
= 5 -0.0229
= 4.9771
Therefore, pH = 4.9771
≈ 4.98
39. Find solubility of PbSO4 in water, having solubility product 1.3 x 10-8 (Molecular mass
of PbSO4 = 303 gram mol-)
Ans. Let, the solubility of PbSO4 be S mol/litre
PbSO4(s)
Pb+2(aq) + SO42-(aq)
S
S
+2
2Solubility product, Ksp = [Pb ][SO4 (aq)]
= [S][S]
= [S2]
But, Ksp = 1.3 x 10-8(given)
Ksp = S2 = 1.3 x 10-8
 S =1.140 x 10-4 mol/litre
Now, molecular weight of PbSO4 = 303 gm/litre
Solubility (in gram/litre) = mol/litre x M.W.
= 1.140 x 10-4 x 303
= 345.47 x 10-4 gm/litre
= 3.45 x 10-2 gm/L
 S = 3.45 x 10-2 gm/litre
40. Ksp of CaF2 is 1.7 x 10-10. What will be the volume of saturated solution of 10 milligram
salt? (Atomic mass : Ca = 40 and F = 19)
Ans. Let, molar solubility of CaF2 be S mol/litre
Solubility Equilibrium,
CaF2
Ca2+(aq) + 2F-(aq)
S
2S (Because 2 moles of F- are formed)
Ksp(given) = 1.7 x 10-10
Ksp = [Ca2+][F-]2
= s[2s]2
= 4S3
From above data, Ksp = 4S3 = 1.7 x 10-10
 4S3 = 1.7 x 10-10
 S3+ = 0.425 x 10-10
= 42.5 x 10-12
S = 3.489 x 10-4 mol/litre
[ S3 = 42.5 x 10-12
S = (42.5 x 10-4)1/3
(42.5)1/3 = (log 42.5)/3
= (1.6284)/3
= 0.5428
= antilog(0.5428)
= 3.489
3.49]
-4
 S 3.49 x 10 mol/L
Where S= Solubility of CaF2
Now, molecular weight of CaF2 = Ca + 2(F)
= 40 + 2(19)
= 78 g/mol
Now, molar Solubility =
Wt.(gm) x 1
Mol. Wt. (gm/mole) x v(in litre)
 Solubility of CaF2 = Wt. of CaF2(gm)
x
m.w of CaF2 (gm/mole)
Substituting the values in above reaction
W= 10 mg
= 10 x 10-3 gm
3.49 x 10-4 = 10 x 10-3 x 1000
78 x V(ml)
 V(ml) =
10 x 1
78 x 3.49 x 10-4
 V(ml) = 367.34 ml
1000
V(ml)
41. Ksp of CaF2 at 298 K temperature is 1.7 x 10-10. If a person drinks 2.5 litre of water
saturated with this salt every day, how much CaF2 will enter in his body every day?
(Molecular mass of CaCl2 =78 g mol-)
Ans. Suppose, Solubility of CaF2 be S mol/L
CaF2
Ca2+(aq) + 2F-(aq)
S
2S
Ksp = [Ca2+][F-]2 = s[2s]2
= 4S3
Now, Ksp(given) = 1.7 x 10-10
= 4S3
 S3 = 0.425 x 10-10
= 42.5 x 10-12
 S = (42.5)1/3 x 10-4 mol/L
 S = 3.489 x 10-4 mol/L
 3.49 x 10-4 mol/L
[ S = (42.5)1/3 = (log(42.5))/3
= (1.6248)/3
= 0.5428
= antilog(0.5428)
= 3.489 ]
Now, person drinks 2.5 litre of water saturated with CaF2 salt every day
 V = 2.5 L
Molecular mass = 7.8g/mol
S = 3.49 x 10-4 mol/litre
Solubility Product (s) = Wt (gm) x 1
M.W(gm/mole) x V(L)
3.49 x 10-4 = W(gm) x 1
78 x 2.5
 W(gm) = (3.49 x 10-4) x 78 x 2.5
= 680.55 x 10-4
= 0.068055 gm
 W(gm) = 0.068055 gm
= 6.8055 x 10-2 gm
42. pH of saturated aqueous solution of Ca(OH)2 is 12.25. Find Ksp of Ca(OH)2
Ans. Suppose, the solubility product of Ca(OH)2 be S mol/litre
Equilibrium state
Ca(OH)2(s)
Ca2+(aq) + 2OH-(aq)
S
2S
Now, pH = 12.25
PH + pOH = 14
 pOH = 14 - pH
= 14 – 12.25
= 1.75
OH
Now, p = -log10[OH-]
= 1.75
 -log10[OH-] = 1.75
Shifting ‘mius sign’ to the right side and taking ‘antilog’ on both sides
 [OH-] = -antilog(1.75)
= antilog(2.25)
= 1.778 x 10-2
 [OH-] = 1.778 x 10-2 M
But, according to reaction
[OH-] = 2S = 1.778 x 10-2M ….(i)
 S = 0.889 x 10-2 M …. (ii)
Now, Ksp = [Ca2+][OH-]2
= [S][2S]2 (from reaction)
= 4S3
= 4(0.889 x 10-2)3 [from equation (i)]
 Ksp = 2.810 x 10-6 M3
43. What will be the value of Ksp if the concentration in saturated solution of Mg(OH)2 is
8.2 x 10-4 %W/V. (Molecular Mass of Mg(OH)2 = 58.3 g.mol-).
Ans. 8.2 x 10-4% W/V means 100 ml saturated solution contains 8.2 x 10-4 gm Mg(OH)2
100 ml
8.2 x 10-4 gm Mg(OH)2
1000 ml(1L)
(?)
-4
= 8.2 x 10 x 1000
100
= 8.2 x 10-3 gm/litre
M.W. of Mg(OH)2 = 58.3 gm/mol
 Molarity of saturated solution of Mg(OH)2
= 8.2 x 10-3 gm/litre
58.3 gm/mole
-4
 S = 1.4065 x 10 mol/litre
Mg(OH)2(s)
Mg2+(aq) + 2OH-(aq)
S
2S
2+
- 2
Now, Ksp = [Mg ][OH ]
= [S][2S]2
= 4S3
= 4(1.4065 x 10-4)3
= 11.130 x 10-12 M3
 Ksp = 1.113 x 10-11 M3
44. How many grams of Zn(OH)2 can be dissolved in 2 litres of 0.02 M NaOH solution ?ksp
for Zn(OH)2 = 4.5 x 10-17.
Ans. Suppose solubility of Zn(OH)2 is S mol/litre
Zn(OH)2(s)
Zn2+(aq) + 2OH-(aq)
S
2S
NaOH(aq)
Na+(aq) + OH-(aq)
0.02M
0.02M
0.02M (given)
So, total concentration of [OH ] in the solution
= [OH-] from Zn(OH)2 + [OH-] from NaOH
= 2S + 0.02
(But here, 0.02 >>> S, hence [OH-] = 0.02 M )
(because Zn(OH)2 is sparingly soluble)
TotL [OH ] = 0.02 M
Now, Ksp for Zn(OH)2 = [Zn2+][OH-]2
= [S][0.02]2
 Ksp = S x 4 x 10-4
But Ksp = 4.5 x 10-17(given)
Ksp = 4.5 x 10-17 = S x 4 x 10-4
S = 4.5 x 10-17
4 x 10-4
S = 1.125 x 10-13 mol/litre
Mol. Wt of Zn(OH)2 = Zn + 2(O) = 2(H)
= 65 + 2(16) + 2(1)
= 99 gm/mol
 Solubility of Zn(OH)2 in o.o2M NaOH solution
= (1.125 x 10-13) x 99 gm/litre
= 1.11375 x 10-11 grams/litre
But, given solution is of 2 litres
 Solubility = 1.11375 x 10-11(gm/L) x 2 L
Solubility of Zn(OH)2 = 2.2275 x 10-11 gm
45. Solubility product of Mg(OH)2 is 1.2 x 10-11. Calculate its solubility in pure water as
well as in aqueous solution of 0.05 M NaOH
Ans. Suppose, solubility of Mg(OH)2 be S mol/L
Mg(OH)2(s)
Mg2+(aq) + 2OH-(aq)
S
2S
NaOH(aq)
Na+(aq) + OH-(aq)
0.05M
0.05M
0.05M
(a) Solubility in pure water
Ksp = [Mg2+][OH-]2
= [S][2S]2 = 4S3
But, Ksp = 1.2 x 10-11 (given)
4S3 = 1.2 x 10-11
 S3 = 0.3 x 10-4
= 3 x 10-12
1/3
S = (3) x 10-4
S= 1.442 x 10-4 mol/litre
[ (3)1/3 = (log3)/3
= (0.4771)/3 = 0.1590
= antilog(0.1590)
= 1,442]
S = 1.442 x 10-4 mol/litre……(a)
(In pure water)
(b) Solubility in 0.05M NaOH
Total concentration of [OH-]
= 2S + 0.05
[ But, here 0.05 >>> S as Mg(OH)2 is sparingly soluble)
[OH-] = 0.05 M
Concentration of Mg2+ = [S] = ?
-
Ksp = [Mg2+][OH-]2
Ksp = 1.2 x 10 = [S][0.05]2
= 1.2 X 10-11 = S x 25 x 10-4
S = 1.2 x 10-11 = 0.048 x 10-7 = 4.8 x 10-9
25 x 10-4
S = 4.8 x 10-9 mol/litre
= 4.8 x 10-19 M
46. At 298 K temperature, 2.901 litre saturated solution can be prepared by dissolving 0.08
gram CaF2. Calculate Ksp of salt. (Molecular mass of CaF2 = 78.0 g.mol-)
Ans. W = 0.08 gm
Molecular wt = 78 gm/mol
V(L) = 2.901 litre
Now, concentration = 3.5354 x 10-4 M
Now,
CaF2
Ca2+
+
2F(3.53 x 10-4)
(2 x 3.53 x 10-4)
2+
- 2
Now, Ksp =[Ca ][F ]
= (3.53 x 10-4)(2 x 3.53 x 10-4)2
= 4 x (3.53 x 10-4)3
= 4 x 44.191 x 10-12 M3
= 176.767 x 10-12 M3
= 1.7676 x 10-10 M3
Ksp = 1.768 x 10-10 M3
47. Equal volumes of 2 x 10-4 M BaCl2 and 5.0 x 10-3 M H2SO4 are mixed. Will precipitation
occur or not? (Ksp of BaSO4 = 1.1 x 10-10)
Ans, If equal volumes of solutions are mixed, concentration of each salt will be halved
[C1V1 = C2V2
But if V2 = 2V1  C1V1 = C2(2V1)
 C2 = (C1)/2]
 Concentration of BaCl2 in the mixture
= 2 x 10-4 M = 1 x 10-4 M ….(i)
2
Concnetration of H2SO4 in the mixture = (5 x 10-3)/2 M = 2.5 x 10-3 M …(ii)
BaCl2
Ba2+
+
2Cl-4
-4
(1 x 10 )
1 x 10 M
(2 x 1 x 10-4) M
+
H2SO4
2H
+
SO4-2
(2.5 x 10-3)M
(2 x 2.5 x 10-3)M
2.5 x 10-3 M
When, they are mixed, salt formed is BaSO4
BaCl2 + H2SO4
BaSO4 + 2HCl
2+
BaSO4(aq)
Ba (aq) + SO42-(aq)
Ionic Product of BaSO4, Ip = [Ba2+][SO42-]
= (1 x 10-4)(2.5 x 10-3)
Ip = 2.5 x 10-7 M2
But, Ksp = 1.1 x 10-10(given)
Comparing above both values
Ip > Ksp
-11
 Precipitation will occur
48. By addition of how many grams of FeSO4 to 500 ml solution of 0.02 M NaOH, Fe(OH)2
will be precipitated? (Ksp of Fe(OH)2 = 1.5 x 10-15 )
Ans. NaOH
Na+ + OH0.02M
0.02M
Fe(OH)2
Fe+2(aq) + 2OH-(aq)
S
S
2S
[Suppose, molar solubility of Fe(OH)2 be S mol/L
Now, Fe(OH)2 is sparingly soluble salt
 S <<< 0.02 M
 [OH-] = 0.02 M
Now, Ksp = [Fe2+][OH-]2 …(i)
But Ksp = 1.5 x 10-15 (given) …..(ii)
For, precipitation to occur, Ip > Ksp …(iii)
 From equation (i), (ii), (iii)
Ip = Ksp = 1.5 x 10-15 = [Fe2+][OH-]2
S[0.02]2
 1.5 x 10-15 = S x 4 x 10-4
 S = 3.75 x 10-12 M
 S = [Fe2+] = 3.75 x 10-12 M
Now, l mole [Fe2+] = 1 mol FeSO4
[ because FeSO4
Fe2+(aq) + SO42-(aq)]
 [Fe2+] = [FeSO4] = 3.75 x 10-12 mol/litre
For l litre, FeSO4 = 3.75 x 10-12 moles
Hence for 500 ml = (3.75 x 10-12)/2
= 1.875 x 10-12 moles/500 ml
Mol. Wt of FeSO4 = Fe + S + 4(O)
= 56 + 32 + 4(16)
= 152 gm/mol
 Wt of FeSO4 = 1.875 x 10-12 x 152
= 285 x 10-12
= 2.85 x 10-10 gm
For precipitation, Ip > Ksp
 Wt (FeSO4) > 2.85 x 10-10 gm.
