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1. A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature.
The volume of the container is suddenly increased.
a) What is the initial effect of the change on vapour pressure?
b) How do rates of evaporation and condensation change initially?
c) What happens when equilibrium is restored finally and what will be the final vapour
pressure?
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Solution:
a. The vapour pressure will decrease due to increase in volume.
b. If the volume of the container is increased but the surface area remains the same then the rate of
evaporation remains the same while the rate of condensation decreases.
c. When equilibrium is restored finally, the vapour pressure will have the same value as vapour
pressure depends only on the temperature and not on the volume of the vessel.
2. What is Kc for the following equilibrium when the equilibrium concentration of each
substance is: [SO2] = 0.60 M, [O2] = 0.82 M and [SO3] = 1.90 M?
2SO2(g) + O2(g) f 2SO3(g)
Solution:
Kc =
=
= 12.23
∴ Kc = 12.23.
Pa = 2.67 × 104 Pa.
.in
on
Kp =
ti
Solution:
po = 105 Pa
pI2(g) = 2.6 × 109 Pa (Experimentally)
ca
3. At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40% by
volume of I atoms I2(g) f 2I(g) Calculate Kp for the equilibrium.
4. Write the expression for the equilibrium constant, Kc for each of the following
reactions: (i) 2NOCl(g) f 2NO(g) + Cl2(g) (ii) I2(S) + 5F2 f 2IF5
Solution:
(i) Kc =
(ii) Kc =
5. Find out the value of Kc for each of the following equilibrium from the value of kp:
(i) 2NOCl(g) f 2NO(g) + Cl2(g); Kp = 1.8 × 10-2 at 500 k
(ii) CaCO3(S) f CaO(S) + CO2(g); Kp = 167 at 1073 K
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Solution:
(a) 2NOCl(g)
2NO(g) + Cl2(g)
Kp = 1.8 × 10-2 at 500 K
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For the reaction, ∆n = (2+1) – 2 = 1
Kp = Kc (RT)n
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KC =
=
= 4.4 × 10-4
(b) CaCO3(s) CaO(s) + CO2(g)
Kp = 167 at 1073K
Kc =
The concentration of CaO(s) and CaCO3(s) may be assumed to be constant i.e.
[CaO(s)] – 1 and [CaCO3(s)] = 1 and hence neglected so that Kc = [CO2(g)]
Since the partial pressure of a gaseous component is proportional to its concentration
Kp = p[CO2(g)]
Kc =
=
= 1.90.
6. For the following equilibrium, Kc = 6.3 × 1014 at 1000
K NO(g) + O3(g) f NO2(g) + O2(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular
reactions. What is Kc, for the reverse reactions?
= 1.6 × 10-15.
on
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∴ Kc (reverse) =
ca
Solution:
Kc for the reverse reaction is inverse of Kc for forward reaction.
7. Explain why pure liquids and solids can be ignored while writing the equilibrium
constant expression?
.in
Solution:
Concentration of solids and liquids are constant and taken as unity
[Solids] = 1; [liquids] = 1.
Therefore pure liquids and solids can be ignored while writing the equilibrium constant expression.
8. Reaction between N2 and O2- takes place as follows:
2N2(g) + O2(g) f 2N2O(g)
If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and
allowed to form N2O at a temperature for which Kc = 2.0 × 10-37, determine the composition
of equilibrium mixture.
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Solution:
2N2(g) + O2(g)
2N2O(g)
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Initial moles: 0.482 + 0.933
Moles at equilibrium: (0.482 – x) (0.933 – x/2) x
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Since the value of K is very small (2.0 × 10-37) it indicates that the equilibrium is very much in favour
of reactants and thus the value of x is very small. Therefore, the approximate concentrations at
equilibrium may be written as
= 0.0482 mol L-1
[N2] =
= 0.0933 mol L-1
[O2] =
[N2O] =
Kc =
or 0.1 x mol L-1
=
= 2.0 × 10-37
x = 6.6 × 10-20
[N2] = 0.0482 mol L-1
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[O2] = 0.0933 mol L-1.
ca
∴ [N2O] = 0.1 × 6.6 × 10-20 = 6.6 × 10-21 mol L-1
9. Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:
.in
2NO(g) + Br2(g) f 2NOBr(g)
When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant
temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount
of NO and Br2.
Solution:
Amount of NO Br, formed at equilibrium = 0.0518 mol.
Initial amount of NO = 0.087 mol
Initial amount of Br2 = 0.0437 mol.
The reaction is
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2NOBr 2NOBr 2NOBr
2NO + Br2(g)
2 mols of NOBr is formed from 2 mols of NO
Therefore 0.0518 mol of NOBr is formed from 0.0518 mol of NO
So the amount of NO at the equilibrium = 0.087 – 0.0518 = 0.0352 mol.
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Amount of Br2 at the equilibrium
According to equation
Number of moles of Br2, reacting with 0.0518 mol of NoBr =
= 0.0259 mol.
Therefore Amount of Br2 at the equilibrium = 0.0437 – 0.0259 = 0.0178 mol.
10. A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the
partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?
2HI(g) f H2(g) + I2(g)
Solution:
2HI(g)
H2(g) + I2(g)
Initial Pressure 0.2 atm - -
At equilibrium 0.2 p p/2 p/2
(0.04 atm) 0.8 atm 0.8 atm
=
= 400.
ca
Kp =
Solution:
[H2] =
M = 0.096 M
[N2] =
M = 0.079 M
[NH3] =
.in
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11. A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a
20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the
reaction N2(g) + 3H2(g) f 2NH3(g) is 1.7 × 102. Is the reaction mixture at equilibrium? If
not, what is the direction of the net reaction?
M = 0.407 M
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Qc =
=
= 2370
since Qc > Kc (equilibrium constant) the reaction is said to proceed in the reverse direction.
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12. The equilibrium constant expression for a gas reaction is,
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Ke =
replace [H2O] by [H2O]6 replace [O2] by [O2]5
Write the balanced chemical equation corresponding to this expression.
Solution:
Ke =
Hence chemical equation is
4NO + 6H2O
4NH3 + 5O2.
13. One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At
equilibrium 40 % of water (by mass) reacts with CO according to the equation,
H2O(g) + CO(g)
H2(g) + CO2(g)
Calculate the equilibrium constant for the reaction.
