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4
Type I
:
Remember
:
(1)
EQUILIBRIUM
Chemical Equilibrium
For any hypothetical reaction…
aA + bB
cC
Ke
=
c
Kf
Kr
=
+ dD
d
[C] [D]
[A]a [B]b
If reaction will occur in homogeneous solution… Ke = Kc =
CC c C D d
CA a CB b
If reactants and products are in gaseous phase… Ke = Kp =
pCc p Dd
p Aa pBb
Similarly…
X Cc X Dd
If reactants & products are expressed in mole fraction… Kx =
X A a X Bb
(2)
Kp
If
If
If
=
∆n
∆n
∆n
∆n
Kc (RT)
= 0 then
> 0 then
< 0 then
Kp = Kc
Kp > Kc
Kp < Kc
∆n
(3)
Kp = Kx (P)
(4)
(5)
Kc = Kx (V)– n
If equilibrium constant of forward reaction is Kc, then equilibrium constant of
reverse reaction of the original reaction…Kc’ = (1/Kc).
If equilibrium constant of a reaction is Kc, and if the whole chemical reaction is
multiplied by ‘n’ then the equilibrium constant of new chemical reaction Kc’ = (Kc)n.
If equilibrium constant of two chemical equation is Kc1 and Kc2 respectively, and if
both these chemical equations are added then equilibrium constant of resultant
equation… Kc’ = Kc1× Kc2
If equilibrium constant of two chemical equation is Kc1 and Kc2 respectively, and if
one of the equation is subtracted from other then equilibrium constant of resultant
equation… Kc’ = Kc1/Kc2
Direction in which chemical reaction will occur at any stage will be calculated
using equilibrium quotient (Qc). Equilibrium quotient is calculated in a similar way
as equilibrium constant.
If Qc < Kc → Chemical reaction will occur in forward direction.
If Qc > Kc → Chemical reaction will not occur in forward direction; but it will
occur in reverse direction.
If Qc = Kc → Chemical reaction will be in equilibrium state.
(6)
(7)
(8)
(9)
1
∆
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
1.
2 HI(g) in a closed vessel at 400
The value of Kc for the reaction H2(g) + I2(g) −2
K is 4.0 × 10 . If at equilibrium, the concentration of hydrogen is 0.6 mole/lit
and that of iodine is 0.8 mole/lit., calculate the concentration of hydrogen iodide.
(Text. Book Illus. - 1 Page 98)
2 HI(g)
H2(g) + I2(g)
Equilibrium constant
2.
∴
4.0 × 10−2
=
∴
[HI]2
∴
∴
[HI]
[HI]
=
=
=
=
Initial mole
Equili. mole
Total mole
Conc. mole/lit
Kc
=
=
=
∴ Kc
+
CO(g)
1
1
1−0.4
1−0.4
1−0.4 + 1−0.4 + 0.4 + 0.4 = 2
0.6/10
0.6/10
H2(g)
+
CO2(g)
0
0.4
0
0.4
0.4/10
0.4/10
[H 2 ][CO 2 ]
[H 2 O][CO]
[0.4 / 10][0.4 / 10]
[0.6 / 10][0.6 / 10]
[0.4][0.4]
[0.6][0.6]
= 0.44
The equilibrium constant Kp of the following reaction at 700 K is 1.80 × 10−3
atmosphere. Calculate its Kc in mole/lit at the same temperature.
2 SO2(g) + O2(g)
2 SO3(g)
Kp
1.80 × 10−3
∴
Kc
Kc
2
[HI]2
[0.6][0.8]
4.0 × 10−2 × 0.6 × 0.8
192 × 10−4
13.86 × 10−2
1.386 × 10−1
When a mixture of 1 mole of water vapour and 1 mole of CO gas is heated at 725°°
K in a 10 lit container, 40% (w/w%) water undergoes the following reaction with
CO, when the reaction reaches equilibrium. Calculate the chemical equilibrium
constant. (Text. Book Illus. - 2 Page 99)
H2O(g)
3.
Kc =
[HI]2
[H 2 ][I2 ]
= Kc
× (RT)∆n
= Kc
× (0.082 × 700)1
1.80 × 10 −3
=
0.082 × 700
= 3.136 × 10−5
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
4.
Solid ammonium carbamate (NH4COONH2) dissociates into CO2(g) and NH3(g).
The total pressure of the system in the state of equilibrium is 27 atmosphere.
Calculate chemical equilibrium constant of the system.
2 NH3(g)
+
CO2(g)
NH4COONH2(s)
Initial mole
1
0
0
Euili. Mole
1−x
2x
x
Here ammonium carbamate is solid, hence it’s mole is neglected in calculation of
equilibrium constant.
Total mole
2x + x = 3x
2x
x
Mole fraction
3x
3x
Partial pressure of NH3 = Mole fraction × Total Pressure
2x
=
× 27 = 18 atm
3x
Partial pressure of CO2 = Mole fraction × Total Pressure
x
=
× 27 = 9 atm
3x
[NH 3 ] 2 [CO 2 ]
Equilibrium constant Ke
=
[NH 4 COONH 2 ]
Kp
= (pNH3)2 × pCO2
= (18)2 × 9
= 2916 (atm)3
5.
5 moles of NH3 gas was heated in a closed vessel of 200 ml capacity at a
temperature of 600 K. At equilibrium, 30% of NH3 remained unreacted. Calculate
the value of Kc for the reaction.
N2 (g)
+
3 H2 (g)
2 NH3 (g)
Initial mole
: 5
0
0
At equilibrium : 5 − 3.5
1.75
5.25
−1
Conc. Mole lit : 1.5/0.2
1.75/0.2
5.25/0.2
(Because, 30% NH3 is unreacted at equilibrium, means 70% NH3 is reacted.
Therefore 70% of 5 mole is 3.5 mole)
Thus, equilibrium constant Kc
=
[N 2 ][H 2 ]3
[NH 3 ]2
=
[1.75/0.2][5.25/0.2]3
[1.5/0.2]2
[8.75][26.25]3
[7.5]2
= 2813.67 (mole/lit)2
=
Kc
3
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
6.
At a temperature of 327°°C, if the value of Kp is 3.2 × 102 atm for the reaction
2 NH3(g) + CO2(g) , calculate the value of Kc.
NH2COONH4(s) According to the relationship between Kp and Kc…
Kp
= Kc × (RT)∆n
∴ Kc
= Kp × (RT)−∆n
= 3.2 × 102 × (0.082 × 600) −(3−0)
= 320 × (49.2)−3
320
=
[49.2]3
∴ Kc
= 2.686 × 10−3
7.
On heating solid ammonium carbamate at 400 K the value of Kp was found to be
600 atm3 at equilibrium. Calculate the total pressure in the system at
equilibrium.