Thus, if FeSO4 added in slightly greater than 2.85 x 10-10 gm, Ip > Ksp and Fe(OH)2 will
precipitate.
49. Will precipitates of PbI2 be obtained by mixing 20 ml 3 x 10-3 M pb(NO3)2 and 80 ml.
2 x 10-3 M NaI Solutions ? (Ksp for PbI2 = 6.0 x 10-9)
Ans. Pb(NO3)2(aq)
Pb2+(aq) + 2NO3-(aq)
-3
3 x 10 M
3 x 10-3M
2(3 x 10-3M)
[Pb2+] = 3 x 10-3 M….(i)
NaI(aq)
Na+(aq) +
I-(aq)
2 x 10-3M
2 x 10-3M
2 x 10-3M
-3
[I ] = 2 x 10 M….(ii)
20 ml Pb(NO3)2 and 80 ml NaI solution is mixed.
 Total volume, V2 = (20 + 80) ml
= 100 ml
V2 = 100 ml
Now, concentrations of Pb2+ and I- ions in mixture according to M1V1 = M2V2.
For Pb2+
M1 = 3 x 10-3M
V1 = 20 ml
V2 = 100 ml.
M1V1 = M2V2  M2 = M1 x V1
V2
-3
 M2 = (3 x 10 ) x 20 = 6 x 10-4
100
2+
 [Pb ] 6 x 10-4 M…(iii)
For IM1 = 2 X 10-3 M
V1 = 80 ml
V2 = 100 ml
M1V1 = M2V2  M2 = M1 x V1
V2
= (2 x 10-3) x 80 = 1.6 x 10-3 M
100
[I-] = 1.6 x 10-3 M….(iv)
Now, Ksp = 6 x 10-4 (given)…(v)
For PbI2,
Ip = [Pb2+][I-]2
[ PbI2(S)
Pb2+(aq) + 2I-(aq)]
From equation (iii) and (iv)
Ip = (6 x 10-4) x (1.6 x 10-3)2
Ip = 15.36 x 10-10
 Ip = 1.536 x 10-9 ….(vi)
From equation (v) and (vi)
Ip(1.536 x 10-9) < Ksp(6.0 x 10-9)
Ip < Ksp
 PbI2 will not precipitate
50. The concentration of F- in a sample of water is 2 x 10-5 M. How many minimum grams
of solid CaCl2 will have to be added for precipitation of F-?( Ksp for CaF2 = 1.7 x 10-10,
molecular mass of CaCl2 = 111 grammolAns.
[F-] = 2 x 10-5 M
CaCl2
Ca2+(aq) + 2Cl-(aq)
Suppose, x mole of CaCl2 should be added to 1 litre of water
concenetration of Ca2+ in water will be x M
 [Ca2+] = x M
CaF2(aq)
Ca2+(aq) + 2F-(aq)
xM
(3 x 10-5) M
-10
Ksp = 1.7 x 10
(given)
Now, Ksp = [Ca2+][F-]2
= (X)(3 x 10-5M)2
1.7 x 10-10 = X x 9 x 10-10
 X = (1.7)/9 = 0.1888 M
 X = 0.189 M …..(i)
Now, 1 mol of Ca2+ = 1 mole of CaF2
= 1 mole of CaCl2
1 mole of CaF2 = 1 mole of CaCl2.
0.189 moles of CaF2 = 0.189 moles of CaCl2 [from equation (i)]
If little more than 0.189 moles of CaCl2 is added i.e 0.20 moles, then CaF2 will precipitate.
 moles = 0.20
Molar mass of CaCl2 = 111 gm/mol
 0.20 mol of CaCl2 will contain
= 111 gm/mol x 0.20
= 22.2 gm solid CaCl2
22.2 gms solid CaCl2 is added for precipitation of F- (minimum value)
Section –D
1. Explain dynamic nature of chemical and physical equilibrium.
Ans. Dynamic Nature of Equilibrium
The most important matter in the case of equilibrium is that there is a continuous transformation
of reactant to product and product to reactant. This state appears to be steady but it is not so. This
type of reaction which takes place in both the directions is called reversible reaction and it is
expresses by the symbol of two half arrows(↔). This symbol indicates that such reaction occurs
simultaneously in both (forward and reverse) directions. Generally, the change of reactant to
product is called forward reaction and the change of product to reactant is called reverse reaction.
Thus, in reversible reactions, forward and reverse reactions continuously occur and we find it as
equilibrium state. The misture of reactants and products obtained at equilibrium time is called
equilibrium misture. The decomposition reaction of solid calcium carbonate in a cold vessel, at
high temperature can be shown as below.
∆
CaCO3(s) ↔CaO(s) + CO2(g)
The equilibrium is dynamic and not steady or static as the forward and the reverse
reactions occur with the same velocity at the equilibrium time in equilibrium reactions in closed
vessels. In the above reaction obtaining CaO and CO2 by decomposing of CaCO3and obtaining
CaCO3 by combination of CaO and CO2 continuously take place. Suppose, if we deposit, some
amount in our bank account and withdraw the same amount, then balance in the account appears
steady or static. But this can be considered operative or dynamic and not closed or static. It is
very difficult to determine the dynamic nature of equilibrium, even then with the help of
radioactive isotope, it can be proved, viz. 14CO2 gas containing radioactive isotope, 14C and
CaCO3 are taken in two different flasks and CO2 obtained by decomposition is connected with
vessel containing 14CO2 gas, after sometime, Ca14CO3 will be formed in the vessel of CaCO3 and
CO2 will be obtained in the vessel containing 14CO2. Thus if the equilibrium would have been
steady, there must not be exchange of 12C and 14C. With the help of suitable counter, the
radioactivity can be measured and the proof for the dynamic nature of equilibrium can be
obtained through the proportions of concentrations of reactants and products remain constant.
The reaction can be fast or slow depending upon the nature of the reactant and the experimental
conditions.
Equilibrium reactions can be divided into following three categories:
(i) Reactions which are almost at the extent of completion and concentration of
reactants may be negligible. It is not possible to detect this experimentally.
(ii) Reactions in which the products are formed in very less proportions and most part
of the reactant remains unchanged at the equilibrium.
(iii) Reactions in which the concentrations of reactants and products are in comparable
proportions at equilibrium.
Dynamic Nature of Chemical Equilibrium
The dynamic nature of chemical equilibrium can be demonstrated by taking example of the
production of ammonia. By keeping known quantities of dinitrogen and dihydrogen at high
temperature and pressure in a closed vessel, the amount of ammonia formed can be determined at
constant intervals by a series of experiments. The quantities of unreacted dinitrogen and
dihydrogen also can be determined. From this it is concluded that even if the reactants and
products are in different proportions, their concentrations are same at equilibrium. This
constancy in composition indicates dynamic nature of equilibrium. For this, in synthesis of
ammonia, deuterium (D2) instead of dihydrogen, (H2) is used and ammonia gas is produced by
Haber process and it is studied. The results obtained are similar to those obtained above. In the
mixture proportions of N2, Dand ND3 instead of N2, H2, NH3 can be determined and equilibrium
can be obtained. If D2 is added after the formation of ammonia by reaction of N2 and H2, the
reaction may not occur but H in NH3 is displaced by D and ND3 can be determined by mass
spectrometer. Thus, it is proved that in the reaction
That the reactions from reactants to products and products to reactants that is forward and
reverse reactions continuously occur with the same rates and so ND3 instead of NH3is obtained.
By the use of radioactive isotope, the dynamic nature of equilibrium can be proved viz. For the
reaction, H2(g) + I2(g)
2HI(g)
Radioactive isotope 131I of iodine can be used to study the dynamic nature of chemical
equilibrium. As the equilibrium is dynamic, certain properties or factors are found similar. E.g.
Intensity of colour, constant pressure, constant concentration etc.
2. Explain equilibrium solid in solution and gas in solution.
Ans. Equilibrium Involving Dissolution of Solid in Solution
For the study of this type of equilibrium, the example of sugar and its solution in water can
be taken. At constant temperature, take some water. Add some sugar into it and stir. In the
beginning sugar easily dissolves, but as more and more amount of sugar is added, it dissolves
according to its solubility and then some amount of sugar remains in solid form without
dissolution. We know this state as saturated solution but in this system the equilibrium is also
establishes sugar(solid) and liquid(solution of sugar). This can be expressed as below:
Sugar(s)
Sugar solution(aq)
Solid
liquid
As studies earlier, equilibrium is dynamic because the forward and the reverse reactions
continuously occur in each system. In this system, the amount of sugar that dissolves in water,
the same amount of sugar separates from solution of sugar. Hence, the number of molecules of
sugar and number of aqueous molecules of sugar in the solution remain constant in this system.
Equilibrium involving Gas and Solution
At constant temperature and pressure, carbon dioxide can be dissolved in water in a closed
vessel(system), so that a system containing gas and a solution of carbon dioxide can be formed.
As temperature and pressure are constant, carbon dioxide dissolves according to its pressure and
temperature and forms solution of carbon dioxide and the excess carbon dioxide gas remains in
equilibrium with it. Equilibrium is dynamic and so molecules have carbon dioxide gas that
dissolve in water is the same as the number of molecules of the gas that release from the solution
in the system. Thus, in this closed system, the total number of molecules(amount) of carbon
dioxide in gaseous form and those that have dissolved in water remain constant. This equilibrium
can be shown as below:
CO2(g)
CO2(soln)
All the reactions studies above are processes in which only physical change takes place and so
they all are examples of physical process equilibrium.