Solution:
The equilibrium reaction involved is
H2 (g) + CO2 (g)
n.
in
Initial moles of H2O and CO taken = 1 mole each.
% dissociation of H2O = 40%
Number of moles of H2O dissociated =40/100 ×1= 0 .4 moles
Number of moles of CO at equilibrium = 1- 0.4 = 0.6
Number of moles of H2O at equilibrium = 1 - 0.4 = 0.6
No. of moles of H2 formed = 0.4
No. of moles of CO2 formed = 0.4
Therefore [H2O] = 0.6/10 =0.06mol/L
[CO] = 0.6/10 = 0.06mol/L
[H2] = 0.4/10 = 0.04mol/L
[CO2] = 0.4/10 = 0.04mol/L
io
at
H2O (g) + CO (g)
Hence,
= 4/9 = 0.44
14. At 700 K, equilibrium constant for the reactions:
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H2(g) + I2(g) f 2HI(g) is 54.8. If 0.5 mol L-1 of HI(g) is present at equilibrium at 700 K,
what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g)
and allowed it to reach equilibrium at 700 K?
Solution:
For the equilibrium,
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H2(g) +I2(g)
2HI(g)
The value of equilibrium constant K is 54.8.
Hence, if we start with HI and allow the reaction to reach equilibrium then the equilibrium constant
K' = 1/K = 1/54.8
K' = [H2][I2] /[HI]2 = 0.0182
Let the concentration of both H2 and I2 be x
Then K' = 0.0182 = x2/ (0.5)2
x2 = 0.0182 × 0.25 = 0.00456
x = 0.067mol/litre.
15. What is the equilibrium concentration of each of the substances in the equilibrium
when the initial concentration of ICl was 0.78 M?
2ICl(g)
I2(g) + Cl2(g) ; Kc = 0.14
Solution:
2Icl(g)
I2(g) + Cl2(g)
Initial Concentration 0.78M - -
At equilibrium (0.78-x)M x/2M x/2M
ti
ca
Kc =
=
[ICl] = 0.78 – 0.33 = 0.45M
[I2] =
in
∴ x = 0.33M
.
on
0.14 =
= 0.165M
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[Cl2] =
= 0.165M.
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16. The ester, ethyl acetate is formed by the reaction of ethanol and acetic acid and the
equilibrium is represented as:
CH3COOH + C2H5OH
CH3COOC2H5 + H2O
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(i) Write the concentration ratio, Q, for this reaction. Note that water is not in excess and is
not a solvent in this reaction.
(ii) At 293 K, if one starts with 1.0 mol of acetic acid and 0.180 mol of ethanol, there is
0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium
constant.
(iii) Starting with 0.50 mol of ethanol and 1.0 mol of acetic acid and maintaining at 293 K,
0.214 mol of ethyl acetate is formed after some time. Has equilibrium been reached?
Solution:
(i)
(ii) At equilibrium, concentration of acetic acid is = 1.0 - 0.171 = 0.829 mol
and concentration of ethanol is
= 0.180-0.171
= 0.009
K = (0.171)/(0.829 × 0.009) = 22.19
io
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(iii) Initial conc. of ethanol = 0.5mol
Initial conc. of acetic acid = 1.0mol
After some time, conc. of ethanol = 0.5 - 0.214 = 0.286 mol
conc. of acetic acid = 1.0 - 0.214 = 0.786 mol
Hence, Q = (0.214)/(0.786 × 0.286)
= 0.9520
Since the value of Q is lesser than that of K, the reaction must be proceeding in the forward direction
readily.
17. A sample of pure PCl5 was introduced into an evacuated vessel at 473K. After
equilibrium was attained, concentration of Pcl5 was found to be 0.5 × 10-1 mol L-1 . If value
of Kc is 8.3 × 10-3, what are the concentrations of PCl3 and Cl2 at equilibrium?
Kc =
in
PCl5
PCl3 + Cl2
Let the concentration of PCl5 and Cl2 be x at equilibrium.
n.
Solution:
Or 8.3 × 10-3 =
Or 4.15 × 10-4 = x2
Or x = 2 × 10-2 = 0.02 mol L-1
Hence the concentration of PCl5 and Cl2 each at equilibrium = 0.02 mol L
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18. One of the reaction that takes place in producing steel from iron ore is the reduction
of iron (II) oxide by carbon monoxide to give iron metal and CO2.
FeO(s) + CO(g) f Fe(s) + CO2(g); Kp = 0.265 atm at 1050K
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What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial
pressures are: PCO = 1.4 atm and PCO2 = 0.80 atm?
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Solution:
FeO(s) + CO(g)
Fe(s) + CO2(g)
Kp = 0.265 atm at 1050 K
FeO(s) + CO(g)
Fe(s) + CO2(g)
Initaial pressure 1.4 atm 0.8 atm
Equilibrium pressure (1.4 – x) atm (0.8 + x) atm
Kp =
= 0.265 atm
0.8 + x = 0.265 (1.4 – x) = 0.371 – 0.265 x
x + 0.265x = 0.371 – 0.8
1.265x = -0.429
∴ Equilibrium pressure of CO = 1.4 – (-0.34) = 1.74 atm
on
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Equilibrium pressure of CO2 = 0.8 + (-0.34) = 0.46 atm.
ca
x = -0.34 atm
.in
19. Equilibrium constant, Kc for the reaction N2(g) + 3H2(g)
2NH3(g) at 500 K is
0.061. At a particular time, the analysis shows that composition of the reaction mixture is
3.00 mol L-1 of N2, 2.00 mol L-1 of H2 and 0.500 mol L-1 of NH3. Is the reaction at
equilibrium? If not, in which direction does the reaction tend to proceed to reach
equilibrium?
Solution:
N2(g) + 3H2(g)
2NH3(g)
Kc =
=
=
=
= 0.01
Since Qc < Kc the reaction is not at equilibrium. The reaction proceeds to form more product.
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Edumass
Edumass
20. Bromine mono-chloride, BrCl decomposes into bromine and chlorine and reaches the
equilibrium.
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2BrCl(g)
Br2(g) + Cl2(g)
for which Kc = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.30 × 103
mol L-1 what is its molar concentration in the mixture at equilibrium.