2 NH3(g)
+
CO2(g)
NH4COONH2(s)
Initial mole
1
0
0
Euili. Mole
1−x
2x
x
Here ammonium carbamate is solid, hence it’s mole is neglected in calculation of
equilibrium constant.
Total mole
2x + x = 3x
2x
x
Mole fraction
3x
3x
Partial pressure of NH3 = Mole fraction × Total Pressure
2x
=
×P
(Total pressure is assumed to be P)
3x
Partial pressure of CO2 = Mole fraction × Total Pressure
x
=
×P
(Total pressure is assumed to be P)
3x
[NH 3 ] 2 [CO 2 ]
Equilibrium constant Ke
=
[NH 4 COONH 2 ]
Kp
= (pNH3)2 × pCO2
2x
x
∴ 600
= (
× P)2 ×
×P
3x
3x
4 2 1
=
×P × ×P
9
3
600×
27
∴ P3
=
4
∴ P3
= 150 × 27
∴ P
= 15.9
8.
On heating solid ammonium carbamate at 400 K the value of Kp was found to be
600 atm3 at equilibrium. Calculate the total pressure in the system at
equilibrium. Reaction : (NH4)2CO3(s) → 2 NH3(g) + CO2(g) + H2O(g)
4
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
H2(g ) + I 2(g ) 2HI (g ) at 700 K
temperature is 54.8. The concentration of HI at equilibrium is 0.5 mol lit–1.
Suppose reaction is carried by taking HI(g) find concentrations of H2(g) and I2(g)
at equilibrium. (Text. Book Illus. - 3 Page 100)
9.
The equilibrium constant of the reaction
10.
PCl5(g) can be obtained by chemical reaction between PCl3(g) and Cl2(g) in closed
vessel. The concentrations of PCl3, Cl2 and PCl5 in this reaction at 500 K are 1.59
m, 1.59 m and 1.41 m respectively, calculate equilibrium constant of this reaction.
(Text. Book Illus. - 4 Page 100)
11.
Ethyl acetate is obtained by reaction of ethanol with acetic acid in presence of H+
ions. Suppose 1 mole acetic acid and 0.18 mole ethanol are taken in this reaction.
At equilibrium 0.171 mole Ethyl acetate is obtained. Calculate equilibrium
constant of this esterification reaction. (Text. Book Illus. - 5 Page 101)
12.
The value of equilibrium constant for the reaction 2NOCl(g) 2NO(g) + Cl2(g) is
obtained as 0.033 atm at 1060 K temperature, then what will be the value of Kc for
this reaction ? (Text. Book Illus. - 6 Page 101)
13.
The value of Kc was found to be 1.6 × 103 at 127°C at equilibrium for the
following reaction. If the reaction is carried out in a closed vessel of capacity 5
liter taking 50 mole of each reactants, calculate the concentration of CO and CO2
at the equilibrium.
H2O2(g)
+ CO(g)
CO2(g) + H2O(g)
H2 O 2(g)
Initial mole
:
Equili. Mole
:
−1
Conc. Mole lit :
x/5
+
50
50−x
(50−x)/5
CO(g)
50
50−x
(50−x)/5
CO 2(g)
0
x
x/5
+
H2 O(g)
0
x
Equilibrium constant Kc for the above reaction is…
Kc
=
∴ Kc
=
=
40
∴
5
=
∴
∴
40 (50−x) =
2000
=
∴
x
=
∴
∴
x
x/5
=
=
[CO 2 ][H 2O]
[H 2 O 2 ][CO]
[x/5][x/5]
[(50 - x)5][(50 - x)/5]
x2
(50 − x) 2
x
(50 − x)
x
x + 40x
2000
41
48.78 mole
48.78/5 = 9.756 mole/lit
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
∴ 50 − x
∴ (50–x)/5
=
=
50 − 48.78 = 1.22 mole
1.22/5
= 0.244 mole/lit
14.
The value of Kc was found to be 1.6 × 103 at 127°C at equilibrium for the
following reaction. If the reaction is carried out in a closed vessel of capacity 5
liter taking 50 mole of each components (i.e. reactants and products), calculate
the concentration of CO and CO2 at the equilibrium.
+ CO(g)
CO2(g) + H2O(g)
H2O2(g)
15.
The equilibrium constant of the reaction CO(g) + H2O(g) CO2(g) + H2(g), Kc is
4.25 at 810 K. Suppose, if this reaction begins with reactants having
concentrations 0.1 mol lit–1 or M, then calculate the concentrations of CO(g),
H2O(g) , CO2(g) and H2(g) at equilibrium at 810 K. (Text. Book Illus. - 7 Page 102)
16.
The value of Kc for the reaction H2(g) + I2(g) 2 HI(g) is found to be 57.0 at 700
K temperature. Suppose in reaction mixture the concentrations of reactants
and products at some definite time are as follows : [H2] = 0.05M. [I2 ] = 0.10M
and [HI] = 0.20 M. Determine in which direction the reaction will proceed ?
(Text. Book Illus. - 8 Page 103)
17.
6.9 gram N2O4 is taken in a closed vessel of 0.5 litre volume. Temperature is 400
K. If N2O4(g) undergo following reaction… N2O4(g) 2 NO2(g) and attain
equilibrium state. At equilibrium total pressure is 9.15 bar, calculate Kc , K p
and partial pressures at equilibrium. R = 0.082 lit atm mol–1 deg–1. (Text. Book
Illus. - 9 Page 104)
18.
The value of equilibrium constant in reaction of phosphorylation of glucose
during glycolysis was obtained 3.8 × 10–3 at 298 K temperature. Calculate the
value of ∆G° for this reaction and mention your opinion about this reaction.
∆G °
= − 2.303 R T log K
= − 2.303 × 8.314 × 10−3 × 298 × log (3.8 × 10–3)
= − 2.303 × 8.314 × 10−3 × 298 ×[ log 3.8 + log 10–3]
= − 2.303 × 8.314 × 10−3 × 298 ×[ 0.5798 +(–3)]
= − 2.303 × 8.314 × 10−3 × 298 ×[ –2.4202]
∴ ∆G °
= +13.809 KJoule/mol
Since value of ∆G° is positive, the reaction will not be spontaneous.
19.
2.0 moles of PCl5 were introduced in a 2 lit container and heated at 625 K to
establish equilibrium when 60% of PCl5 was dissociated into PCl3 and Cl2. Find
the value of equilibrium constant. ( Equilibrium constant K = 0.9)
20.
At a certain temperature and total pressure of 105 Pa, iodine vapour contains 40%
by volume of I atoms. Calculate Kp for the following equilibrium.
I2(g) 2 I(g)
21.
Reaction between N2 and O2 takes place as follows.
2N2 (g) + O2 (g) 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.