3. Mention the characteristics of chemical equilibrium
Ans. Characteristics of Chemical Equilibrium
(viii) During chemical equilibrium, the properties like colour, concentration, pressure or
temperature of the system remain constant and they are similar in the total area of the
system.
(ix)
When chemical equilibrium is attained, the rates of forward and reverse reactions
become equal.
(x)
If the factors like concentration, pressure, temperature etc. which affect the chemical
equilibrium are changed, they produce effect on equilibrium.
(xi)
Even if the initial concentrations are different, the equilibrium constant remains
constant at constant temperature.
(xii) The value of equilibrium constant changes if the temperature changes.
(xiii) For attainment of equilibrium, the reaction can be carried out from left to right
(reactants to products) or from right to left (products to reactants)
(xiv) There is no effect of catalyst on the equilibrium constant and so the proportions of
products remain same but the rate of reaction to attain equilibrium increases.
4. Give chemical equilibrium (active masses) law and derive its formula
Ans. Law of Chemical Equilibrium and Equilibrium Constant
The mixture of reactants and products at equilibrium is called equilibrium mixture. We shall
study the relation between concentrations of reactants and products at equilibrium state.
Let us take a simple reversible reaction as follows:
A+B
C + D
In this reaction A and B are reactants and C and D are products. This means that in this reaction
moles of reactants and products are one each but in all reactions this may not happen. Hence, it is
necessary that their moles are expressed. Balanced reaction determines their moles. Viz.
N2(g) + 3H2(g)
2NH3(g)
From the experimental studies of many reversible reactions scientists of Norway, Guldberg and
Waage mentioned in 1864 that the concentrations of substances in equilibrium mixture can be
expressed by following equilibrium equation.
Where Kc is equilibrium constant and [ ] bracket expresses concentration of reactant or product
on mol/lit or M. The equilibrium equation is also known as law of active masses because in the
early years of chemistry, concentration was said to be ‘active mass’.
Now, we shall derive the equation for equilibrium constant of a general reaction. Suppose, if a
reaction takes places, as given below in which the reactants and products are shown in balanced
form with their proper moles (a, b, c or d)
aA + bB
cC + dD
On the basis of Guldberg and Waage’s law the rate of forward reaction
Vf [A]a [B]b or
Vf = Kf [A]a [B]b
Where Kf is the proportionality constant for forward reaction.
The rate of reverse reaction
Vr [C]c [D]d
Vr= Kr [C]c [D]d
Where Vr is the proportionality constant for reverse reaction.
At equilibrium the rates of forward and reverse reaction will be equal and so Vf = Vr
That is, Kf[A]a [B]b = Kr [C]c[D]d
Thus, when equilibrium is attained if we determine the concentration of the reactants and the
products in any reaction and their stoichiometric multiples, the equilibrium constant Kc can be
obtained.
5. What is meant by Kc and Kp? Deduce relation between Kc and Kp.
Ans. Relation between Kp and Kc
As seen earlier the equilibrium constant of a gaseous reaction can be written as
…. 1
But we know that according to simple gas equation PV = nRT. Hence, it can be written as
 P = (n/V)RT = CRT
(where n/V = C = concentration in mol lit-1)
Substituting the values of p in the above equation 4.10, it can be written as
….2
ng
…3
= Kc X (RT) …. 4
Where ∆ng = (c+d) – (a+b)
Means number of total moles of gaseous products minus number of total moles of gaseous
reactants.
Hence, it can be written as
Kp = Kc X (RT)ng…. 4
6 Deduce the equilibrium constant Kp pCO2 for decomposition of solid CaCO3 in closed vessel
Ans. Kc = [CaO][CO2]
Now, Density of CaCO3 = Constant
[CaCO3]
Density of CaO
 Kc x Density of CaCO3 = [CO2]
Density of CaO
 Kp = PCO2
7. Deduce the formula Kp = (P2)/4 for decomposition of solid NH4HS in closed vessel.
Ans. NH4HS(s)
NH3(g) + H2S(g)
x
x
Total moles = 2x
Total pressure = Patm P/2
P/2
Kp = [PNH3][PH2S] = (P/2)(P/2) = (P2/4) atm2
8. Derive the formula for decomposition constant of solid ammonium carbonate and prove
that Kp = (4/27)P3.
Ans. NH4COONH2(s)
2NH3(g) + CO2(g)
Moles = 2x
x
Total moles = 3x
Kp = [PNH3]2[PCO2] = (2P/3)2(P/3)2 = (4P3/27) atm3
9. Give Le-Chatelier’s principle and explain in detail the effect of concentration on
equilibrium.
Ans. Equilibrium constant is independent of initial concentration of reactant but, if change is
carried out in concentration of pressure, there is an effect on equilibrium, the equilibrium tries to
nullify this effect. Reaction can be endothermic or exothermic. Hence, the heat absorbed or
evolved, functions like that of the reactant and its effect is on equilibrium which tries to nullify
the effect. Le Chatelier studies the effect of concentration and temperature and his presentation is
called Le Chatelier’s principle whose statement can be written as below:
“ If from the factors determining the equilibrium state, any one factor is changed, there
will be such a change in the system that the effect will be nullified or made negligible so that the
value of equilibrium constant at that temperature will remain constant.” This principle can be
applied to both physical and chemical equilibrium.
We shall study in detail the factors like (1) change in concentration (2) change in pressure
(3) addition of inert gases (4) change in temperature and (5) use of catalyst affecting the
equilibrium
(1)Effect of change in concentration:
If we add or remove the reactant or the product from the reaction in equilibrium, its effect on
equilibrium according to Le Chatelier’s principle will be as follows:
(a)If the concentration of reactant or product is increased by addition of reactant or product, the
reaction will occur in such a way that the increase in concentration will be taken for use, i.e. the
increase in concentration of reactant, the concentration of product will increase and if
concentration of product is increased the reaction will result in the direction of increase in
concentration of the reactant.
(b)If the concentration of reactant or product is decreased by removing reactant or product, the
reaction will occur in such a way that product or reactant will be established again.
Hence, if any change in concentration of reactant of product is carried out then the equilibrium
will try to make this effect minimum and equilibrium will be established accordingly. If we take
this as an example, in the reaction
If concentration of reactants dinitrogen or dihydrogen is increased, the reaction will proceed
towards right hand side and the product ammonia obtained will be more. If the concentration of
nitrogen or hydrogen is decreased, the reaction will proceed towards left hand side and reactants
of nitrogen or hydrogen will be obtained back i.e. production of ammonia will decrease. Here, it
is necessary to remember that 1 mole dinitrogen combines with 3 moles of dihydrogen in this
reaction and forms 2 moles of ammonia. Hence, increase in concentration of dihydrogen rather
than dinitrogen, will give more product. We take another example of heterogeneous equilibrium.
If solid CaCO3(s) is heated in closed vessel, the following decomposition reaction will occur.
CaCO3(s)
CaO(s) + CO2(g)
Hence, if more CaO(s) is to be obtained then, the CO2(g) formed in the reaction should be removed
because CO2(g), gas can combine with solid CaO(s) and carry out reverse reaction then proportion
of product will decrease. Hence, by removing CO2(g) from the reaction vessel, more CaO(s) can
be obtained.
10. Explain Arrhenius acid-base theory and mention it’s drawbacks
Ans. Arrhenius concept about Acid and Base:
According to Arrhenius concept, substances which dissociate in water and give hydrogen ion
(H+) are called acids and substances which dissociate in water and give hydroxyl ions (OH-) are
called bases. E.g;
(1) HCl(g)
+H2O
+H2O
H+ + Cl-
Acid
(2) NaOH(s)
Na+ + OHBase
The limitations of this concept are as follows:
(i) proton (H+) is highly unstable.
(ii) It can not exist independently.
(iii) It immediately combines with molecules of solvent water and gives H+3O ions. Its addition
in certain bases OH- is not present even then they show properties of base, viz., NH3. Similarly
compounds like BF3 do not possess H+, even then they act as acid.
(iv) this concept can only be applied to aqueous solution because salt like NH4Cl reacts as acid in
solvent like liquid NH3. Hence, it is difficult to accept Arrhenius concept as the universal one.
Because ionization is given importance. (In addition it is necessary to know that ionic radius of
H+ is about 10-15 meter and so it is very small in size9. Hence it easily gets combined with
molecules of water and forms H3+O ion which is called as hydronium ion. One estimate is that
one H+ can be combined with four molecules of water showing H+ + 4H2O  H9O4+ ion.
11. Explain with suitable example, Lowry-Bronsted acid-base theory(proton transfer
theory)
Ans. Concept of Bronsted – Lowry for acid and base:
Danish chemist Bronsted, an English chemist Lowry presented the concept of acid and base.
They made H+ (proton) as a base. According, to their concept , the substance which gives a
proton or donates a proton is called the acid and the substance which receives the proton or
accepts a proton is called the base. Thus, acid is a proton donor and base is proton acceptor. Let
us take the dissociation reaction of hydrogen chloride in water.
Base
Acid
H3+O (aq) + Cl-(aq)
HCl(g) + H2O(l)
Acid
Similarly,
Base
Acid
NH3(g) + H2O(l)
Base
NH4+(aq)+ OH-(aq)
Base
Acid
We shall understand in detail the first from the above reactions :
HCl
H+
+ ClAcid-1 proton conjugate base-1
As it gives proton, HCl is an acid
H2O + H+
H3O+
Base-2
Proton
Conjugate acids-2
As it accepts a proton, H2O is a base.
Total reaction:
HCl + H2O
H3O+(aq) + Cl-(aq).
Acid-1
Base-2
Acid-2
Base-1.
In the above reaction giving-taking of proton is not shown. Hence, it can be said that only
transfer of proton takes place, it is not obtained free. Every acid will lose proton and so its
conjugate base will be formed and every base will accept a proton and so its conjugate acid will
be formed. Hence, this concept is known as proton transfer or conjugate acid-base concept.
We have earlier seen in limitations of Arrhenius concept that OH- is not present in NH3 even
then it acts as a base but according to Bronsted-Lowry concept it can be explained.
NH3 + H2O
NH+4(aq) + OH-(aq)
Base-1 Acid -2
Acid-1
Base-2
In the reaction, base NH3 accepts a proton and forms conjugate acid NH4+ ion and acid H2O loses
proton and forms conjugate base OH-.
Thus the concept of Bronsted - Lowry is found to be more applicable and acceptable than
Arrhenius concept. Even then its limitations are also known. This difficulties are observed in the
study of reactions in organic chemistry and complex salts. BF3 has no proton even then it acts as
an acid. Hence, the third concept has come in to existence, which is known as Lewis acid- base
concept. Proton is given importance in Bronsted – Lowry concept.