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Solution:
2BrCl(g)
Br2(g) + Cl2(g)
3.30 × 10-3 0 0
KC = 32 at 500K
From the equation 2 mols of BrCl on decomposition gives = 1 mol of Br and 1 mol of Cl2
Therefore, 3.30 × 10-3 mol L-1 of BrCl gives =
= 1.65 × 10-3 mol of Br2 and Cl2
Let x be the molar concentration of BrCl at the equilibrium
Molar concentration of Br2 = 1.65 × 10-3 mol.
Molar concentration of I2 = 1.65 × 10-3 mol
Therefore Kc =
32 =
x2 =
x2 =
ca
x=
= 3 × 10-4 mol L-1
Therefore Molar concentration of BrCl = 3 × 10-4 mol L-1.
C(s) + CO2(g)
2CO(g)
Calculate the Kc for this reaction at the above temperature.
.in
Solution:
Let total mass of the gaseous mixture be = 100g
on
ti
21. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with
solid carbon has 90.55% CO by mass.
Mass of CO = 90.55 g
Mass of CO2 = 9.45g
Edumass
Edumass
9
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Edumass
Edumass
= 3.234
Moles of CO2 =
= 0.215
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Moles of CO =
Total number of moles of gases = 3.234 + 0.215 = 3.449
× 1 = 0.938 atm
PCO =
× 1 = 0.0623 atm
Kp =
Kp = Kc(RT)∆
= 14.12
n
= Kc(RT) Q ∆ n = (2-1) = 1
Kc =
= 0.153.
22. Calculate a) ∆ G°° and b) the equilibrium constant for the formation of NO2 from NO
and O2 at 298 K
NO2(g)
Where
∆ fG0 (NO2) = 52.0 kJ/mol
∆ fG0 (O2) = 0 kJ/mol
Solution:
NO2(g)
in
.
on
∆ fG0 (NO) = 87.0 kJ/mol
NO(g) + ½ O2(g)
ti
ca
NO(g) + ½ O2(g)
∆ fG0 (NO2) = 52.0 kJ/mol-1
∆ fG0 (NO) = 87.0 kJ/mol-1
∆ fG0 (O2) = 0 kJ/mol-1 T = 298 K
Edumass
Edumass
10
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Edumass
Edumass
∆ G° = ∆ fG0 (NO2) – [∆ fG0 (NO) + ∆ fG0 (O2)]
= ∆ fG0 (NO2) – ∆ fG0 (NO)
= (52-87) kJ mol-1
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We have, ∆ G° = -2.303 RT log Kc
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-35 kJ mol-1 = -2.303 × 8.314 JK-1 × 298 × log Kc
log Kc =
=
=
log Kc = 6.134
Kc = 1.361 × 106.
23. Does the number of moles of reaction products increase, decrease or remain the
same when each of the following equilibria is subjected to a decrease in pressure by
increasing the volume?
PCl3(g) + Cl2(g)
Fe3O4(s) + 4H2(g)
Solution:
a) No. of moles of the products increases.
b) No. of moles of the products decreases.
.in
on
c) 3Fe(s) + 4H2O(g)
CaCO3(s)
ti
b) CaO(s) + CO2(g)
ca
a) PCl5
c) No. of moles of the products decreases.
24. Which of the following reactions will get affected by increase of pressure? Also
mention, whether change will cause the reaction to go into the right or left direction?
Edumass
Edumass
11
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Edumass
Edumass
(i) CH4(g) + 2S2(g)
(ii) CO2(g) + C(s)
CS2(g) + 2H2S(g)
2CO(g)
4NO(g) + 6H2O(g)
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(iii) 4NH3(g) + 5O2(g)
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Solution:
The following reactions will get effected by increase of pressure
(ii) CO2(g) + C(s)
2CO(g)
The increase of pressure will shift the equilibrium to left hand direction.
(iii) 4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
The increase of pressure will shift the equilibrium to left hand direction.
25. The equilibrium constant for the following reaction is 1.6 × 105 at 1024K H2(g) +
Br2(g)
2HBr(g) Find the equilibrium pressure of all gases if 10.0 bar of HBr is
introduced into a sealed container at 1024 K.
Solution:
Kp = Kc =
Let (pH2)eq = (pBr2)eq = x bar
ca
1.6 × 105 =
ti
x2 =
.in
on
x=
-1
or x = 0.25 × 10
-2
or 2.5 × 10
Therefore, (pH2)eq = 2.5 × 10-2
(pBr2)eq = 2.5 × 10-2 bar
and (pHBr)eq = 10.0 bar.
26. Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per
the following endothermic reaction:
CH4(g) + H2O(g)
Edumass
Edumass
CO(g) + 3H2(g)
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a) Write as expression for Kp for the above reaction.
b) How will the values of Kp and composition of equilibrium mixture be affected by
(i) increasing the pressure
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
09
(ii) increasing the temperature
(iii) using a catalyst?
Solution:
(a) Kp =
(b) (i) The value of Kp remains unchanged on increasing the pressure. When pressure is increased
then according to Le-Chatelier’s principle the equilibrium shifts in the direction where there is less
number of moles of gases i. e., backward direction in case of the given reaction.
(ii) In case of endothermic reactions the value of Kp increases with increase in temperature. With
increase in temperature, the equilibrium shifts in the endothermic directions i. e., forward direction in
case of the given reaction.
(iii) Kp will remain undisturbed. Equilibrium composition will remain unchanged. However, in the
presence of catalyst, the equilibrium would be attained quickly.
27. Describe the effect of:
ti
ca
a) addition of H2
b) addition of CH3OH
c) removal of CO
on the equilibrium of the reaction:
CH3OH(g)
Solution:
2H2(g) + CO(g)
in
2H2(g) + CO(g)
.
on
d) removal of CH3OH
CH3OH(g)
a) Effect of addition of H2
This will favour forward reaction i. e., formation of CH3OH because increase in the concentration of
the reactants will result in the formation of more products.
Edumass
Edumass
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b) Addition of CH3OH
This will shift the equilibrium towards backward direction because when the product concentration
increases, backward reaction is favoured.
c) Removal of CO
This favours the increase in formation of CO i. e., backward reaction is favoured.
w
09
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
d) Removal of CH3OH
This will result in the increased production of CH3OH i. e., forward reaction is favoured.
All these explanations are according to Le Chatelier’s principles.