6
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
22.
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) 2NH3(g) is 1.7×102. Is the reaction mixture at
equilibrium ? If not, what is the direction of the net reaction ?
23.
At 700 K, equilibrium constant for the reaction… H2(g) + I2(g) 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 ?
24.
A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After
equilibrium was attained, concentration of PCl5 was found to be 0.05 mol L–1. If
value of Kc is 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium ?
PCl5(g)
PCl3(g)
+ Cl2(g)
25.
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) Fe(s) + CO2(g)
; Kp = 0.265 atm at 1050 K
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 ?
26.
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. Calculate Kc for this reaction at the
above temperature.
C(s) + CO2(g) 2 CO(g)
27.
Calculate a) ∆G° and b) the equilibrium constant for the formation of NO2 from
NO and O2 at 298 K for the reaction…NO(g) + ½ O2(g) NO2(g). Where ∆Gf°(NO2)
= 52.0 KJ/mol, ∆Gf°(NO) = 87.0 KJ/mol.
28.
7
The equilibrium constant for the following reaction is 1.6×105 at 1024 K
H2(g)
+
Br2(g)
2 HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a
sealed container at 1024K.
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
Type II
:
Remember
:
Strong acid and Strong base
(1) Strong acid : HCl, HNO3, H2SO4
(2) Strong base : NaOH, KOH
(3) Strong acid and strong base are completely dissociate.
(4) (i)
(ii)
(iii)
(iv)
(v)
[A]
[H3O+][OH–] = Kw = 10–14
pH = – log [H3O+] = – log [H+]
pOH = – log [OH–]
pH + pOH = 14
Wt in gm
1000
Molarity (M) =
×
V in ml
Mol. wt
Where… M
V
W
m
Molarity
→
= Molarity of solution in mole/lit
= Volume of solution in ml
= Weight of solute in gm
= Molar mass of solute in gm/mole
pH / pOH
1 gm NaOH is dissolved in 100 ml solution. Calculate…(i) [OH–] , (ii) [H+]
(iii) pH and (iv) pOH of solution.
Wt in gm
1000
Ans : Molarity of NaOH solution… M =
×
V
Mol. wt.
1000
1
=
×
100
40
∴ Molarity (M) = 0.25 M
NaOH is strong base, therefore completely ionized
29.
NaOH
0.25 M
Na+
+
0.25 M
→
OH—
0.25 M
∴ [OH–] = 0.25 M
∴
pOH
∴
pOH = 0.6021
Now… [H3O+] [OH–]
∴
8
[H+]
= –log [OH–]
= – log [2.5×10–1]
= – [0.3979 + (–1)]
= 10–14
=
10 −14
OH − 
=
10 −14
[0.25]
=
100
× 10–14
25
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
∴
Now…
∴
[H+]
= 4 × 10–14 M
pH
= –log [H3O+]
= – log [4.0 × 10–14]
= – [0.6021 – 14]
pH
= 13.3979
30.
Calculate [H+], [OH–] pH & pOH of 0.06 M H2SO4 solution.
Ans : H2SO4 is strong acid, therefore it is completely ionized
∴ H2SO4
→
0.06 M
∴
2H+ +
SO42–
2(0.06)
0.06 M
[H+]
Now, [H3O+] [OH–]
[OH–]
∴
pH
∴
= 0.12 M
= 10–14
=
10 −14
H 3O + 
=
10 −14
[0.12 ]
= 8.33 × 10–14
= – log [H3O+]
= – log [0.12]
= – [ 1 .0792]
= 1.0 – 0.0792
∴ pH
=
0.9208
∴
pOH
= 14 – 0.9208
∴
pOH
= 13.0792
31.
Calculate [H+], [OH–], pH & pOH of 0.002 M HNO3 solution.
32.
Calculate [H+], [OH–], pH & pOH of an aq. solution of 0.08 M H2SO4 solution.
33.
(3.65 × 10–2) gm Hydrochloric acid is dissolved in 500 ml of solution. Find out of
its pH & pOH.
34.
In one drink, the concentration of hydrogen ion is found to be 4 × 10–3. What
will be the value of its pH ? Also calculate the value of pOH. (Text. Book Illus. 11 Page 112)
35.
Calculate pH and pOH of solution containing 0.03 M NaOH. (Text. Book Illus. 12 Page 112)
36.
Calculate pH and pOH of following solutions. (Text. Book Exe – 4 (1) Page 126)
(a) 0.1 M HCl, 0.1 M H2SO4, 0.1 M HNO3
(b) 0.1 M NaOH, 0.1 M KOH, 0.1 M Ba(OH)2
37.
Calculate pH of following solutions. (Text. Book Exe – 4 (2) Page 126)
(a) 3.65 gram HCl in 250 ml solution
(b) 9.80 gram H2SO4 in 500 ml solution
(c) 1.6 gram NaOH in 250 ml solution
(d) 11.2 gram KOH in 500 ml solution
9
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
[B]
pH / pOH →
Molarity
38.
Calculate weight of H2SO4 required to prepare 100 ml aq. solution having 13.6
pOH.
Ans :
pOH
= 13.6
∴
pH
= 0.4
∴ – log [H3O+] = 0.4
log [H3O+]
∴
=
=
– 0.0 – 1.0 + 1.0 – 0.4
1 .6
∴
[H3O+]
= antilog ( 1 .6)
∴
[H3O+]
= 3.981 ×10–1
∴
[H3O+]
= 0.3981 M
H2SO4 is strong electrolyte, therefore it is completely ionized.
H2SO4
∴
?
0.3981
=
2
∴ [H2 SO4]
→
2 H+
←
0.3981
=
0.3981
2
+ SO42–
= 0.19905 M
Now,
Molarity of H2SO4 solution… M
0.3981
2
39.
:
Remember
:
(2)
10
Wt in gm
∴
Wt in gm
=
Wt in gm
1000
×
V
Mol. wt.
Wt in gm
1000
×
100
98
0.3981
100
×
× 98
2
1000
100
= 0.19905 ×
× 98
1000
= 1.9507 gm
=
Calculate quantity of NaOH required to prepare 100 ml aq. solution having pH = 10.
(Ans : [OH–] = 10–4 M, [NaOH] = 10–4 M , W = 4.0 × 10–4 gm )
Type III
(1)
∴
=
pH and pOH of diluted solution
When any acid solution is diluted by adding water, [H3O+] ion decreases but pH of
solution increases.
When any base solution is diluted by adding water, [OH–] ion decreases hence
pOH of solution increases; but pH decreases.
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
40.
10 ml 0.1 M NaOH solution is diluted to 50 ml by adding water. (1) Calculate pH of
original solution (2) pH of diluted solution (3) Change in pH.