12. Explain in detail Lewis acid-base theory.
Ans. Lewis Concept of Acid and Base :
Lewis, in presenting this concept, in 1923, mentioned that acid means a substance which can
accept a pair of electrons and base means a substance which can donate a pair of electrons. Thus,
instead of the concepts of proton, ionization, conjugate acid or base, he made electrons
associated with all reactions and substances and its arrangement as the base of the concept. As
seen earlier BF3 can be said acid or not and NH3 can be said a base or not can be solved by this
concept. It will be clear from the following reaction:
Thus, BF3 accepts the pair of electron and so it is acid and NH3 donates a pair of electron, so it is
acid and NH3 donates a pair of electron, so it is base. Electron deficient substances or ions like
AlCl3, Co3+, Mg+2, will act as acid and substances like H2O, NH3, OH-, F- will act as base. They
are respectively called Lewis acid and Lewis base.
13. Derive the formula of dissociation constant of Weak acid
Ans. Ionization Constant of Weak Acid (Ka) :
In aqueous solution of weak monobasic acid HA, there is a partial ionization and so the
equilibrium is obtained as below :
HA(aq) + H2O(l)
H3+O(aq) + A-(aq).
Suppose the initial concentration of weak acid (HA) is C mollit-1 and degree of ionisation
is α, then following can be written :
Reaction :
HA(aq) + H2O(l)
H3+O(aq) + A-(aq).
Initial
Concentration(M)
C
0
0
Degree of ionisation(α)
1-α
α
α
Concentration at equilibrium(M) (1 - α)C
αC
αC
Equilibrium constant Ke=
[H3O+][A-]
[HA][H2O]
= (αC)(αC)
…1
(1 - α)C[H2O]
But [H2O] is accepted as constant and so
Ke[H2O] = (αC)(αC)
(1 - α)C
= α2C = Ka
…2
(1 - α)
Where Ka is the ionization constant or dissociation constant of the s\acid HA. (At this stage we
consider both the terms same) Hence, for any weak monobasic acid, following can be written
[H3O+][A-]
Ke=
[HA]
…3
The unit of Ka will be mollit-1.
As the values of Ka depend on [H3+O] the values of ka will be different for different.
Lesser the value of Ka, weaker will be the acid. The value of Ka will be constant at definite
temperature.
14. Derive the formula of dissociation constant of weak base
Ans. ionisation constant (Kb) of weak base:
The ionization of monoacidic weak base MOH will take place in aqueous solution as follows:
H2O
MOH(aq)
M+(aq) + OH-(aq)
As base is weak, incomplete ionization will occur and so equilibrium will be obtained and it can
be expressed as below:
[M+] [OH-] …………4 and
Kc = [MOH] [H2O]
[M+] [OH-]
=Ke ………. 5
Kex[H2O]=
[MOH]
Where Kb is the ionization or dissociation constant of monoacidic weak base. If we know the
initial concentration of weak base and its degree of ionization, we can calculate the value of Kb
as we have studied earlier. [M+] = [OH-].
15. Obtain relation between ionic equilibrium constant Ka, initial concentration Co and
concentration of H3O+ in aqueous solution of weak formic acid.
Ans. As formic acid is weak acid, it is partially ionized in water. The equilibrium between the
undissociated species of CH3COOH and the ions H+(aq) can be represented as
CH3COOH(l) + H2O(l)
H3O+(aq) + CHCOO-(aq)
The equilibrium constant K for this reaction is as follows :
K = [H3O+][CH3COO-]
[CH3COOH][H2O]
Here the change in concentration of water on dissolving the weak acid will be quite negligible
compared to its original concentration, so considering the value of [H2O] as constant and joining
it with equilibrium constant K we get a new constant Ka as follows.
K[H2O] = Ka = [H3O+][CH3COO-]
[CH3COOH]
Here formic acid will ionize to much less extent and so the concentration of its unionized part
will remain nearly same as its original concentration.
[CH3COOH] = Co So, we have
Ka = [H3O+][CH3COO-]
Co
Again, the concentration of the positive ions [H3O+] and that of the negative ions [CH3COO-]
will be the same i.e. [H3O+] = [CH3COO-]
[CH3COO-] can be replaced by [H3O+], Ka = [H3O+]2
Co
+ 2
+
1/2
[H3O ] = Ka.Co
[H3O ] = (KaCo)
16. Obtain relation between ionic equilibrium constant Kb, initial concentration Co and
concentration of OH- in aqueous solution of weak base aniline.
Ans. As aniline is weak acid, it is partially ionized in water. The equilibrium between the
undissociated species of C6H5NH2 and the ions H+(aq) can be represented as
NH3(g) + H2O(l)
NH4+(aq) + OH-(aq)
The equilibrium constant K for this reaction is as follows :
K = [NH4+][OH-]
[NH3][H2O]
Here the change in concentration of water on dissolving the weak base will be quite negligible
compared to its original concentration, so considering the value of [H2O] as constant and joining
it with equilibrium constant K we get a new constant Ka as follows.
K[H2O] = Kb = [NH4+][OH-]
[NH3]
Here aniline will ionize to much less extent and so the concentration of its unionized part will
remain nearly same as its original concentration.
[NH3] = Co So, we have
Ka = [NH4+][OH-]
Co
Again, the concentration of the positive ions [NH4+] and that of the negative ions [OH-] will be
the same
[OH-] can be replaced by [NH4+],
Kb = [OH-]2
Co
[OH-]2 = Kb.Co
[OH-] = (KbCo)1/2
17. What is meant by ionic product of water? Deduce its equation.
Ans. Ionic Product of Water
Water is an amphoteric oxide when acid is added to it; it accepts the proton and act as a base and
when base is added to it, it donates a proton and acts as acid. When reaction between two
molecules of water takes place, one molecule donates proton and other molecule receives proton
and shows conjugate acid-base reaction.
H2O(l) + H2O(l)
H3O+(aq)
+
OH-(aq)
Acid-1
Base-2
Conjugate acid-2
Conjugate base-2
If we express the equilibrium constant of above reaction them.
[H3O+][OH-]
Ka =
[H2O][H2O]
Where Ka is the dissociation constant of acid. There is no significant change in concentration of
water (55.5M) because H2O is a weak acid (possesses about 10-7M H+). Thus, if H2O is
considered constant.
Ka x [H2O]2 = [H3+O] [OH-] = Kw.
Where Kw is ionic product of water. Water is neutral and so [H3+O] and [OH-] in it are 10-7 M.
Hence, Kw = [H3+O] [OH-] = (10-7)(10-7) = 10-14 which is constant and equilibrium constant
remains constant at constant temperature; so the value of Kw will be constant at constant
temperature, viz., the value of Kw is 1 x 10-14 at 298 K. If we find the ratio of concentrations of
dissociated and un dissociated water, it will remain on left hand side, i.e. the number of un
dissociated molecules or its concentration will be more.
18. Discuss the factors affecting strength of acid.
Ans. Factors Affecting strength of acid:
From the study of dissociation constant, pH value etc., it is found that their values are different.
The reason for this is that [H3+O] available can be more or less. What may be the reason for this?
If the acid is strong, its value of Ka will be high and the value of pH will be low. The dissociation
of acid will depend on strength of acid and the polarity of H-A bond. As the strength of H-A
bond decreases, the energy required for breaking that bond will decrease and HA will be
stronger. When difference between electro negativities of A and B will increase, apparently
ionization will occur and will be easy to break the structure of the bond. Hence, the acidity will
increase.
On comparing the elements of the same group of periodic table, the strength of H-A bond will be
more important factor than polar nature. As we go down in the group the size of A will increase
and so strength of H-A bond will decrease and hence acid strength will increase.
Increase in size
HF<< HCl <<HBr<<HI
Increase in acid strength
For this reason, H2O is stronger acid than H2S, but if we discuss the elements in the same period
of periodic table, the polarity of H-A bond will determine the strength of acid. As the
electronegativity of A increases, the strength of acid will increase.
Increase in electro negativity
CH4<< NH3<< H2O << HF
Increase in acid strength
19. Write a short note on buffer solutions.
Ans. Buffer solutions
The pH of the fluids like blood in our body and urine is almost constant. If there is change in this
pH, it affects biochemical reaction in the body. The pH of chemical and biochemical reactions in
our body are constant, biochemical reactions in our body are constant, viz. the pH of human
saliva is 6.4. In addition, hydrochloric acid is present in human stomach which helps in
digestion. The pH of cosmetics are also kept constant. Hence, the question arises that how pH in
any solution can be kept constant. Such solutions are called buffer solutions. Its definition can be
given as below:
“ The solution which resists the change in pH carried out by addition of acid or base in small
proportion to them or are being diluted, and the values of their pH remain constant are called
buffer solution.”
Buffer solutions can be acidic or basic. If pKa of weak acid and pKb of weak base are known,
buffer solutions of known pH can be prepared. Buffer solutions can be of three types as follows:
(i) Acidic buffer solution: Acidic buffer solution can be prepared by mixture of weak acid and its
salt with strong base.
(ii) Basic buffer solution:Basic buffer solution can be prepared by mixture of weak base and its
salt with strong acid.
(iii) Neutral buffer solution: Neutral buffer solution can be prepared by neutralization of weak
acid and weak base. These type of buffer solutions are shown below:
Type
Substances
Value of pH
Acidic
CH3COOH + CH3COONa
<7
Basic
NH4OH + NH4Cl
>7
Neutral CH3COOH + NH4OH
=7
Buffer solution of known pH can be prepared by using the following HendersonHaschelback equation.
For acidic solution,
[Salt]
pH = pKa + log [Acid] …………. 1
where [acid] is concentration a weak acid and its dissociation constant is Ka and [salt] is
concentration of the salt of this weak acid with strong base. For a buffer solution, it can be
written as
pH= pKa + log [CH3COONa]
[CH3COOH]
similarly, for basic buffer solution e.g., NH4OH + NH4Cl can be written that,
[NH4Cl]
pH = pKa + log [NH4OH]
Such buffer solution can be used in chemical and biochemical reactions and especially in
analytical chemistry. In human body buffer solutions containing [HCO-3] and [CO2-3] as well as
[H2PO-4] and [HPO2-4] are present.
20. Write a short note on pH scale.
Ans. pH Scale.
If we express the concentration of hydronium ion [H3+O] in molarity then values like 10-12 to 10-2
are possible. It is difficult to express these values on simple graph paper. Hence, scientist
Sorenson found a scale which is called pH scale. According to him pH = -log10[H3+O]. The
values 10-12 to 10-2 shown above can be converted to +12 to +2 if calculated on the basis of this
relation and plotting of graph can be easy. The definition of pH can be given like this, “pH of a
solution is the negative logarithm to the base 10, of the concentration of hydrogen or hydronium
ion”. According to thermodynamics, activity is more proper word instead of concentration but in
dilute solutions activity and concentration can be considered can be considered to be the same.
Now as seen earlier a solution containing 10-7M [H3+O] and [OH-] is neutral. Hence,
pH = -log10[H3+O] = -log1010-7M = 7 and for acidic solution [H3+O]> 10-7M, pH < 7
Similarly, for basic solution [H3+O]< 10-7M, Hence, pH > 7. Therefore, it can be written as :
pH< 7 Acidic solution
pH> 7 Basic solution
pH = 7 Neutral solution
AS seen above,
Kw = [H3+O] [OH-]
Putting the values, Kw= (10-7)(10-7) = 10-14and –log Kw = -log(10-14)
Therefore, pKw = 14
Therefore, pH + pOH =pKw = 14
Temperature affects the values of pH, pOH, pKw. The above discussion can be shown in the
following table :
Concentration (M)
Acidic
Neutral
Basic
[H3+O]
More than 10-7
10-7
Less than 10-7
-7
-7
[OH ]
Less than 10
10
More than 10-7
pH
Less than 7
7
More than 7
pOH
More than 7
7
Less than 7
pH paper, litmus paper or universal indicator can be used to test whether the solution is acidic or
basic but the exact values of pH can be determined with the help of instrument called pH meter.