28. At 473 K, equilibrium constant, Kc for decomposition of phosphorus pentachloride,
PCl5 is 8.3 × 10-3 . If decomposition is depicted as
= 124.0 KJ mol -1
(a) Write an expression of Kc for the reaction.
(b) What is the value of Kc for the reverse reaction at the same temperature?
(c) What would be the effect on Kc if (i) more PCl5 is added (ii) the pressure is increased
(iii) the temperature is increased?
Solution:
PCl5(g)
PCl3(g) + Cl2(g)
(a) Kc =
(b) Reverse reaction
=
=120.48
ti
Kc =
PCl5(g)
ca
PCl3(g) + Cl2(g)
.in
on
(c) (i) More PCl5 is added: The equilibrium will be shifted to the right. This will continue till the entire
extra amount of PCl5 has been used up.
(ii) The pressure is increased:
PCl5(g)
1 mole
PCl3(g) + Cl2(g)
1 mole
1 mole
Since backward reaction takes place with decrease in number of moles, an increase in pressure will
favour combination of PCl3 and Cl2 molecules to produce back PCl5.
(iii) The temperature is increased: The reaction is endothermic. Therefore, an increase in
temperature will shift the equilibrium to the right.
Edumass
Edumass
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29. Predict which of the following reaction will have appreciable concentration of
reactants and products:
2Cl(g) Kc = 5 × 10-39
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
09
a) Cl2(g)
b) Cl2(g) + 2NO(g)
c) Cl2(g) + 2NO2(g)
2NOCl(g) Kc = 3.7 × 108
2NO2Cl(g) Kc = 1.8
Solution:
a) Will have appreciable concentration of reactants.
b) Will have appreciable concentration of products.
c) Will have appreciable concentration of reactants and products.
30. The value of Kc for the reaction 3O2(g)
2O3(g) is 2.0 × 10-50 at 25°° C. If the
equilibrium concentration of O2 in air at 25°° C is 1.6 × 10-2, what is the concentration of O3?
Solution:
3O2(g)
2O3(g)
Kc = 2 × 10-5 T = 25° C = 298 K
= 2 × 10-50 × (1.6 × 10-2)3
= 2 × 10-50 × 10-6 × 4.096
= 8.192 × 10-56
[O3] = √ 81.92 × 10-12 = 2.86 × 10-28.M
.in
on
ti
[O3]2 = Kc × [O2]3
ca
Kc =
31. The reaction, CO(g) + 3H2(g) f CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1L
flask. It also contain 0.30 mol of CO, 0.10 mol of H2and 0.02 mol of H2O and an unknown
equilibrium constant, Kc for the reaction at the given temperature is 3.90.
Edumass
Edumass
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Solution:
CO(g) + 3H2(g)
CH4(g) + H2O(g)
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
09
Kc =
3.9 =
[CH4] =
= 0.0585 mol.
32. The concentration of sulphide ion in 0.1 M, HCl solution saturated with hydrogen
sulphide is 1.0 × 10-19 M. If 10 ml of this is added to 5 ml of 0.04 M solution of the following:
FeSO4, MnCl2, ZnCl2 and CdCl2. in which of the solutions will precipitation take place. (Please
consult table 8).
Solution:
= 6.3 x 10-13
Ksp MnS
........................
= 2.5 x 10-13
Ksp ZnS ..............................
= 1.6 x 10-24
Ksp CdS
= 8.0 x 10-27
......................
Fe2+ + SO42-
FeSO4
H 2S
..........................
= 2H+ = S2-
The total volume after
mixing
[S2-]
.........................
[Fe2+] [S2-] .................
= 1.33 x 10-2
= 0.6 x 10-19
in
[Fe2+] ........................
= (10 + 5) = 15 ml
.
on
Fe2+ + S2FeS
ti
ca
Ksp FeS
.........................
= 0.798 x 10-21
Similarly [Zn2+] [S2-] and [Cd2+] [S2-] are also 0.798 x 10-21
Since in the case of ZnCl2 and CdCl2 Solutions, the [Zn2+] [S2-] and [Cd2+] [S2-] > Ksp of ZnS and Cds
respectively.
Edumass
Edumass
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Therefore ZnCl2 and CdCl2 give precipitation of their respective sulphides.
33. Which of the followings are Lewis acids?
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
09
H2O, BF3 , H+ and NH
Solution:
BF3, H+, NH4+.
34. What will be the conjugate bases for the Bronsted acids: HF, H2SO4 and HCO3?
Solution:
The conjugate base of HF is F- that of H2SO4 is HSO4- and that of HCO3- is CO32-.
35. Write the conjugate acids for the following Bronsted bases: NH2-, NH3 and HCOO-
Solution:
Bronsted Bases
Conjugate Acids
NH2-
NH3
NH3
NH4+
HCOO-
HCOOH
36. The spieces: H2O, HCO3-, HSO4- and NH3 can act both as Bronsted acids and bases.
For each case give the corresponding conjugate acid and base.
Solution:
H 2O
H 3O+
HCO3-
H2CO3
HSO4-
H2SO4
NH3
NH4+
Conjugate Base
OH-
.in
on
ti
Conjugate Acids
ca
Bronsted Acid/Base
CO32-
SO42NH2-
37. Classify the following species into Lewis acids and Lewis bases and show how these
act as Lewis acid/base: (a) OH- (b) F- (c) H+ (d) BCl3.
Solution:
a) OH- ion is a Lewis base as it has lone pairs of electrons on oxygen which it can donate
Edumass
Edumass
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b) F- ion has four lone pairs of electrons. Hence, it can act as a Lewis base by donating any one of
these lone pair.
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
w
09
c) H+ acts as a Lewis acid as it has a vacant orbital and hence can accept a pair of electrons from
Lewis bases.
d) BCl3 acts as a Lewis acid. In BCl3, there are only six electrons in the valence shell of boron. It can
accept a pair of electrons from bases such as NH3 to complete its octet.
38. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10-3 M. What is its
pH?
= - log 3.8 + (-log 10-3)
.in
pH = 2.42
on
= -log 3.8 + 3 log 10
ti
= - log (3.8 × 10-3)
ca
Solution:
PH = -log [H3O+]
39. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in
it.
Solution:
€pH = 3.76
Edumass
Edumass
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pH = -log [H3O+]
3.76 = -log [H3O+]
log [H3O+] = -3.76
w
log [H3O+] = 4.2400
09
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
[H3O+] = antilog (4.2400) = 1.7 x 10-4
40. The ionization constant of acetic acid is 1.74 × 10-5. Calculate the degree of
dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion
in the solution and its pH.