Ans. pH of original solution :
NaOH is strong electrolyte, therefore completely ionized.
NaOH
0.1
∴
→
Na+
0.1
+ OH–
0.1
∴
pOH
= – log [OH–]
∴
pOH
= – log [0.1]
= – [–1.0]
∴
pOH
= 1
∴
pH
= 13.0
pH of diluted solution :
M1 V1
= M2 V2
0.1 × 10
M2
∴
M2
= M2 × 50
0.1 × 10
=
50
= 0.02 M
Now, NaOH is strong electrolyte.
NaOH
0.02
∴
→
pOH =
=
=
=
∴
pH
∴
∴€Change in pH
Na+
0.02
+ OH–
0.02
– log [OH–]
– log [0.02]
– log [2.0 × 10–2]
– [0.3010 – 2.0]
= 12 . 3010
= 13.0 – 12.3010 = 0.6990
41.
What would be the change in pH of 0.02 M H2SO4, which is diluted to 100 times.
(Ans. Change in pH = 1.3979 – 3.3979 = 2)
42.
Calculate pH of solution containing 3.65 gm HCl per lit. How can you increase pH
of this solution by 1 unit using water.
43.
Calculate pH of solution containing 4.9 gm H2SO4 per lit. How much water should
be added to increase pH of this solution by 3 unit ?
Type IV
:
pH and pOH of Highly diluted solution
44.
Calculate pH of solution prepared by diluting 1 ml 0.1 M HCl solution to 10000 lit.
Ans :
1 ml = 10–3 lit
10000 lit = 104 lit
11
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
Molarity of HCl solution after dilution :
M1V1
=
M2V2
0.1 × 10–3
=
M2 × 104
0.1 × 10 −3
= 10–8 M
10 4
HCl is strong electrolyte, hence it will completely ionized.
M2
∴
=
→
H+(aq)
+
Cl–(aq)
HCl
10–8 M
10–8 M
This concentration of H+ ion (i.e. [H3O+]) is very less, hence [H3O+] ion produced due
to self ionization of water should be also considered.
→
H3O+(aq)
2 H2O
According to self ionization of water…
[H3O+]
∴
=
[OH–]
+
OH–(aq)
[H3O+][OH–] = 1.0 × 10–14
= 1.0 × 10–7M
∴
Total [H3O+]
=
=
(1.0 × 10–8) + (1.0 × 10–7)
(1.0 × 10–8) + (10.0 × 10–8)
∴
Total [H3O+]
=
=
11.0 × 10–8 M
1.1 × 10–7 M
∴ pH
=
=
=
– log [H3O+]
– log [1.1 × 10–7]
– [0.0414 – 7.0000]
∴ pH
=
6.9586
Therefore,
45.
Calculate pH of solution prepared by diluting 1ml 0.02 M NaOH solution to 10000
lit.
46.
0.1 ml of 0.001 M HCl solution is diluted with water to make 10 litre solutions.
Calculate pH of this dilute solution. (Text. Book Illus. - 13 Page 113)
Type V
:
pH and pOH of mixture of Strong acid and Strong base
47.
Find out the pH of a mixture of 50 ml 0.03 M HCl and 60 ml of 0.02 M NaOH.
Ans : Mole of H3O+ before mixing :
∴
1000 ml solution contains
→
0.03 mole HCl
50 ml solution contains
→
?
50 × 0.03
1000
= 1.5 × 10–3 mole HCl
=
Now, HCl is strong electrolyte…
HCl
→
1.5 × 10–3
12
H+
+
Cl–
1.5 ×10–3
1.5 ×10–3
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
Mole of OH– before mixing :
∴
1000 ml solution contains
→
0.02 mole NaOH
60 ml solution contains
→
?
60 × 0.02
1000
= 1.2 × 10–3 mole HCl
=
Now, NaOH is strong electrolyte…
→
Na+ +
OH–
NaOH
1.2 × 10–3
1.2 ×10–3
1.2 ×10–3
Concentration of H3O+ OR OH– ion in Mixture :
When acid and base are mixed, neutralization takes place according to following
reaction. According to given reaction, 1 mole of OH– is neutralized by 1 mole H+ ion.
Hence 1.2 × 10–3 mole OH– ion is neutralized by 1.2 × 10–3 mole H+ ion, when NaOH is
mixed with HCl.
+
H+(aq)
1 mole
1.5 ×10–3
OH–(aq) →
1 mole
1.2 ×10–3
H2O(1)
1 mole
1.2 ×10–3
∴ Un neutralized mole of H+ = (1.5 × 10–3) – (1.2 × 10–3)
= (0.3 × 10–3)
= 3.0 × 10–4 mole H+ ion
Now,Total volume of solution = (50 + 60) ml = 110 ml
∴ 110 ml mixture of acid and base contains
→
3.0 × 10–4 mole
∴ 1000 ml mixture of acid and base contains
→
?
1000 × 3.0 × 10 −4
110
= 2.727 × 10–3 mole/lit
=
OR
Molarity =
Mole
lit
=
3 × 10 −4
0.110
= 2.727 × 10–3 mole/lit
pH of Mixture :
∴
pH
= – log [H3O+]
= – log [2.727 × 10–3]
= – [0.4357 – 3.0]
∴
pH
= 3 – 0.4357
∴
pH
= 2.5643
48.
Calculate pH of a mixture of solution containing 24 ml of 0.1 M HCl and 25 ml 0.1
M NaOH.
49.
If 20 ml 0.15 M NaOH solution is added to a 10 ml 0.1 M H2SO4 solution, what
will be the pH of the resulting mixture ? (Text. Book Exe – 4 (13) Page 127)
13
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50.
What would be pH of solution when following solution are added to 10.0 ml 0.1M
HCl solution ?
(a) 5.0 ml 0.1 M NaOH (pH = 1.4771)
(b) 9.0 ml 0.1 M NaOH (pH = 2.2788)
(c) 9.9 ml 0.1 M NaOH (pH = ? )
(d) 10.0 ml 0.1 M NaOH (pH = 7.0)
(e) 10.1 ml 0.1 M NaOH (pH = ? )
(f) 10.2 ml 0.1 M NaOH (pH = 10.9956)
51.
If 20 ml 0.15 M HCl solution is added to 10 ml 0.1 M Ba(OH)2 solution what will
be the pH of the resulting mixture ? (Text. Book Exe – 4 (14) Page 127)
52.
Calculate the pH of the resultant mixtures.
(a) 10 mL of 0.2 M Ca(OH)2 + 25 mL of 0.1 M HCl
(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
Type VI
:
Remember
:
(1)
(2)
(3)
Weak acid and Weak base
Weak acid : CH3COOH, HCOOH, C6H5COOH, C6H5OH etc.