21. Explain the calculation of pH of a solution from the dissociation constant of weak acid.
Ans. In aqueous solution of weak monobasic acid HA, there is a partial ionization and so the
equilibrium is obtained as below :
HA(aq) + H2O(l)
H3+O(aq) + A-(aq).
Suppose the initial concentration of weak acid (HA) is C mollit-1 and degree of ionisation
is α, then following can be written :
Reaction :
HA(aq) + H2O(l)
H3+O(aq) + A-(aq).
Initial
Concentration(M)
C
0
0
Degree of ionisation(α)
1-α
α
α
Concentration at equilibrium(M) (1 - α)C
αC
αC
[H3O+][A-]
Equilibrium constant Ke= [HA][H2O]
= (αC)(αC)
…1
(1 - α)C[H2O]
But [H2O] is accepted as constant and so
Ke[H2O] = (αC)(αC)
(1 - α)C
= α2C
= Ka
…2
(1 - α)
Where Ka is the ionization constant or dissociation constant of the s\acid HA. (At this stage we
consider both the terms same) Hence, for any weak monobasic acid, following can be written
[H3O+][A-]
Ke=
[HA]
…3
-1
The unit of Ka will be mollit .
As the values of Ka depend on [H3+O] the values of ka will be different for different.
Lesser the value of Ka, weaker will be the acid. The value of Ka will be constant at definite
temperature. IN table 4.3 the values of ionisation constants of some weak acids are given.
According to the relations seen earlier,
pH = -log10[H3+O]
pOH = -log10[OH-]
Similarly, pKa = -log10[Ka]
PA- = -log10[A-] can be written. Where [A-] is the concentration of negative ion.
It is apparent from above relations that if the values of initial concentration [H3+O] and Ka are
known, then at equilibrium, [H3+O] concentration can be determined and pH can be calculated
22. Explain effect of common ion on solubility of sparingly soluble salt.
Ans. Effect of Common Ion on Solubility of Sparingly Soluble Salt
The sparingly soluble salt that has dissolved in solution that is completely dissociated and so it is
ionic in form. Hence, it is a strong electrolyte. Earlier, in chemical equilibrium, effect of
concentration, application of Le Chatelier’s principle etc. have been studied. Solubility product is
ionic equilibrium and the effect of concentration can be studied. As it is equilibrium constant, it
will depend on temperature, but its value will be constant at constant temperature.
What will happen if we add soluble ionic substance like KCl in the solution of a sparing soluble
salt like AgCl?
AgCl(s)
Ag+(aq) + Cl-(aq)
KCl(s)
K+(aq) + Cl-(aq)
The Cl from AgCl in equilibrium and Cl- ion obtained by complete ionization of KCl, the
concentration of Cl- will increase. Hence, according to Le Chatelier’s principle, the equilibrium
will shift towards left side so as to nullify the effect of Cl- i.e. more AgCl will be formed. In
other words, there will decrease in solubility of AgCl. Hence, it can be said that because of the
effect of common ion on sparingly soluble alt its solubility decreases and sparingly soluble salt
precipitates more.
The use of effect of common ion can be made to separate one ion from the other in presence of
other ion in qualitative analysis. It can also be used for decrease in solubility of the components
in the mixture. In qualitative analysis, the solubility products of sulphides of metal ions of
second group are less in comparison to solubility products of sulphides of metal of III B group
ions, therefore, HCl is added before adding H2S water to test the second group ions.
H2S(aq)
2H+(aq) + S2-(aq)
HCl(aq)
H+(aq) + Cl-(aq)
The common ion available from HCl creates common ion effect on the equilibrium and
decreases concentration of S2- ions. Hence, the sulphides of the ions included in second group
can only be precipitated because their solubility products are less. In the same way, for
precipitation of ions of group III A ammonium chloride (in excess) is added along with NH4OH.
The concentration of OH- available from ionization of NH4OH gets decreased due to common
ion effect due to NH+4 available from NH4Cl. Hence, the hydroxides of the ions of III A group
only will be precipitated because the values of solubility products of the hydroxides of III A
group are low. Above this, if HCl gas is passed through saturated solution of NaCl, NaCl
becomes insoluble and separates as precipitates due to the common ino effect of Cl- ions.
Its is necessary to note that under certain situations the solubility increases instead of decreasing.
The solubility of salt like phosphates increase when acid is added to their solutions or pH of the
solution decreases. The reason for this is that, phosphate ion combines with H+ available from
acid and converts into acid by accepting proton. Hence, solubility of phosphate salt increases.
23. Explain with suitable example the use of common ion effect in qualitative analysis.
Ans. The use of effect of common ion can be made to separate one ion from the other in presence
of other ion in qualitative analysis. It can also be used for decrease in solubility of the
components in the mixture. In qualitative analysis, the solubility products of sulphides of metal
ions of second group are less in comparison to solubility products of sulphides of metal of III B
group ions, therefore, HCl is added before adding H2S water to test the second group ions.
H2S(aq)
2H+(aq) + S2-(aq)
HCl(aq)
H+(aq) + Cl-(aq)
The common ion available from HCl creates common ion effect on the equilibrium and
decreases concentration of S2- ions. Hence, the sulphides of the ions included in second group
can only be precipitated because their solubility products are less. In the same way, for
precipitation of ions of group III A ammonium chloride (in excess) is added along with NH4OH.
The concentration of OH- available from ionization of NH4OH gets decreased due to common
ion effect due to NH+4 available from NH4Cl. Hence, the hydroxides of the ions of III A group
only will be precipitated because the values of solubility products of the hydroxides of III A
group are low. Above this, if HCl gas is passed through saturated solution of NaCl, NaCl
becomes insoluble and separates as precipitates due to the common ino effect of Cl- ions.
Its is necessary to note that under certain situations the solubility increases instead of decreasing.
The solubility of salt like phosphates increase when acid is added to their solutions or pH of the
solution decreases. The reason for this is that, phosphate ion combines with H+ available from
acid and converts into acid by accepting proton. Hence, solubility of phosphate salt increases.
24. What is meant by common ion effect? In qualitative analysis, NH4Cl is added before
addition of NH4OH in precopotation of ions of III A group. Explain giving reason.
Ans. Let us take the example of weak acid, acetic acid ( )
CH3COOH(aq)+ H2O(l)
H3+O(aq) + CH3COO-(aq)
OR
HAc + H2O(l)
H3+O(aq) + Ac-(aq)
Where HAc and Ac- are the short forms of CH3COOH and CH3COO- ion.
[H3+O][Ac-]
Ka = [HAc]
………………..1
Suppose, we add CH3COONa or HCl, to the solution of HAc in equilibrium, then what will
happen? As studied earlier in chemical equilibrium if HCl is added, [H3+O] will increase and if
CH3COONa is added, [Ac-] will increase. Hence, according to Le Chatelier’s principle, the
equilibrium will make negligible change and will keep the same value of equilibrium constant.
This means that the equilibrium will be shifted towards left and concentration of HAc will
increase i.e. the amount of un dissociated acid will increase and there will be decrease in [H3+O]
and hence, there will be increase in pH. By addition of HCl due to increase in [H3+O] similar
result will be obtained. This effect is known as effect of common ion effect on dissociation
constant of acid.
In the same way, in the case of ionization of weak base NH3, if we increase [NH4+] by adding
salt like NH4Cl then, according to Le Chatelier’s principle, as [NH4+] increases the equilibrium
will shift towards left and hence, un dissociated NH3 will increase, i.e. [OH-] will decrease. As a
result, pH will increase.
In qualitative analysis, to precipitate the radicals of III A viz Fe2+, Fe3+, Al3+, Cr3+ as their
hydroxides, NH4Cl is added to the solution before adding NH4OH. Consequently, the reverse
reaction is favoured in the equilibrium, resulting in decrease of OH- ion concentration. Because
of this, the hydroxides of III B, IV groups and that of Mg2+ are not precipitated alon with those
of group III A.
NH4Cl(aq)
NH4+(aq) + Cl-(aq)
NH4OH(aq)
NH4+(aq) + OH-(aq)
The hydroxides of the radicals of III A group are extremely less soluble compared to those of
succeeding groups. So, in presence of NH4Cl the hydroxides of III A group are precipitated by
NH4OH.
(25). 2 moles of PCl5 is heated in closed vessel of 4 liter a t definite temperature. At
equilibrium state PCl5 remained undecomposed. Calculate Kc of the reaction.
Solution: Reaction,
PCl5(g)
PCl3(g) +Cl2(g)
Now, 55% PCl5 remains undecomposed.
Therefore, 45 % PCl5 goes under reaction.
= 45 x 2 = 0.9 moles PCl5
100
As ratio of PCl5 and (PCl3 and Cl2 ) is 1 : 1 in number of moles, 0.9 moles of PCl5 will yield 0.9
moles of PCl3 and Cl2.
PCl5(g)
PCl3(g) +Cl2(g)
In the beginning:
2 moles
0
0
Used moles at
equilibrium:
0.9 moles
0.9
0.9
Unused moles at
equilibrium:
1.1 moles
Now, Kc = [concentration of products]
[concentration of reactants] ……..(i) Kc = equilibrium constant
Now, at equilibrium, moles of reactants = 1.1 moles
and moles of products = 0.9 moles ( in solution)
Now, this vessel consists of 4 liter solution
Therefore, concentration = 1.1 = 0.275 M/L (for reactant)
4
= 0.9 = 0.225 M/L (for products)
…………(ii)
4
From equation (i) Kc = [PCl3] [Cl2]
[PCl5]
From equation (ii) Kc = (0.225) (0.225)
= 0.050625
(0.275)
0.275
Therefore, Kc = 0.184 mol/lit
Kc = 1.84 x 10-1 mol/lit.
(26). The value of Kp obtained at 1060 kelvin temperature is 0.033 mollit-1 for the reaction
2NOCl(g)
2NO(g) + Cl2(g). What will be the value of Kc for the reaction.
Solution:
2NOCl(g)
2NO(g) + Cl2(g).
∆n(g) = np – nr (for gaseous molecules)
=(2+1)–2=1
Data:
Kp = 0.033 atm
T = 1060K
0.082 atm lit mol-1 K-1
Kp = Kc x (RT)∆n(g)
 Kc = Kp
(RT)∆n(g)
Substituting the values from data,
Kc = 0.033
= 0.033
(0.082 x 1060)
86.92
= 3.796 x 10-4 M
Therefore, Kc = 3.796 x 10-4 M
(27). At definite temperature and 3 atm pressure, 75 % PCl5 is decomposed into PCl3 and
Cl2. Calculate Kp of this reaction. PCl5(g)
PCl3(g) +Cl2(g).
Solution: 75% PCl5 is decomposed into PCl3 and Cl2.
At equilibrium,
moles PCl5
decomposed
100
75
1
x [ suppose initial moles of PCl5 is 1.]
x= 75 x 1
= 0.75 moles
100
0.75 moles of PCl3 and Cl2 are formed and unreacted moles of PCl5 = ( 1 – 0.75) = 0.25 moles
PCl5(g)
PCl3(g) +Cl2(g).