Solution:
Given data: Kα of acetic acid = 1.74 × 10-5
α=?
Concentration of acetic acid = 0.05 M
∴ α = 0.0187
[CH3COO-] = Cα = 0.05 × 0.0187 mol L-1 = 0.00094 mol L-1
[CH3COO-] = 9.4 × 10-4 mol L-1.
= - log5 + 2 log 10
∴ pH = 1.301.
.in
= -0.6990 + 2 = 1.301
on
= -log (5 × 10-2)
ti
∴ pH = -log [H3O+]
ca
[H3O+] = [CH3COO] = 0.05M = 5 × 10-2 M
41. The ionisation constants of HF, HCOOH and HCN at 298 K are 6.8 × 10-4, 1.8 × 10and 4.8 × 10-4 respectively. Calculate the ionisation constants of the corresponding
conjugate base.
4
Solution:
Edumass
Edumass
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We know that Ka x Kb = Kw = 1 x 10
-14
Kb = Kw/Ka.
i) Ionisation constant of HF (Ka) = 6.8 x 10-4
Ionisation constant of F- (Kb) = Kw/Ka = 1.5 x 10-11
w
ii) Ionisation constant of HCOOH (Ka) = 1.8 x 10-4
09
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
Ionisation constant of HCOO- (Kb) = 5.6 x 10-11
iii) Ionisation constant of HCN (Ka) = 4.8 x 10-9
Ionisation constant of CN- (Kb) = 2.08 x 10-6.
42. The ionisation constant of phenol is 1.0 × 10-10. What is the concentration of
phenolate ion in 0.05M solution of phenol? What will be its degree of ionisation if the
solution is also 0.01 M in sodium phenolate.
Solution:
= 1.0 × 10-10
Ionisation constant of phenol
Degree of ionisation
=α=
[Phenolate ion]
=
=
= 4.47 × 10-5
= cα =
=
= 2.2 × 10-6
= -log (2.2 × 10-6 M/M)
= 5.65
..........................
Degree of ionisation
=
=
=
.in
on
Degree of ionisation of the solution is also 0.01M
ti
ca
PH
..................
=
=
= 1 × 10-4
43. The first ionization constant of H2S is 9.1 × 10-8. 10-8. Calculate the concentration
of HS- ion in its 0.1 M solution and how will this concentration be effected if the solution is
0.1 M in HCl also. If the second dissociation constant of H2S is 1.2 × 10-13, calculate the
Edumass
Edumass
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concentration of S2- under both conditions.
Solution:
H 2S
H+ + HS- Ka1 = 9.1 × 10-8
HS-
H+ + S2- Ka2 = 1.2 × 10-13
w
09
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
HCl
H+ + ClIn the case of HCl, the dissociation is complete.
So [H+] = 0.1 M.
Let [H+] from H2S be x and also x be the [HS-].
In the solution of 0.1M HCl
[H+] ..........................
[HS] ..........................
[H2S] ..........................
= 0.1 + x
=X
= [0.1 – x]
Ka ..........................
Ka1 ..........................
Since x is very small
=
= 9.1 × 10-8
9.1 × 10-18 ..........................
=
x
..........................
=
= 9.1 × 10-8
In 0.1 M HCl solution, concentration of [HS ] = 9.1 × 10-8 M / litre
2nd dissociation
HSH+ + S2Ka2 ..........................
Ka2
= 1.2 × 10-13
..........................
1.2 × 10-13 ..................
=
-20
= 1.09 × 10
M / Litre
ti
S2- ..........................
ca
=
44. Assuming complete dissociation, calculate the pH of the following solutions:
.in
on
a) 0.003 M HCl b) 0.005 M NaOH
c) 0.002 M HBr d) 0.002 M KOH
Solution:
a) 0.003 M HCl
HCl + H2O
H3O+ + Cl
0.003 M
pH ..........................
= -log [H3O]+
-3
= -log(3 × 10 )
= (3-log 3)
= (3-0.4771)
= 2.5229 = 2.52
Edumass
Edumass
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b) 0.005 M NaOH
Na+(aq) + OH-(aq)
NaOH
0.005M
[OH-] ..............................
w
[H3O+]
= 0.005 M = 5 × 10-3 M
..........................
=
=
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
= 0.2 × 10-11 = 2 × 10-12
09
PH ..........................
= - log [H3O+]
= - log (2 × 10-12)
= [12 – log 2]
= [12 - 0.3010 ] = 11.70
c) 0.002 M HBr
HBr + H2O
Br-(aq) + H3O+(aq)
0.002 M
[H3O+] ..........................
pH ................................
= 0.002 M = 2 × 10-3 M
= - log[H3O+]
= - log(2 × 10-3)
= (3 – log2) = (3-0.3010)
= 2.699 = 2.70
d) 0.002 KOH
KOH → K+(aq) = OH-(aq)
[OH-]
..........................
=
=
= 0.5 × 10-
11
M
............................
= -log[H3O+]
= -log [5 × 10-12]
45. Calculate the pH of the following solutions:
.in
= (12 – log 5)
= 12 – 0.699
= 11.3
on
pH
ti
[H3O+] ...........................
= 0.002 M = 2 × 10-3 M
ca
0.002 M
a) 2 g of TIOH dissolved in water to give 2 litres of solution.
b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution.
c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
Edumass
Edumass
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d) 1 ml of 13.6 M HCl is diluted with water to give 1 litre of solution.