Weak base : NH3, NH4OH, CH3 – NH2, C6H5 NH2 etc.
For any weak acid HA…
HA(aq) + H2O(l)
∴
Ka
=
∴ [H3O+]
=
∴
=
α
[H3 O + ][A − ]
=
[HA]
αC
=
H3O+(aq) + A–(aq)
[H3 O + ]2
=
C
[αC][αC ]
=
[1 − α ]C
[α 2 C]
[1 − α ]
Ka C
Ka
C
Wt in gm
1000
×
V in ml
Mol. wt
C = Molarity of weak acid solution in mole/lit
Where… C = M =
α = Degree of dissociation
Ka = Dissociation constant
(4)
For any weak base MOH…
MOH (aq) + H2O(l)
Kb
=
∴ [OH–]
=
∴
=
∴
14
α
[M + ][OH − ]
=
[MOH]
αC
=
M+(aq) + OH–(aq)
[OH − ]2
C
=
[αC][αC ]
[1 − α ]C
=
[α 2 C]
[1 − α ]
Kb C
Kb
C
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Wt in gm
1000
×
V in ml
Mol. wt
C = Molarity of weak base solution in mole/lit
Where… C = M =
α = Degree of dissociation
Kb = Dissociation constant
pH
= – log [H3O+] = – log [H+]
pOH = – log [OH–]
(5)
53.
If the value of ionization constant Ka = 1.28 × 10– 10 for an aq. solution of 0.05 M
C6H5OH. Calculate the concn. of H3O+ and pH of solution.
Ans :
C6H5OH + H2O
[H3O+]
54.
=
Ka C
=
1.28 × 10 −10 × 0.05
=
0.064 × 10–10
=
6.4 × 10–12
=
6.4 × 10–6
∴
[H3O+]
∴
pH
=
=
=
=
∴
pH
= 5.5969
C6H5O–
+ H3O+
= 2.5298 × 10–6 M
– log [H3O+]
– log [2.5298 × 10–6]
– [0.4031 – 6.0]
6.0 – 0.4031
Find out pH of 0.02M methyl amine solution Ionization constant of methyl
amine is 5.0 × 10–4.
CH3 – NH2 +
∴
[OH–]
H2O
=
Kb C
=
5.0 × 10–4 × 0.02
=
0.1 × 10 −4
=
10 × 10 −6
=
10 × 10 −3
∴
[OH–]
= 3.1623 × 10–3 M
∴
pOH
= – log [OH–]
= – [0.5000 – 3.0000]
= 2.5000
pOH
∴
15
pH
CH3 – NH3+ + OH–
= 11.5
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
55.
At equilibrium, 0.02 M CH3–NH2 ionizes to
ionization constant.
Ans :
CH3 NH2
Initial mole
At equili
∴ [OH–]
+
CH3 – N+H3
+
OH–
0.02
0.02 – 0.0003
0
0
= 0.0197 ≈ 0.02
0.0003
0.0003
= 0.0003 M = 3.0 × 10–4 M
[OH − ]2
C
Kb
=
∴
Kb
 3.0 × 10 −4 
=
0.02
= 4.5 × 10–6
OR
α
= 1.5 % =
∴
Kb
=
[α 2 C]
[1 − α ]
=
[(0.015)2 × 0.02]
[1 − 0.015]
∴
H2O
1.5 % Calculate the value of its
2
1. 5
100
= (0.015)2 × 0.02
= 0.015
But… α < < 1
∴ 1–α ≈ 1
= 4.5 × 10–6
56.
If 0.02 M CH3COOH ionizes up to 2% at 25 ° C. Calculate its ionization constant.
57.
10 lit. of water contains 1.7 g NH3. Calculate its pOH Kb = 1.77 × 10–5.
58.
Calculate ionization constant of the acid when a solution of 0.043 M of HOOH has a
pH = 4.37
59.
NH3 + H2O NH4+ + OH– equilibrium exists in aqueous solution of ammonia.
The equilibrium constant Kb of NH3 is 1.77 × 10–5. Calculate the concn. of NH3,
NH4+, OH– and H3O+ in 0.01 M solution of ammonia.
(Ans : [OH–] = 4.207 × 10–4 M,
[NH4+] = 4.207 × 10–4 M, [NH3] = 9.5793 × 10–3 M,
+
–11
[H3O ] = 2.378 × 10 M)
60.
Calculate [H3O+], [CH3COOH], pH, pOH and pKw of 0.1 M CH3-COOH solution.
Ka of acetic acid is 1.74 × 10–5. (Text. Book Illus. - Page 114)
61.
pH of 0.1 M monobasic acid HA is 4.40. Calculate the value of [H3O+], [A–], [HA],
Ka and pKa in the solution at equilibrium.
62.
Calculate the values of [NH4+], [OH–], pH and pOH of 0.1 M NH4OH solution at
equilibrium. The ionization constant of NH4OH is 1.8 × 10–5. (Text. Book Illus. - 14
Page 115)
63.
The ionization constant of dimethyl amine [(CH3)2NH] is 5.4 × 10–4. If the initial
concentration of dimethyl amine is 0.02 M, calculate [OH–], [(CH3)2NH], [H3O+],
pH and pOH in solution at equilibrium. (Text. Book Illus. - 15 Page 116)
64.
Calculate pH of 250 ml solution containing 6 gram acetic acid. Dissociation
constant of acetic acid is 1.8 × 10–5. (Text. Book Exe – 4 (3) Page 126)
16
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65.
Calculate pH of 500 ml ammonium hydroxide solution containing 7 gram
ammonium hydroxide. Dissociation constant of ammonium hydroxide is 1.8 ×
10–5. (Text. Book Exe – 4 (4) Page 126)
66.
The dissociation constant of benzoic acid is 6.5 × 10–5 at 298 K. What will be the
pH of its 0.15 M solution and its [H3+O] ion concentration ? (Text. Book Exe – 4 (5)
Page 126)
67.
The dissociation constant of dimethyl amine is 5.4 ×10–5 at 298 K. What will be
the pH of its 0.25 M solution and [H3O+] ion concentration ? (Text. Book Exe –
4 (6) Page 127)
68.
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.
69.
It has been found that the pH of a 0.01M solution of an organic acid is 4.15.
Calculate the concentration of the anion, the ionization constant of the acid and
its pKa.
70.
The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate
the pH of the solution and the pKa of bromoacetic acid.
71.
The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization
constant and pKb.
Type VII
:
Remember
:
(1)
(2)
Relation between Ka and Kb
Ka × Kb
= Kw
Where … Ka = Dissociation constant of acid OR conjugate acid
Kb = Dissociation constant of conjugate base OR base
pKa + pKb = pKw = 14
The ionization constant of acetic acid is 1.7 × 10–5. Calculate ionization constant
of its conjugate base ?Also calculate pKa and pKb of acid and its conjugate base.