Initial moles:
1
0
0
Used moles at
equilibrium:
0.75
0.75
0.75
Unused moles
at equilibrium:
0.25
Now, total moles at equilibrium = 0.25 + 0.75 + 0.75
= 1.75 moles
Total pressure in system = 3 qtm
Now, partial pressure at equilibrium: moles of 1 element x total pressure
Total moles
Hence, partial pressure of PCl3 ( Ppcl3) = 0.75 x 3 = 3 x 3
1.75
7
= 9 atm
7
………… (i)
Similarly , partial moles of Cl2
Pcl2 = 0.75 x 3
= 3 x3
=9
1.75
7
7
atm …………..(ii)
Partial pressure of PCl5
Ppcl5 = 0.25 x 3 = 1 x3
= 3 atm
1.75
7
7
……………….. (iii)
Now, Kp = (Ppcl3) x (Pcl2)
(Ppcl5)
From equation (i) , (ii) and (iii)
Kp = 9/7 x 9/7
3/7
Kp = 3.857 atm.
(28). 1 mole N2 and 3 moles H2 are heated at 473 K temperature and 100 atmosphere
pressure. If 0.50 mole of NH3 are formed at equilibrium time calculate equilibrium
constant Kp for this reaction. N2(g) + 3H2(g)
2NH3(g)
Solution: N2(g) + 3H2(g)
2NH3(g)
Now, the ratio of production of moles of this reaction is 1 : 3 : 2
Initial moles of N2 = 1mole
Data
Initial moles of 3H2 = 3 moles
Therefore, from the ratio, moles of NH3 formed = 2 moles
But from the data, moles of NH3 fromed = 0.50 moles.
Therefore, ¼ th of NH3 is formed [ because 2 /0.50 = 4]
Hence, 1/4th of reactants must be used.
For N2 , ¼ x 1 = ¼ moles
For H2 , ¼ x 3 = ¾ moles
For NH3 = 0.50 moles ……(i)
Above these are the moles formed in solution
Therefore, unused moles at equilibrium
For N2 , 1 – ¼ = ¾ moles ….(ii)
For H2 , 3 – ¾ = 9/4 moles ……(iii)
N2(g) + 3H2(g)
2NH3(g)
Initial moles:
1
3
2
Used moles at
equilibrium
¼
¾
0.50
Unused moles
at equilibrium
¾
9/4
-
Partial pressure = moles of 1 element x total pressure
Total number of moles
Total pressure = 100 atm
Total number of moles = ¾ + 9/4 + 0.50
= 3.50 moles
Partial pressure of N2
PN2 = ¾ x 100
3.50
= 21.428 atm …………..(iv)
Partial pressure of H2
PH2 = 9/4 x 100
3.50
= 64.285 atm ……….(v)
Partial pressure of NH3
PNH3 = 0.50 x 100
3.5
= 14.2857 atm …….(vi)
Now, from equation N2(g) + 3H2(g)
2NH3(g)
Kp = (PNH3)2
=
(14.2857)2
(PN2) x (PH2)3
21. 2428 x (64.285)3
Kp ≈ 3.616 x 10-5 atm -2
(29). On heating 0.5 mole solid calcium carbonate in a vessel of 500ml volume, at 400 K
temperature , Kc 0.9 mol lit-1 is obtained at equilibrium state, then calculate mole of CO2 at
equilibrium. How many percent the reaction would have been completed?
CaCO3(s)
CaO(s) + CO2(g)
Solution:
CaCO3(s)
CaO(s) + CO2(g)
Initial moles:
0.5
Used moles at
equilibrium:
x
x
x [as ratio is 1 : 1]
unused moles at
equilibrium :
(0.5 – x)
concentration
(0.5 – x)
x
(mole/ liter)
0.5
0.5
Kc = [CaO][CO2]
[CaCO3]
= (x/0.5) (x/0.5)
= (2x) (2x)
((0.5 –x)/0.5)
2(0.5 – x)
Kc = 4x2
1 – 2x
……….(i)
From data, Kc = 0.9 mol/lit
Therefore, from equation (i),
0.9 = 4x2
=> 9
= 4x2
1-2x
10
1 – 2n
9 – 18 x=40x2
 40x2 + 18x -9 = 0
x
0.5 [ volume in solution is 500ml (from data)]
Comparing with quadrative equation, ax2 + bx + c = 0
a = 40
b = 18
………(ii)
c= -9
Solution, x = - b ±√ b2 – 4ac
2a
Substituting values of a, b ,c from equation (ii)
x = -18 ± √ (18)2 – 4(40) (-9)
2( 40)
x= -18 ± √ 324 + 1440
80
x = -18 ± √ 1764
= -18 + 42
80
80
x = -18 + 42
or
x = -18 -42
80
80
x= 24 or
x= -60
80
80
(not possible as solution of x is negative value)
Therefore
3
=> x = 0.3 moles
10
Therefore, moles of Co2 at equilibrium = x = 0.3 moles
Therefore, from data ,
0.5 ml of CaCO3  0.3 moles of CO2 and 0.3 moles of CaO
Therefore, moles of CO2 formed = 0.3 moles
CaCO3(s)
CaO(s) + CO2(g)
0.5
0.3
100
(?)
0.3 x 100 = 60
0.5
Therefore, % of reaction completed = 60%
(30). Mg(HCO3)2(s) is decomposed as follows in a closed vessel. If mole fraction of CO2 is 0.8
and Kp = 64 atm3 then calculate total pressure of mixture at equlibrium.
Mg(HCO3)2
MgCO3(s) + CO2(g) + H2O(g)
Solution:
Mg(HCO3)2
MgCO3(s) + CO2(g) + H2O(g)
Now, gaseous mole in system are CO2 and H2O
Mole fraction of CO2 = 0.8
Total moles = 1
Therefore, mole fraction of H2O = 1 -0.8
= 0.2 …….(i)
Therefore, mole fraction Xco2 = 0.8
Mole fraction XH2O = 0.2
Now, partial pressure = moles of element x
total pressure
Total moles
= mole fraction x total pressure
Therefore, partial pressure of CO2 (PCO2) = Xco2 x total pressure
= 0.8 x ……….(ii) [let total pressure be x]
Partial pressure of H2O (PH2O) = XH2O x total pressure
= 0.2 x ………………(iii)
Now, Kp = [PCO2] [PH2O]
But, from data, Kp = 64 atm2 and from equation (ii) and (iii)
64 = (0.8x) ( 0.2x)
Therefore, 64 = 0.16x2
Therefore, x2 = 64
0.16
2
x = 400
therefore, x = 20 atm
Therefore, total pressure in system = 20 atm
(PCO2 = 16 atm PH2O = 4 atm)
(31). If 3.65 x 10 -2 gram HCl is dissolved in 500 ml solution, then find its pH (pH = 2.699)
Solution: Data,
For HCl, weight (W) = 3.65 x 10-2 gram
M = 36.5 gram/ mole
V = 500 ml = 0.5L
Moles of HCl = W = 3.65 x 10-2
M
36.5
= 1 x 10-3 mole
Concentration of [H+] in HCl = 1 x 10-3
0.5
= 2 x 10-3 M/L
pH= -log[H+]
= -log[2 x 10-3]
= - [log 2 + log (10-3)] [ log(axb) = log a + log b]
= 3 – log2
= 3 – 0.3010
pH= 2.6990.
(32). Find pH of mixture of 50ml 0.03 M HCl and 60 ml 0.02M NaOH .
Solution: 0.03 ml HCl  1000ml HCl solution contains 0.03 mol HCl
Therefore, 50 ml solution contains = 0.03 x 50
1000
= 1.5 x 10-3 moles HCl ……..(i)
Similarly,
0.02 M NaOH  1000ml NaOH contains 0.02 mol NaOH
Therefore, 60 ml solution contains = 0.02 x 60
1000
= 1.2 x 10-3 moles NaOH …..(ii)
NaOH + HCl neutralisation
NaCl + H2O
-3
Therefore, 1.2 x 10 moles of NaOH will neutralize 1.2 x 10-3 moles of HCl out of 1.5 x 10-3
moles of HCl
Therefore, moles of HCl that remains unneutralised are = (1.5 – 1.2) x 10-3
= 0.3 x 10-3
= 3 x 10-4 moles ………(iii)
Total volume of the mixture = 110ml
Therefore, molarity of HCl after neutralization = moles
= moles x 1000
Volume(L)
V (ml)
-4
= 3 x 10 x 1000
[from equation (iii)]
110
Therefore, molarity of HCl = 2.727 x 10-3 M
Now, HCl is strong acid and hence complete ionization takes place
Therefore, HCl(aq) + H2O(l)  Cl-(aq) + H3O+(aq)
2.727 x 10-3 M
2.727 x 10-3 M
pH= - log[H3O+]
= - log [ 2.727 x 10-3]
Therefore, pH = [ log 2.727 + ( -3) log 10]
pH = - log 2.727 + 3log10
therefore, pH = -0.4357 + 3
therefore, pH=2.5643
(33). Find weight of CH3COOH in 100ml aqueous solution of CH3COOH having pH 3.0 Ka
for CH3COOH = 1.75 x 10-5
Solution: CH3COOH + H2O
H3O+ + CH3COOAs CH3COOH is weak acid, partial ionization takes place
Therefore, Ka = 1.75 x 10-5
Therefore, for weak acid, Ka = [H3O+]2
Co
Where,Co = initial concentration of weak acid i.e. CH3COOH
 Co = [H3O+]2
Ka ………….. (i)
Now, pH = - log10 [H3O+]
= 3 [given]
Therefore, -log[H3O+] = 3
Therefore, [H3O+] = 1 x 10-3 M …………………..(ii)
Substituting the value of [H3O+] from equation (ii) in equation (i)
Therefore, Co = [H3O+]2 = ( 1 x 10-3)2
Ka
1.75 x 10-5
-2
Co = 5.7142 x 10 M
Now, concentration = mole
Liter
Therefore, 1 liter solution conatin 5.7142 x 10-2 moles of acetic acid
1L
5.7142 x 10-2 moles
-1
100ml(10 L)
?
= 5.7142 x 10-2 =
5.714 x 10-3 moles
-1
10
Now, mole =
weight
Molecular weight
M.W. of CH3COOH = 60 gram / mole
Therefore, moles= W
60
Therefore, W = 60 x moles
= 60 x 5.7412 x 10-3
W = 3.428 x 10-1 gram
= 0.342 gram.
(34).0.02 M CH3NH2 is ionized to 15% at equilibrium .Find its ionization constant.
Solution: CH3NH2(g) + H2O(l)
CH3NH+3(aq) + OH-(aq)
Methyl amine ionizes to 15% at equilibrium
If 100 moles of CH3NH2 are taken, then 15 moles of OH- are formed
CH3NH2
OH100
:
15
0.02
:
(?)
= 15 x 0.02
= 0.3 x 10-2 M
100
Therefore, [OH ] = 3 x 10-3 M
Now, CH3NH2 is weak base
Therefore, weak base , Kb = [OH-]2
Co
Where, Kb = ionization constant of weak base
Co = initial concentration of base = 0.02 M
Therefore, Kb = ( 3 x 10-3)2
0.02
= 9 x 10-6
2x 10-2
Therefore, Kb = 4.5 x 10-4
(35). For NH3 Kb = 1.77 x 10-5 , what will be the pH of 500 ml solution containing 1.7 gram
ammonia?