Solution:
a) Molar mass of TlOH = 221.4
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
09
nTlOH = 2/221.4 = 9.03 × 10−3
[OH-] = n
TlOH_
/V
lit
= 9.03 × 10-3/ 2 = 4.51 × 10-3 M
pOH = −log (4.51 × 10-3 ) = 3€−€log 4.51 = 3€−€0.6542 = 2.35
pH + pOH = 14 ; pH = 14€− 2.35 = 11.65
b) Molar mass of Ca(OH) = 74
2
n
Ca(OH)2
= 0.3 / 74 = 4.05 × 10−3
[OH−] = 4.05 × 10−3/ 0.5 = 0.0081 M
2[OH] = 0.0081 × 2 = 16.2 × 10-3 (Ca(OH)2 contains 2 OH groups.)
pOH = −log (16.2 × 10-3) = 3 – log 16.2 = 1.79
pH = 14 – 1.79 = 12.21
(c) n NaOH = 0.3 / 40 = 7.5 × 10-3M
[OH-] = 7.5€×€10−3/0.2 = 0.0375
pH = 14€ −€1.43 = 12.57
(d) To get molarity(concentration) we can use V1M1 = V2M2
M2 = 1× 13.6/1000 = 0.0136 = [H+]
in
pH =€−log(1.36 × 10-1) = 1 − log 1.36 = 1.87
.
on
1 × 13.6 = 1000 × M2
ti
ca
pOH = −log(0.0375) = −log(3.75 × 10−2) = 2 −€log3.75 = 1.43
46. The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the
pH of the solution and the pKa of bromoacetic acid.
Solution:
Given Data:
C = 0.1 M
Edumass
Edumass
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α = 0.132
[H3O+] = Cα = 0.1 × 0.132 = 0.0132
pH = - log[H3O+]
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
09
= - log 0.0132
= - log (1.32 × 10-2)
= - log 1.32 + 2 log 10
= -0.1206 + 2
∴ pH = 1.88
α=
Squaring both sides,
α
2
=
Ka = Cα2 / 1−α = 0.1 × (0.132)2 / 1€−€0.132
∴ Ka = 2.01 × 10-3
pKa = - log (2.01 × 10-3) = 3 − log 2.01 = 2.70
pKa = 2.77
ti
ca
pKa = - log Ka
.
on
47. The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization
constant and pKb.
Solution:
in
pH = 9.95 ; [H+] = antilog (-9.95) = 1.12 X 10-10M
[OH-] = 1X10-14/1.12 X 10-10 = 8.91 X 10-5M
Kb = [M+] [OH-] / [MOH] = (8.91 X 10-5)2 /0.005 = 1.59 X 10-6
pKb = -log Kb = -log(1.59 X 10-6) = 5.80
Edumass
Edumass
24
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Edumass
Edumass
48. What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be
taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also
calculate the ionization constant of the conjugate acid of aniline.
Solution:
pH = 0.001M
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
09
From the table, the ionization constant of aniline is
Kb = 4.3 × 10-10
α
b
=
∴α
b
=
= 0.002.
Kb × Kα = Kw
= 2.33 × 10-5.
Kα =
∴ Kα = 2.33 × 10-5.
49. Calculate the hydrogen ion concentration in the following biological fluids whose pH
are given below:
Solution:
a) pH
................................
+
= 6.83
= -pH = -6.83
[H3O+] ...............................
= antilog(- 6.83)
= 10-7 × 100.17
= 1.48 × 10-7 M
b) pH .................................
= 1.2
+
log [H3O ] .........................
= -pH = -1.2
[H3O+] ...............................
= antilog(-1.2)
in
[H3O+] ...............................
.
on
log [H3O ] ..........................
ti
ca
a) Human muscle – fluid = 6.83
b) Human stomach fluid = 1.2
c) Human blood
= 7.38
d) Human saliva
= 6.4
= 10-2 × 100.8
= 6.31 × 10-2 M
c) pH
................................
Edumass
Edumass
= 7.38
25
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DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES
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Edumass
Edumass
log [H3O+] ..........................
= -pH = -7.38
[H3O+] ...............................
= Antilog(-7.38)
[H3O+] ...............................
= 10-7.38
w
= 10-8 × 10.62
= 4.17 × 10-8
= 6.4
d) pH
...............................
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
log [H3O+]
.......................
= -pH = -6.4
09
+
[H3O ] .............................
= Antilog(-6.4)
= 3.98 × 10-7 M
50. The pH of milk, black office, tomoto juice, lemon juice and egg white are 6,8, 5.0,
4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
Solution:
i) pH
log[H3O+]
[H3O+]
=
=
=
=
6.8
-pH = -6.8
Antilog [-6.8]
1.5 × 10-7 M
ii) pH = 5.0
log[H3O+]
= -pH =-5.0
[H3O+]
= Antilog(-5.0) = 10-5 M
=
=
=
=
2.2
-pH = -2.2
Antilog(-2.2)
6.31 × 10-3
v) pH
log[H3O+]
=
=
=
=
7.8
-pH = - 7.8
Antilog (-7.8)
-8
1.58 × 10 M
[H3O+]
.in
iv) pH
log[H3O+]
[H3O+]
on
4.2
-pH = -4.2
Antilog(-4.2)
-5
6.31 × 10 M
ti
=
=
=
=
ca
iii) pH
log [H3O+]
[H3O+]
51. What is the minimum volume of water required to dissolve 1g of calcium sulphate at
298K. For Calcium sulphate Ksp is 9.1 × 10—6.
Solution:
CaSO4 → Ca2+ + SO42Let the solubility of CaSO4 be x moles/litre. Then the solution will contain x moles of Ca2+ and x
moles SO42- ions respectively per litre. Hence, the solubility product, Ksp of CaSO4 is
Ksp
.......................
We know Ksp
.................
Edumass
Edumass
= [Ca2+] [SO42-]
= x × (x) = x2
= 9.1 x 10-6
26
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Edumass
Edumass
Hence x2 .......................
Or x .............................
1 mole of CaSO4 .............
3.01 × 10-3 of CaSO4 ......
9.1 × 10-6
3.01 × 10-3 mol L-1
40 + 32 + 64 = 136g
136 × 3.01 × 10-3
136 × 0.00301
0.409 g
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
09
=
=
=
=
=
=
0.41 g needs 1 L of water for complete dissolution
1 g needs =
= 2.38 L.
52. The solubility of Sr(OH)2 at 298 K is 19.23g/L of solution. Calculate the
concentrations of strontium and hydroxyl ions and the pH of the solution.
Solution:
Solubility of Sr(OH)2 at 298 K = 19.23 g/L
Sr(OH)2
Sr2+(aq) + 2OHMol. mass of Sr(OH)2 = 87 + 34 = 121
Molar concentration of Sr(OH)2
=
Sr(OH)2 .........................
0.16M 2(0.16) ......................
[Sr2+] .........................
[OH-] .........................
[H3O+] .........................
=
=
=
=
=
=
.........................
pH .........................