Ans :
CH3-COOH + H2O CH3-COO– + H3O+
Acid
Base
con. base
con. acid
Ka × Kb = Kw
Where … Ka = Dissociation constant of acid
Kb = Dissociation constant of conjugate base of acid
Kw
1.0 × 10−14
∴
Kb
=
=
= 5.882 × 10–10
Ka
1.7 × 10−5
Now…
pKa = – log Ka
= – log (1.7×10–5)
= – [ log 1.7 + (–5)]
= – [0.2304 – 5]
= 5 – 0.2304
= 4.7696
72.
17
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Now, relation between pKa and pKb of acid and its conjugate base (vice versa) is …
pKa + pKb
= 14
∴
pKb
= 14 – pKa = 9.2304
73.
The ionization constant of formic acid (HCOOH) is 1.8 × 10–4. Calculate
ionization constant of its conjugate base format ion (HCOO–) ?Also calculate pKa
and pKb of acid and its conjugate base. (Text. Book Illus. – 16 Page 116)
74.
The ionization constant of NH3 is 1.8 × 10–5. Calculate ionization constant of its
conjugate acid ?Also calculate pKa and pKb of base and its conjugate acid. (Text.
Book Illus. - Page 116)
75.
If dissociation constant of aniline is 4.27 × 10–10 at 298 K, what will be the
dissociation constant of its conjugate acid ? Kw = 1.04 × 10–14. (Text. Book Exe – 4
(15) Page 127)
76.
If dissociation constant of acetic acid is 1.76 × 10–5 at 298 K, what will be
dissociation constant of its conjugate base ? Kw = 1.04 × 10–14. (Text. Book Exe –
4 (16) Page 127)
77.
The ionization constant of HF, HCOOH and HCN at 298K are 6.8×10–4, 1.8×10–4
and 4.8×10–9 respectively. Calculate the ionization constants of the corresponding
conjugate base.
78.
What is the pH of 0.001 M aniline solution ? The ionization constant of aniline
4.27 × 10–10. Calculate the degree of ionization of aniline in the solution. Also
calculate the ionization constant of the conjugate acid of aniline.
Type VIII
:
Common ion Effect on Dissociation constant of Weak
acid and Weak base
79.
What will be the change in pH of 0.1 M CH3COOH acid if 0.1 M CH3COONa is
added to its solution ? (pKa of CH3COOH = 4.74) (Text. Book Illus. – 17 Page 118)
Ans : pH of solution in absence of common ion :
pKa = 4.74
∴
– log Ka = 4.74
log Ka = –4.74
= –4.0 – 0.74
= –4.0 –1.0 + 1.0 –0.74
= –5.0 + 0.26
∴
Ka
= antilog ( 5 .26)
= 1.82 × 10–5
Now,
[H3O+] =
Ka C
∴
18
=
1.82 × 10 −5 × 0.1
=
1.82 × 10–3
= 1.349 × 10–3
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
= – log [H3O+]
= – log [1.349 × 10–3]
= – [ log 1.349 + (–3)]
= – [0.1300 – 3]
= 2.87
_____(1)
Now, if 0.1 M CH3COONa is added then it is strong electrolyte.
∴
CH3COONa
→
CH3COO– + Na+
0.1
0.1
0.1
While CH3COOH is weak electrolyte and since value of Ka is very small, degree of
dissociation is very less. So after addition of 0.1 M CH3COONa in aq. solution of
CH3COOH its equilibrium will be as follows.
∴
CH3COOH + H2O CH3COO–
+ H+
Initial mol
0.1
0.1
0
Equi. mol
0.1 – x
0.1+x
x
≈ 0.1
≈ 0.1
Therefore, Ka of CH3COOH will be…
[CH 3COO − ][H3O + ]
∴
Ka
=
[CH 3COOH]
∴
pH
[0.1][H 3O + ]
[0.1]
+
∴
[H3O ]
= 1.82 × 10–5
∴
pH
= – log [1.82 × 10–5 ]
= – [log 1.82 + (–5)]
= – [0.2601 – 5]
= 5 – 0.2601
= 4.74
_____(2)
From (1) and (2) it is clear that pH of solution is increased from 2.87 to 4.74; hence
concentration of [H3O+] decreases. Thus, in presence of common ion dissociation of
weak acid decreases.
1.82 × 10–5
=
80.
What will be the change in pH of 0.1 M weak base NH4OH if 0.1 M NH4Cl is
added to the solution ? (Kb of NH4OH = 1.77 × 10–5) (Text. Book Illus. – 18 Page
118)
81.
The ionization constant of phenol is 1.0×10–10. What is the concentration of
phenolate ion in 0.05 M solution of phenol ? What will be its degree of
ionization if the solution is also 0.01 M in sodium phenolate solution ?
82.
The first ionization constant of H2S is 9.1×10–8. Calculate the concentration of
HS– ion in its 0.1M solution. How will this concentration be affected if the
solution is in 0.1 M HCl solution ? If the second dissociation constant of H2S is
1.2×10–13, calculate the concentration of S2– under both conditions.
83.
Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74.
How is the degree of dissociation affected when its solution also contains…
(a) 0.01 M (b) 0.1 M HCl ?
19
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84.
The ionization constant of dimethyl amine is 5.4 × 10–4. Calculate its degree of
ionization in its 0.02 M solution. What percentage of dimethyl amine is ionized
if the solution is also 0.1 M in NaOH ?
85.
The ionization constant of propanoic 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 in 0.01 M HCl solution also ?
Type IX
:
Remember
:
(1)
For a salt of Strong base and Weak acid : (e. g. CH3COONa. HCOOK, C6H5COONa,
Na2CO3, K2CO3 etc.
Kh
(2)
Hydrolysis of Salt
K
= w
Ka
=
OH– 
C
2
For a salt of Weak base and Strong acid : (e. g. NH4Cl, C6H5NH3+Cl– etc )
2
86.
H+ 
K
Kh = w =  
Kb
Co
Find out pH of a solution of 0.025 M CH3COONa. Ionization constant of
CH3COOH is 1.75 × 10–5. Also calculate Hydrolysis constant of CH3COONa.
Ans :
∴
Kh
=
Kw
Ka
Kh
=
1.0 × 10 −14
1.75 × 10 −5
=
1.0 × 10 −14
1.75 × 10 −5
OH – 
∴ 
0.025
20
2
=
OH– 
Co
2
= 5.714 × 10–10
1.0 × 10 −14
× 0.025
1.75 × 10 −5
= 0.571 × 10–9 × 0.025
= 0.0143 × 10–9
∴
[OH–]2 =
∴
[OH–]2 = 1.43 × 10–11
∴
[OH–]
=
1.43 × 10 -11
=
1.43 × 10–6
∴
[OH–]
= 3.7815 × 10–6 M
∴
pOH
= – log [OH–]
= – log [3.7815 × 10–6]
= – [0.5777 – 6.000]
∴
pOH
= 6.0 – 0.5777
= 5.4223
∴
pH
= 8.5777
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
87.