Solution: NH3(g) + H2O(l)
NH+4(aq) + OH-(g)
For weak base, Kb = 1.77 x 10-5
V (ml) = 500 ml = 500 x 10-3 L
W = 1.7 gram
M.W. of ammonia = 17 gm/ mole
Concentration = moles
Liter
=
weight of NH3 (gm)
M.W. of NH3 (gm/mole) x volume (L)
=
1.7
17 x 500 x 10-3
= 0.2 M
Therefore, concentration = 0.2 M ……..(i)
For weak base,
Kb = [OH-]2
Co
Where, Co = initial concentration of NH3 = 0.2 M (from equation (i))
Therefore, [OH-]2 = Kb x Co
= 1.77 x 10-5 x 0.2
= 3.54 x 10-6
Therefore, [OH ] = Kb x Co
= 1.77 x 10-5 x 0.2
= 3.54 x 10-6
Therefore, [OH-] = √ 3.54 x 10-6
Therefore, [OH-] = 1.881 x 410-3 M
Now, pOH = -log10[OH-]
= -log10 [ 1.881 x 10-3]
Therefore, pOH = 3 – log(1.881)
= 3 – 0.2744
Therefore, pOH = 2.7256
Now, pH + pOH = 14
Therefore, pH = 14 – pOH
14 – 2.7256
Therefore, pH = 11.2744
(36). pH of 0.05M solution of a weak acid is 3.68 then find Ka of that acid.
Solution: for weak acid,
Initial concentration (Co) = 0.05M
pH = 3.68
Ka = ?
Now ,pH = - log10[H3O+] = 3.68
Therefore, -log[H3O+]=3.68
Shifting minus (-) sign to right side and taking anti –log on both sides
Therefore, [H3O+] = - antilog(3.68)
anti-log (-3.68)
= antilog(-3.68)
-1 +1
= antilog( 4 + 0.32)
antilog (-4 + 0.32)
Therefore, [H3O+] = antilog(0.32) x 10-4
Therefore, [H3O+] = 2.089 x 10-4 M
Now, for weak acid,
Ka = [H3O+]2
Co
Therefore, Ka = (2.089 x 10-4)2
0.05
= 87.278 x 10-8 = 8.727 x 10-7
Therefore, Ka = 8.7278 x 10-7
(37). Find pH of 0.025M CH3COONa solution. Ka of CH3COOH = 1.75 x 10-5.
Solution: CH3COONa  CH3COO(aq)- + Na(aq)+
0.025M
0.025M
0.025M
Now, CH3COO (aq) + H2O(l)
CH3COOH(aq) + OH-(aq) . ……….(i)
[ A- + H2O
HA + OH-(aq)]
Weak acid
H2O(l)
H(aq)+ + OH-(aq)
Now, hydrolysis constant, Kn = Kw
Ka
[CH3COONa is salt of strong base (NaOH) and weak acid (CH3COOH) and hence, Kn = Kw
Ka
Therefore, Kn = Kw = [CH3COOH][OH-] and from reaction (i)
]
Ka
[CH3COO-]
As concentration of [OH ] produced by the self- ionization of water is negligible in comparison
with the concentration of OH- produced by hydrolysis reaction.
Therefore, [CH3COOH] = [OH-] (At equilibrium)
Therefore, Kn = Kw =
[OH-]2
Ka
[CH3COO-]
Kw = ionic product of water = 1x 10-14
Ka = 1.75 x 10-5
[CH3COO-] = initial concentration of CH3COONa
= Co = 0.025 M
Substituting , these values in above equation
Kw = [OH-]2
=> 1 x 10-14 = [OH-]2
Ka
[CH3COO ]
1.75 x 10-5
0.025
Therefore, [OH-]2 = 1.4285 x 10-11
= 14.285 x 10-12
Therefore, [OH-] = 3.7796 x 10-6 M
≈ 3.78 x 10-6 M
Now, pOH = -log10 [OH-]
= -log[3.78 x 10-6]
= -log 3.78 + 6log10
= 6 –log(3.78)
= 6- 0.5775
Therefore, pOH= 5.4225
Now, pH + pOH = 14
 pH= 14 –pOH
= 14 – 5.4225
= 8.5775
Therefore, pH = 8.5775
(38). Kb of NH3 is 1.8 x 10-5.Calculate pH of 0.20M solution of NH4Cl
Solution: NH4Cl  NH4+ + ClNow, NH4Cl is salt of strong acid (HCl) and weak base (NH3)
 Kn = Kw
Kb
Where, Kn = hydrolysis constant
Kw = ionic product = 1 x 10-14
Kb = ionization constant of weak base = 1.8 x 10-5.
Now,
NH+4 + H2O
NH3 + H3O+
Now,
Kn = Kw = [NH3] [H3O] = [H3O+]2
Kb
[NH+4]
[NH+4]
[ because at equilibrium, [NH3] = [H3O+] ]
Therefore, Kw = 1x 10-14 = [H3O+]2
Kb
1.8 x 10-5
0.20
+ 2
Therefore ,[H3O ] = 1 x 0.20 x 10-9
1.8
= 0.111 x 10-9
= 1.111 x 10-10
+
Therefore, [H3O ] = 1.054 x 10-5 M
Now, pH = -log10 [H3O+]
= -log10[1.054 x 10-5]
= -log 1.054 + 5 log10
= 5 – log(1.054)
(log10 = 1)
= 5 -0.0229
= 4.9771
Therefore, pH = 4.9771
≈ 4.98
39. Find solubility of PbSO4 in water, having solubility product 1.3 x 10-8 (Molecular mass
of PbSO4 = 303 gram mol-)
Ans. Let, the solubility of PbSO4 be S mol/litre
PbSO4(s)
Pb+2(aq) + SO42-(aq)
S
S
+2
2Solubility product, Ksp = [Pb ][SO4 (aq)]
= [S][S]
= [S2]
But, Ksp = 1.3 x 10-8(given)
Ksp = S2 = 1.3 x 10-8
 S =1.140 x 10-4 mol/litre
Now, molecular weight of PbSO4 = 303 gm/litre
Solubility (in gram/litre) = mol/litre x M.W.
= 1.140 x 10-4 x 303
= 345.47 x 10-4 gm/litre
= 3.45 x 10-2 gm/L
-2
 S = 3.45 x 10 gm/litre
40. Ksp of CaF2 is 1.7 x 10-10. What will be the volume of saturated solution of 10 milligram
salt? (Atomic mass : Ca = 40 and F = 19)
Ans. Let, molar solubility of CaF2 be S mol/litre
Solubility Equilibrium,
CaF2
Ca2+(aq) + 2F-(aq)
S
2S (Because 2 moles of F- are formed)
Ksp(given) = 1.7 x 10-10
Ksp = [Ca2+][F-]2
= s[2s]2
= 4S3
From above data, Ksp = 4S3 = 1.7 x 10-10
 4S3 = 1.7 x 10-10
 S3+ = 0.425 x 10-10
= 42.5 x 10-12
S = 3.489 x 10-4 mol/litre
[ S3 = 42.5 x 10-12
S = (42.5 x 10-4)1/3
(42.5)1/3 = (log 42.5)/3
= (1.6284)/3
= 0.5428
= antilog(0.5428)
= 3.489
3.49]
-4
 S 3.49 x 10 mol/L
Where S= Solubility of CaF2
Now, molecular weight of CaF2 = Ca + 2(F)
= 40 + 2(19)
= 78 g/mol
Now, molar Solubility =
Wt.(gm) x 1
Mol. Wt. (gm/mole) x v(in litre)
 Solubility of CaF2 = Wt. of CaF2(gm)
x
m.w of CaF2 (gm/mole)
Substituting the values in above reaction
W= 10 mg
= 10 x 10-3 gm
3.49 x 10-4 = 10 x 10-3 x 1000
78 x V(ml)
 V(ml) =
10 x 1
78 x 3.49 x 10-4
 V(ml) = 367.34 ml
1000
V(ml)
41. Ksp of CaF2 at 298 K temperature is 1.7 x 10-10. If a person drinks 2.5 litre of water
saturated with this salt every day, how much CaF2 will enter in his body every day?
(Molecular mass of CaCl2 =78 g mol-)
Ans. Suppose, Solubility of CaF2 be S mol/L
CaF2
Ca2+(aq) + 2F-(aq)
S
2S
Ksp = [Ca2+][F-]2 = s[2s]2
= 4S3
Now, Ksp(given) = 1.7 x 10-10
= 4S3
 S3 = 0.425 x 10-10
= 42.5 x 10-12
 S = (42.5)1/3 x 10-4 mol/L
 S = 3.489 x 10-4 mol/L
 3.49 x 10-4 mol/L
[ S = (42.5)1/3 = (log(42.5))/3
= (1.6248)/3
= 0.5428
= antilog(0.5428)
= 3.489 ]
Now, person drinks 2.5 litre of water saturated with CaF2 salt every day
 V = 2.5 L
Molecular mass = 7.8g/mol
S = 3.49 x 10-4 mol/litre
Solubility Product (s) = Wt (gm) x 1
M.W(gm/mole) x V(L)
-4
3.49 x 10 = W(gm) x 1
78 x 2.5
 W(gm) = (3.49 x 10-4) x 78 x 2.5
= 680.55 x 10-4
= 0.068055 gm
 W(gm) = 0.068055 gm
= 6.8055 x 10-2 gm
42. pH of saturated aqueous solution of Ca(OH)2 is 12.25. Find Ksp of Ca(OH)2
Ans. Suppose, the solubility product of Ca(OH)2 be S mol/litre
Equilibrium state
Ca(OH)2(s)
Ca2+(aq) + 2OH-(aq)
S
2S
Now, pH = 12.25
PH + pOH = 14
 pOH = 14 - pH
= 14 – 12.25
= 1.75
OH
Now, p = -log10[OH-]
= 1.75
 -log10[OH-] = 1.75
Shifting ‘mius sign’ to the right side and taking ‘antilog’ on both sides
 [OH-] = -antilog(1.75)
= antilog(2.25)
= 1.778 x 10-2
 [OH ] = 1.778 x 10-2 M
But, according to reaction
[OH-] = 2S = 1.778 x 10-2M ….(i)
 S = 0.889 x 10-2 M …. (ii)
Now, Ksp = [Ca2+][OH-]2
= [S][2S]2 (from reaction)
= 4S3
= 4(0.889 x 10-2)3 [from equation (i)]
 Ksp = 2.810 x 10-6 M3
43. What will be the value of Ksp if the concentration in saturated solution of Mg(OH)2 is
8.2 x 10-4 %W/V. (Molecular Mass of Mg(OH)2 = 58.3 g.mol-).
Ans. 8.2 x 10-4% W/V means 100 ml saturated solution contains 8.2 x 10-4 gm Mg(OH)2
100 ml
8.2 x 10-4 gm Mg(OH)2
1000 ml(1L)
(?)