=
- log H3O+
- log 10-12 – log 32
12 + 1.51
13.51
=
.in
on
ti
pH
= 0.1581 M
0.16
Sr2+(aq) + 2OH32 M
0.1581 M
2 × 0.158
0.316 M
ca
=
=
=
=
=
=
=
53. The ionization constant of propionic acid is 1.32 × 10-5. Calculate the degree of
ionization of the acid in its 0.05M solution and also its pH. What will be its degree of
ionization if the solution is 0.01 in HCl also.
Solution:
Ionisation constant of propionic
acid
Concentration of solution
Let the degree of ionisation is α
Edumass
Edumass
= 1.32 × 10-5
= 0.05M
27
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Edumass
Edumass
α ..................................................
=
=
= 1.63 × 10-2
[H+] ..................................
cα
0.05 × 1.63 × 10-2
0.0815 × 10-2
815 × 10-6
-log(815 × 106)
- (log815 + log 10-6)
6-log 815
6-2.9112
3.088
3.09
c
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
=
=
=
=
=
=
=
=
=
=
pH .....................................
09
In the case of molarity of HCl is 0.01 M, it is assumed that it is fully dissociated. Let [H+] be x from
the ionisation of propionic acid. This is (x) also the concentration of propionate ion.
[H+] = 0.01 + X ; [Propionate]
Ka .....................................
= x and [HP] = (C-x)
= [H+] [P-] / [HP]
=
Since x is very small, Ka .......
=
= C. Ka / 0.01
= 0.05 × 1.32 × 10-5 / 0.01
x ..........................................
= 1.60 × 10-5
Degree of ionisation in 0.1 M HCl = x/c
Or x ....................................
= 1.60 × 10-5 / 0.05
= 1.32 × 10_3
io
at
54. The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionisation
constant of the acid and its degree of ionisation in the solution.
Solution:
For weak acid [H+] ..........
C ................................
= c α = 4.57 × 10-3
= 0.1M
α
.................................
2.34
2.34
Antilog [-2.34]
4.57 × 10-3
Ka .................................
=
=α
Ionisation constant Ka
=
= 2.09 × 10-4
Edumass
Edumass
in
=
=
=
=
n.
PH ...................................
- log [H3O+] ....................
or [H3O+] .......................
= 0.0457
2
/C
28
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Edumass
Edumass
55. The ionisation constant of nitrous acid is 4.5 × 10-4. Calculate the pH of 0.04 M
sodium nitrite solution and also its degree of hydrolysis.
Solution:
Ka .................................
w
Hydrolysis constant Kh
=
=h
=
=
× 10-10
09
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
81
Let the degree of hydrolysis
= 4.5 × 10-4
Kh ................................
=
Since h is small 1-h is negligible
Concentration of NaNO3
............................
∴€0.180h2 ......................
= 0.04 M
= 0.04 × h2
= 10-10
h2 ..............................
=
= 2.36 × 10-5
h = .............
PH..................................
= 7.0 +
(pKa + log c)
= 7.0 +
(log 0.04 – log(4.5 × 10-4
= 7.0 +
[-1.40 + 3.35]
= 7.0 +
[1.95] = 7.0 + 0.97 = 7.97
ca
56. A 0.02 M solution of pyridinium hydrochloride has pH = 3.44 M NaOH solution.
Calculate ionization constant of Pyridine.
Solution:
ti
3.44 = 7 −
−3.56 = −
(log 0.02 + pkb)
(−1.70 + pkb)
(log C + pkb),
.in
on
Since pyridinium hydrochloride is a salt of weak base and strong acid pH = 7 −
where kb is the dissociation constant of pyridine
−7.12 = 1.70 − pkb
Pkb = 1.70 + 7.12 = 8.82
Kb = antilog(−8.82) = 1.513 X 10-9
Edumass
Edumass
29
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DIPLOMA CPT CAT SSC AND OTHER EXAMS Also home tutors for other subjects in DELHI AND OTHER MAJOR CITIES
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Edumass
Edumass
57. Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr , NaCN ,NH4NO3, NaNO2 and KF.
c
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
Solution:
Nacl and NaNO2 solutions are neutral. KBr , NaCN and KF solutions are basic. NH4NO3 solution is
acidic.
09
58. The ionisation constant of chloro acetic acid is 1.35 × 10-3. What will be pH of 0.1 M
acid its 0.1 M salt solution?
Solution:
Sodium acetate is a salt of strong base and weak acid.
PH ...............................
Kw ..............................
PH
.............................
=-
[logKw + log Ka – log c]
= 1.0 × 10-14, Ka = 1.35 × 10-3 , c = 0.1 M
=-
[log-14 + log (1.35 + 10-3) – log(0.1)
=-
-14-[-3+0.1303 + 1]
=-
[-14-3+0.1303 + 1]
=[-17 + 1.313]
= 7.94
pH of its sat solution ...........
= 7.94.
Solution:
Ionic product of water
= 2.7 × 10-14
x
x
....................................
pH ...................................
pH ..................................
= 2.7 × 10-14
= 2.7 × 10_14
=
=
=
=
=
=
= 1.64 × 10_7
+
-log [H ]
7 – log 1.64
7 – 0.2148
+6.7696
6.7852.
in
[H+] [OH-] .........................
x2 ....................................
x
n.
H2O → H+ + OH-14
io
at
59. Ionic product of water at 310 K is 2.7 × 10-14. What is the neutral pH of water at this
temperature?
60. Calculate the pH of the resulting mixture of
a) 10 ml of 0.2 M Ca(OH)2 + 25 ml of 0.1 M HCl
Edumass
Edumass
30
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Edumass
Edumass
b) 10 ml of 0.01 M H2SO4 + 10 ml of 0.01 M Ca(OH)2
c) 10 ml of 0.1 M H2SO4 + 10 ml of 0.1 M KOH
Solution:
Ca(OH)2
= 0.2 ×€0.01 = 0.002
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
09
a) n
.nOH-. = 2..× n
Ca(OH)2..=
2.× 0.002 = 0.004
n H+ = nHCl = 0.025€
€× 0.1 = 0.0025
H+ is completely consumed.
∴.nOH-.(left) = 0.004 – 0.0025 = 0.0015
[OH-] =
= 0.0429 M
pOH = -log(0.0429) = 2-0.6325 = 1.3675
pH = 14-pOH = 14-1.3675 = 12.6325
...b) nH2SO4.= .01 ×..01 = 0.0001
nH+...= 2..× .. ) nH2SO4.= 0.0002.....