Calculate pH of an aq. solution of 0.02 M CH3COONa. Ionization constant of
CH3COOH is 1.75 × 10–5.
88.
Calculate pH of an aq. solution of 0.1 M Sodium acetate solution. Ionization
constant of CH3COOH is 1.8 × 10–5. (Text. Book Illus. – 19 Page 119)
89.
Calculate hydrolysis constant and pH of a solution of 0.1 M sodium acetate.
Dissociation constant of acetic acid is 1.8 × 10–5 at 298 K and ionic product of
water is 1.04 × 10–14. (Text. Book Exe – 4 (11) Page 127)
90.
Calculate hydrolysis constant and pH of 0.30 M NH4Cl solution. The dissociation
constant and ionic product of water are 1.8 × 10–5 and 1.04 × 10–14 respectively at
298 K. (Text. Book Exe – 4 (12) Page 127)
91.
Ionization constant of weak base NH4OH is 1.75 × 10–5. Calculate pH of 0.175 M
NH4Cl solution. Also calculate Kh. (Ans : pH = 5, Kh = 5.714 × 10–10)
92.
The ionization 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.
93.
A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the
ionization constant of pyridine.
94.
The ionization constant of chloro acetic acid is 1.35×10–3. What will be the pH of
0.1 M acid and its 0.1M sodium salt solution ?
Type X
:
Remember
:
Buffer solution
(1)
pH
= pKa
+ log
(2)
pOH = pKb
+ log
[ Salt ]
[Acid]
[ Salt ]
[Base]
95.
To prepare one acidic buffer solution 0.125 M sodium acetate is added to 0.25 M
solution of acetic acid. If dissociation constant of acetic acid is 1.8 × 10–5 , what
will be the pH of this buffer solution ? (Text. Book Exe – 4 (7) Page 127)
[ Salt ]
Ans :
pH
= pKa + log
[Acid]
[ Salt ]
∴
= – log Ka + log
[Acid]
[0.125]
= – log (1.8 × 10–5)
+ log
[0.25]
= – [log 1.8 + (–5)]
+ log ( ½ )
= – [0.2553 – 5] + (log 1 – log 2)
= 5 – 0.2553 + 0 – 0.3010
pH
= 4.4437
21
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
96.
To prepare one basic buffer solution, 0.250 M NH4Cl is added to a solution of
0.125 M ammonium hydroxide. If dissociation constant of ammonium hydroxide
is 1.8 × 10–5, what will be the pH of this buffer solution ? (Text. Book Exe – 4 (8)
Page 127)
Type XI
:
Remember
:
Solubility product of Sparingly soluble salts
(1)
Type of Salt
1:1
Solubility Product
Ksp = S2 (mol/lit)2
1:2
2:1
1:3
3:1
2:3
3:2
Ksp = 4 S3 (mole/lit)3
Some example
AgCl, AgI, HgS, ZnS, BaSO4,
CaSO4, PbSO4
CaF2, BaF2, PbCl2
Ag2S, Ag2CrO4
Al(OH)3, Cr(OH)3
Ag3PO4
Bi2S3, As2S3, Sb2S3
Zr3(PO4)2
Ksp = 27 S4 (mole/lit)4
Ksp = 108 S5 (mole/lit)5
Wt in gm
1000
×
V in ml
Mol. wt
Solubility in gm/lit = S × Mol. Wt
Solubility in mg/lit = S × Mol. Wt. × 103
(2)
Solubility (mol/lit) = S =
97.
Find out solubility in water of PbSO4 in g/L whose solubility product is
1.3 × 10–8. Mol. wt of PbSO4 is 303 gm/mole. Also calculate (1) volume of solution
containing 25 mg PbSO4. (2) Wt of PbSO4 dissolved in 234 ml.
Ans :
Ksp
=
∴
S
=
K sp
=
1.3 × 10 −8
=
1.3 × 10–4
S
∴ Solubility in g/L
OR
S
∴ 1.1402 × 10–4
∴ Wt in gm
SO42– (aq)
+
S
S2
∴
∴
22
Pb2+(aq)
S
PbSO4(s) S
= 1.1402 × 10–4 mole/L
= 1.1402 × 10–4 × 303
= 345.47 × 10–4 g/L
= 3.4547 × 10–2 g/L
Wt in gm
1000
×
V in ml
Mol. wt
1000 Wt in gm
=
×
1000
303
= 1.1402 × 10–4 × 303 = 3.4547 × 10–2 g/L
=
e-CHEMTEST / CH-4 - Numericals / Sem-II / Ph : 30004433
(1)
S
=
∴ 1.1402 × 10–4 =
∴
(2)
V
=
S
=
∴ 1.1402 × 10–4 =
Wt in gm =
=
∴
Wt in gm
1000
×
V in ml
Mol. wt
1000
0.025
×
V in ml
303
723.63 ml
Wt in gm
1000
×
V in ml
Mol. wt
Wt in gm
1000
×
234
303
0.00808
8.08 mg
The concentration of a saturated solution of Mg(OH)2 is 8.2 × 10–4 % (W/v).
Calculate its Ksp (Molecular weight of Mg(OH)2 is 58 g/mol)
Ans : Solubility of Mg(OH)2 = 8.2 × 10–4 % (w/v)
= 8.2 × 10–4 gm/100 ml
98.
∴ Solubility in gm/lit = 8.2 × 10–3 gm/1000 ml
8.2 × 10–3
= 0.1414 × 10–3
58
S = 1.414 × 10–4 mole/lit
Wt in gm
1000
S =
×
V in ml
Mol. wt
∴ Solubility in mol/lit =
∴
OR
∴
S =
1000
8.2 × 10 −4
×
100
58
Mg(OH)2
Mg2+(aq)
Now,
S
∴
S
Ksp
Ksp
=
=
=
=
+ 2 OH–(aq)
(1 : 2 type)
2S
S3
4
4 × (1.414 × 10–4)3
11.3086 × 10–12
1.1309 × 10–11 (mole/lit)3
99.
The concentration of Mg(OH)2 in its saturated solution is found to be 8.2 × 10–4
gm lit–1 at 298 K temperature. Calculate Ksp of Mg(OH)2. (Mol. wt of Mg(OH)2 is
58 gm/mol) (Text. Book Illus. – 20 Page 119)
100.
Ksp of Ag2S is 1.0 × 10–51. Molecular weight is 24.8 gm/mole. Calculate solubility
of Ag2S in mg/lit. Calculate solubility of Ag2S in mg in 400 ml solution.