= 8.2 x 10-4 x 1000
100
= 8.2 x 10-3 gm/litre
M.W. of Mg(OH)2 = 58.3 gm/mol
 Molarity of saturated solution of Mg(OH)2
= 8.2 x 10-3 gm/litre
58.3 gm/mole
 S = 1.4065 x 10-4 mol/litre
Mg(OH)2(s)
Mg2+(aq) + 2OH-(aq)
S
2S
2+
- 2
Now, Ksp = [Mg ][OH ]
= [S][2S]2
= 4S3
= 4(1.4065 x 10-4)3
= 11.130 x 10-12 M3
 Ksp = 1.113 x 10-11 M3
44. How many grams of Zn(OH)2 can be dissolved in 2 litres of 0.02 M NaOH solution ?ksp
for Zn(OH)2 = 4.5 x 10-17.
Ans. Suppose solubility of Zn(OH)2 is S mol/litre
Zn(OH)2(s)
Zn2+(aq) + 2OH-(aq)
S
2S
+
NaOH(aq)
Na (aq) + OH-(aq)
0.02M
0.02M
0.02M (given)
So, total concentration of [OH ] in the solution
= [OH-] from Zn(OH)2 + [OH-] from NaOH
= 2S + 0.02
(But here, 0.02 >>> S, hence [OH-] = 0.02 M )
(because Zn(OH)2 is sparingly soluble)
TotL [OH-] = 0.02 M
Now, Ksp for Zn(OH)2 = [Zn2+][OH-]2
= [S][0.02]2
-4
 Ksp = S x 4 x 10
But Ksp = 4.5 x 10-17(given)
Ksp = 4.5 x 10-17 = S x 4 x 10-4
S = 4.5 x 10-17
4 x 10-4
S = 1.125 x 10-13 mol/litre
Mol. Wt of Zn(OH)2 = Zn + 2(O) = 2(H)
= 65 + 2(16) + 2(1)
= 99 gm/mol
 Solubility of Zn(OH)2 in o.o2M NaOH solution
= (1.125 x 10-13) x 99 gm/litre
= 1.11375 x 10-11 grams/litre
But, given solution is of 2 litres
 Solubility = 1.11375 x 10-11(gm/L) x 2 L
Solubility of Zn(OH)2 = 2.2275 x 10-11 gm
45. Solubility product of Mg(OH)2 is 1.2 x 10-11. Calculate its solubility in pure water as
well as in aqueous solution of 0.05 M NaOH
Ans. Suppose, solubility of Mg(OH)2 be S mol/L
Mg(OH)2(s)
Mg2+(aq) + 2OH-(aq)
S
2S
NaOH(aq)
Na+(aq) + OH-(aq)
0.05M
0.05M
0.05M
(a) Solubility in pure water
Ksp = [Mg2+][OH-]2
= [S][2S]2 = 4S3
But, Ksp = 1.2 x 10-11 (given)
4S3 = 1.2 x 10-11
 S3 = 0.3 x 10-4
= 3 x 10-12
S = (3)1/3 x 10-4
S= 1.442 x 10-4 mol/litre
[ (3)1/3 = (log3)/3
= (0.4771)/3 = 0.1590
= antilog(0.1590)
= 1,442]
S = 1.442 x 10-4 mol/litre……(a)
(In pure water)
(b) Solubility in 0.05M NaOH
Total concentration of [OH-]
= 2S + 0.05
[ But, here 0.05 >>> S as Mg(OH)2 is sparingly soluble)
[OH-] = 0.05 M
Concentration of Mg2+ = [S] = ?
Ksp = [Mg2+][OH-]2
-11
Ksp = 1.2 x 10 = [S][0.05]2
= 1.2 X 10-11 = S x 25 x 10-4
S = 1.2 x 10-11 = 0.048 x 10-7 = 4.8 x 10-9
25 x 10-4
S = 4.8 x 10-9 mol/litre
= 4.8 x 10-19 M
46. At 298 K temperature, 2.901 litre saturated solution can be prepared by dissolving 0.08
gram CaF2. Calculate Ksp of salt. (Molecular mass of CaF2 = 78.0 g.mol-)
Ans. W = 0.08 gm
Molecular wt = 78 gm/mol
V(L) = 2.901 litre
Now, concentration = 3.5354 x 10-4 M
Now,
CaF2
Ca2+
+
2F-4
(3.53 x 10 )
(2 x 3.53 x 10-4)
Now, Ksp =[Ca2+][F-]2
= (3.53 x 10-4)(2 x 3.53 x 10-4)2
= 4 x (3.53 x 10-4)3
= 4 x 44.191 x 10-12 M3
= 176.767 x 10-12 M3
= 1.7676 x 10-10 M3
Ksp = 1.768 x 10-10 M3
47. Equal volumes of 2 x 10-4 M BaCl2 and 5.0 x 10-3 M H2SO4 are mixed. Will precipitation
occur or not? (Ksp of BaSO4 = 1.1 x 10-10)
Ans, If equal volumes of solutions are mixed, concentration of each salt will be halved
[C1V1 = C2V2
But if V2 = 2V1  C1V1 = C2(2V1)
 C2 = (C1)/2]
 Concentration of BaCl2 in the mixture
= 2 x 10-4 M = 1 x 10-4 M ….(i)
2
Concnetration of H2SO4 in the mixture = (5 x 10-3)/2 M = 2.5 x 10-3 M …(ii)
BaCl2
Ba2+
+
2Cl-4
-4
(1 x 10 )
1 x 10 M
(2 x 1 x 10-4) M
H2SO4
2H+
+
SO4-2
-3
-3
(2.5 x 10 )M
(2 x 2.5 x 10 )M
2.5 x 10-3 M
When, they are mixed, salt formed is BaSO4
BaCl2 + H2SO4
BaSO4 + 2HCl
BaSO4(aq)
Ba2+(aq) + SO42-(aq)
Ionic Product of BaSO4, Ip = [Ba2+][SO42-]
= (1 x 10-4)(2.5 x 10-3)
-7
2
Ip = 2.5 x 10 M
But, Ksp = 1.1 x 10-10(given)
Comparing above both values
Ip > Ksp
 Precipitation will occur
48. By addition of how many grams of FeSO4 to 500 ml solution of 0.02 M NaOH, Fe(OH)2
will be precipitated? (Ksp of Fe(OH)2 = 1.5 x 10-15 )
Ans. NaOH
Na+ + OH0.02M
0.02M
Fe(OH)2
Fe+2(aq) + 2OH-(aq)
S
S
2S
[Suppose, molar solubility of Fe(OH)2 be S mol/L
Now, Fe(OH)2 is sparingly soluble salt
 S <<< 0.02 M
 [OH-] = 0.02 M
Now, Ksp = [Fe2+][OH-]2 …(i)
But Ksp = 1.5 x 10-15 (given) …..(ii)
For, precipitation to occur, Ip > Ksp …(iii)
 From equation (i), (ii), (iii)
Ip = Ksp = 1.5 x 10-15 = [Fe2+][OH-]2
S[0.02]2
-15
 1.5 x 10 = S x 4 x 10-4
 S = 3.75 x 10-12 M
 S = [Fe2+] = 3.75 x 10-12 M
Now, l mole [Fe2+] = 1 mol FeSO4
[ because FeSO4
Fe2+(aq) + SO42-(aq)]
2+
-12
 [Fe ] = [FeSO4] = 3.75 x 10 mol/litre
For l litre, FeSO4 = 3.75 x 10-12 moles
Hence for 500 ml = (3.75 x 10-12)/2
= 1.875 x 10-12 moles/500 ml
Mol. Wt of FeSO4 = Fe + S + 4(O)
= 56 + 32 + 4(16)
= 152 gm/mol
 Wt of FeSO4 = 1.875 x 10-12 x 152
= 285 x 10-12
= 2.85 x 10-10 gm
For precipitation, Ip > Ksp
 Wt (FeSO4) > 2.85 x 10-10 gm.
Thus, if FeSO4 added in slightly greater than 2.85 x 10-10 gm, Ip > Ksp and Fe(OH)2 will
precipitate.
49. Will precipitates of PbI2 be obtained by mixing 20 ml 3 x 10-3 M pb(NO3)2 and 80 ml.
2 x 10-3 M NaI Solutions ? (Ksp for PbI2 = 6.0 x 10-9)
Ans. Pb(NO3)2(aq)
Pb2+(aq) + 2NO3-(aq)
-3
3 x 10 M
3 x 10-3M
2(3 x 10-3M)
[Pb2+] = 3 x 10-3 M….(i)
NaI(aq)
Na+(aq) +
I-(aq)
-3
-3
2 x 10 M
2 x 10 M
2 x 10-3M
[I-] = 2 x 10-3 M….(ii)
20 ml Pb(NO3)2 and 80 ml NaI solution is mixed.
 Total volume, V2 = (20 + 80) ml
= 100 ml
V2 = 100 ml
Now, concentrations of Pb2+ and I- ions in mixture according to M1V1 = M2V2.
For Pb2+
M1 = 3 x 10-3M
V1 = 20 ml
V2 = 100 ml.
M1V1 = M2V2  M2 = M1 x V1
V2
 M2 = (3 x 10-3) x 20 = 6 x 10-4
100
 [Pb2+] 6 x 10-4 M…(iii)
For IM1 = 2 X 10-3 M
V1 = 80 ml
V2 = 100 ml
M1V1 = M2V2  M2 = M1 x V1
V2
= (2 x 10-3) x 80 = 1.6 x 10-3 M
100
[I-] = 1.6 x 10-3 M….(iv)
Now, Ksp = 6 x 10-4 (given)…(v)
For PbI2,
Ip = [Pb2+][I-]2
[ PbI2(S)
Pb2+(aq) + 2I-(aq)]
From equation (iii) and (iv)
Ip = (6 x 10-4) x (1.6 x 10-3)2
Ip = 15.36 x 10-10
 Ip = 1.536 x 10-9 ….(vi)
From equation (v) and (vi)
Ip(1.536 x 10-9) < Ksp(6.0 x 10-9)
Ip < Ksp
 PbI2 will not precipitate
50. The concentration of F- in a sample of water is 2 x 10-5 M. How many minimum grams
of solid CaCl2 will have to be added for precipitation of F-?( Ksp for CaF2 = 1.7 x 10-10,
molecular mass of CaCl2 = 111 grammolAns.
[F-] = 2 x 10-5 M
CaCl2
Ca2+(aq) + 2Cl-(aq)
Suppose, x mole of CaCl2 should be added to 1 litre of water
concenetration of Ca2+ in water will be x M
 [Ca2+] = x M
CaF2(aq)
Ca2+(aq) + 2F-(aq)
xM
(3 x 10-5) M
-10
Ksp = 1.7 x 10
(given)
Now, Ksp = [Ca2+][F-]2
= (X)(3 x 10-5M)2
1.7 x 10-10 = X x 9 x 10-10
 X = (1.7)/9 = 0.1888 M
 X = 0.189 M …..(i)
Now, 1 mol of Ca2+ = 1 mole of CaF2
= 1 mole of CaCl2
1 mole of CaF2 = 1 mole of CaCl2.
0.189 moles of CaF2 = 0.189 moles of CaCl2 [from equation (i)]
If little more than 0.189 moles of CaCl2 is added i.e 0.20 moles, then CaF2 will precipitate.
 moles = 0.20
Molar mass of CaCl2 = 111 gm/mol
 0.20 mol of CaCl2 will contain
= 111 gm/mol x 0.20
= 22.2 gm solid CaCl2
22.2 gms solid CaCl2 is added for precipitation of F- (minimum value)