The base viz., Ca(OH)2 has same volume and concentration.
c) nH2SO4 = .01 ×..0.1 = 0.001
nH+ = 2 nH2SO4 =2 ×.0.001 = 0.002
nH+ (left) = 0.002 -0.001 = 0.001
pH = -log(5 ×10-2) = 2-log 5 = 1.30
in
[H+] = 0.001 / 0.020 = 0.05 M
.
on
nOH- = 0.01× 0.1 = 0.001
ti
ca
Hence nH+ = nOH- = 0.0002
pH of the mixture = 7
61. Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead
chloride and mercurous iodide at 298 K from their solubility products constants. Determine
also the molarities of individual ions.
Edumass
Edumass
31
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Edumass
Edumass
Solution:
i) Silver chromate Ag2 CrO4, Ksp of Ag2CrO4
= 1.1 × 10-12
Let the solubility of Ag2CrO4 be equal to S,
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
09
Ag2CrO4
2Ag+ + CrO
1.1 × 10-12 .................
or 1.1 × 10-12 ..............
or s3
= (2s)2 (s)
= 4s3
..........................
=
or s ............................
=
∴ Molarity of CrO
.....
Molarity of Ag+ .............
= 0.65 × 10-4 M
= 2 × 0.65 × 10-4 M
= 1.30 × 10-4 M
(ii) Barium chromate : BaCrO4
BaCrO4
Ba2+ + CrO
Ksp of BaCrO4 .................
1.2 × 10-10 ...................
or 1.2 × 10-10 ................
or s ...............................
Molarity of Ba2+ and CrO42-
= 1.2 × 10-10
=s×s
= s2
=
= 1.1 × 10-5 M
= 1.1 × 10-5 M each
Fe(OH)3
Fe3+ + 3OHKsp of Fe(OH)3 ...................
Or 1.0 × 10-38 ...................
Or s4 =
Or .................................
Molarity of OH.............
..............
1.39 × 10-10 M
1.39 × 10-10 M
3 × 1.39 × 10-10
4.17 × 10-10 M
in
Molarity of Fe3+ ...............
=
=
=
=
=
.
on
Or 1.0 × 10-38 = 3s4
= 1.0 × 10_38
= s × 3(s)3
ti
ca
(iii) Ferric hydroxide: Fe(OH)3
iv) Lead chloride : PbCl2
PbCl2
Pb2+ + 2ClKsp of PBCl2 ..................
Edumass
Edumass
= 1.6 × 10_5
32
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Edumass
Edumass
1.6 × 10-5 .....................
or s3 ...........................
or s ...........................
2s3
0.8 × 10-5
(0.8 × 10-5)1/3
(8 × 10-6)1/3
=
=
=
=
1.59 × 10 M
1.59 × 10-2
2 × 1.59 × 10-2
3 .18 × 10-2 M
-2
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
09
Molarity of Pb2+
Molarity of Cl
=
=
=
=
v) Mercurous iodide : Hg2I2
Hg2I2
2Hg2I2
Ksp of Hg2I2 .....................
4.5 × 10-29 .....................
or 4.5 × 10-29 .................
or s4 ..............................
or s
= 4.5 × 10-29
= 2s2 × 2s2
= 4s4
=
× 10-29
...............................
=
= 2.24 × 10-10 M.
Molarity of Hg22+ and I- each = 4.48 × 10-10 M.
13
62. The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10-12 and 5.0 × 10respectively. Calculate the ratio of the molarities in their saturated solution.
[Ag+(aq)] .......................
and [CrO42-(aq)] ...............
Ksp ...........................
=
=
=
=
2s1
s1
[Ag+(aq)]2 [CrO42-(aq)
(2s1)2 × s = 4s13
Similalry AgBr
Let the solubility of AgBr be s2
[Ag+(aq)] ...................
= s2 and [Be-(aq)] = s2
Ksp ....................
= [Ag(aq)] [Be-(aq)]
= s2 × s2 = s12
or s1 .........................
= (Ksp)1/2 = (5.0 × 10-13)1/2
= (0.5 × 10-12)1/2
= 0.707 × 10-6 mol L-1
Edumass
Edumass
in
=
=
= 0.65 × 10-4
= Ag(aq) + Br- (aq)
.
on
or s1 ........................
ti
ca
Solution:
Ag2CrO4 dissolves in water and the equilibrium in the saturated solution is
Ag2CrO4(s) = 2Ag+(aq) + CrO42-(aq)
Let the solubility of Ag2CrO4 be S1
33
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Edumass
Edumass
Raid of the molarities of silver chromate to silver bromide comes to = 0.65 × 10-4 ; 0.707 × 10-6 .
c
du
ne
ha 5
1
ad 3
.b 44
w 1
0
w
w 81
63. Equal volumes of 0.002 M solutions of sodium iodate and copper chlorate are mixed
together. Will it lead to precipitation of copper iodate? For copper iodate Ksp = 7.4 × 10-8.
Solution:
NaIO3
Na+ + IO3-
09
Cu2+ + 2ClO4-
Cu(ClO4)2
Cu2+ + 2IO3-
Cu(IO3)2
Molarity of NaIO3 and Cu(ClO4) each is 0.002 M.
They ionize completely
So [Cu2+] [IO3-] = 0.002 × 0.002
= 4 × 10_6
Since the products of [Cu2+] [IO3-] > KSP of Copper Iodate (7.4 × 10-8)
Copper iodate will be precipitated.
64. What is the maximum concentration of equimolar solution of ferrous sulphate and
sodium sulphide. For iron sulphide Ksp = 6.3 × 10-18.
Solution:
FeSO4 ..............................
Na2S .............................
= 6.3 × 10-18
= [x] and S2- = [x]
= 6.3 × 10-18
n.
= 6.3 × 10-18
x ................................
= (6.3 × 10-18)1/2
= 2.51 × 10-9
Hence the highest molarity for the solution is 5.02 × 10-9 M.
io
at
Fe2+ + S2- → FeS
Solubility product of iron
sulphide
Let Fe2+ ........................
[x] [x]
......................
x2 ...............................
= Fe2+ + SO42= 2Na+ + S2-
in
Edumass
Edumass
34
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