101.
A person drinks 2.5 lit waters every day, which is saturated with CaF2. How
much CaF2 enters in the body every day ? (Mol. wt of CaF2 is 78 gm/mole and Ksp
of CaF2 is 1.7 × 10–10) (Ans : S = 3.489 × 10–4 M and 68.03 mg CaF2 enters in body
every day)
23
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102.
Solubility product of CaF2 is 1.7 × 10–10. Calculate its solubility in water.
Calculate vol. of saturated soln. containing 10 mg CaF2. Calculate wt of CaF2
present in 786 ml.
103.
Calculate Ksp of Fe(OH)3 whose solubility is 1.0 × 10–3 M.
104.
The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the
concentrations of strontium and hydroxyl ions and the pH of the solution.
105.
The solubility product constant of Ag2CrO4 and AgBr are 1.1×10–12 and 5.0×10–13
respectively. Calculate the ratio of the molarities of their saturated solutions.
106.
What is the minimum volume of water required to dissolve 1 g of calcium
sulphate at 298 K ? (For calcium sulphate, Ksp is 9.1×10–6)
Type XII
:
Remember
:
Precipitations
(1)
IP = Ksp →
Solution is exactly saturated
→ No pptn.
(2)
IP > Ksp →
Solution is unsaturated
→ No. pptn.
(3)
IP > Ksp →
Solution concn is more than saturation → pptn. will occur.
107.
Will pptn takes place or not on mixing equal volumes of 2.0 × 10–4 M BaCl2 and
5.0 × 10–3 M H2SO4 ? For BaSO4 Ksp = 1.1 × 10–10.
Whose precipitates will be obtained on addition of equal volumes of 2.0 × 10–2 M
H2SO4 to water solutions containing 2.0 × 10–3 M CaCl2 and 5.0 × 10–3 M BaCl2 ? The
solubility product of CaSO4 and BaSO4 are respectively 2.4 × 10–5 and 1.1 × 10–10 ?
Ans : Calculation for CaSO4.
108.
CaCl2
2 x 10-3
V ml
H2SO4
2 x 10-2
V ml
2 V ml
CaCl2 + H2SO4
→ CaSO4 + 2 HCl
[SO42–] after dilution :
M1 V1
=
M2 V2
[Ca2+] after dilution :
M1 V1
=
M2 V2
∴ 2.0 × 10–3 × V =
M2 × 2V
∴ 2.0 × 10–2 × V =
M2 × 2V
∴
M2
=
1.0 × 10–3 M
∴
M2
=
1.0 × 10–2 M
∴
[CaCl2]
=
1.0 × 10–3 M
∴
[H2SO4]
=
1.0 × 10–2 M
∴
H2SO4 →
1.0 × 10–2
∴
∴
24
CaCl2 → Ca2+(aq)
1.0 × 10–3
1.0 × 10–3
[Ca2+]
=
+ 2 Cl–(aq)
1.0 × 10–3
∴
[SO42–]
=
2 H+(aq)
+ SO42–(aq)
1.0 × 10–2
1.0 × 10–2
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∴
CaSO4
IP
IP
But,
Ca2+
+ SO4
2–
(1 : 1 type)
= [Ca2+] [SO42–]
= [1.0 × 10–3] [1.0 × 10–2]
= 1.0 × 10–5
Ksp of CaSO4 is 2.4 × 10–5
∴ pptn will not occur.
∴ IP < Ksp
H2SO4
2 x 10-2
V ml
BaCl2
5 x 10-3
V ml
2 V ml
[Ba2+] after dilution :
M1 V1
=
BaCl2 + H2SO4 → BaSO4 + 2 HCl
[SO42–] after dilution :
M2 V2
M1 V1
=
M2 V2
∴ 5.0 × 10–3 × V =
M2 × 2V
∴ 2.0 × 10–2 × V =
M2 × 2V
∴
M2
=
2.5 × 10–3 M
∴
M2
=
1.0 × 10–2 M
∴
[BaCl2]
=
2.5 × 10–3 M
∴
[H2SO4]
=
1.0 × 10–2 M
∴
H2SO4 →
1.0 × 10–2
∴
∴
BaCl2 → Ba2+(aq)
2.5 × 10–3
2.5 × 10–3
[Ba2+]
∴
=
BaSO4
IP
IP
+ 2 Cl–(aq)
2.5 × 10–3
∴
Ba2+
+ SO4
[SO42–]
2–
=
2 H+(aq)
+ SO42–(aq)
1.0 × 10–2
1.0 × 10–2
(1 : 1 type)
= [Ba2+] [SO42–]
= [2.5 × 10–3] [1.0 × 10–2]
= 2.5 × 10–5
But, Ksp of BaSO4 is 1.1 × 10–10
∴ IP > Ksp
∴ pptn will occur.
∴ BaSO4 will be precipitate but CaSO4 will not be precipitate.
109.
200 ml 1.0 × 10–3 M Pb(NO3)2 solution is mixed with 100 ml 3.0 × 10–3 M NaCl
solution. pptn of PbCl2 will occur or not ? (Ksp of PbCl2 = 2.0 × 10–4)
(Ans : IP = 6.667 × 10–10, Therefore precipitation will not occur )
110.
Two solutions of (1) 100 ml 2.0 × 10–3 M Pb(NO3)2 and (2) 100 ml 5.0 × 10–3 M H2SO4
are mixed will there be precipitation of PbSO4 ? Ksp of PbSO4 is 1.3 × 10–8.
111.
20 ml 0.04 M NaCl solution is added to 10 ml 0.06 M AgNO3 solution. If
solubility product of AgCl is 1.8 × 10–10 at 298 K, predict whether precipitation
will occur or not ? (Text. Book Exe – 4 (9) Page 127)
112.
20 ml 0.025 M K2CrO4 solution is added to 20 ml 0.05 M BaCl2 solution. If
solubility product of BaCrO4 is 1.2 × 10–10 at 298 K, predict whether precipitation
will occur or not ? (Text. Book Exe – 4 (10) Page 127)
25
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113.
Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are
mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate
Ksp = 7.4×10–8 ).
114.
250 ml solution contains 3.32 g Pb(NO3)2 ; will addition of 174 mg solid K2SO4 to this
solution will precipitate PbSO4 ? (Ksp of PbSO4 is 1.3 × 10–8, Mol. wt of K2SO4 is 174
gm/mole and Pb(NO3)2 is 332 gm/mol)
(Ans : IP = 1.6 × 10–4 , IP > Ksp – pptn will occur)
115.
How many grams FeSO4 should be added to 500 ml of 0.02 M NaOH solution to
precipitate Fe(OH)2 ? Ksp = 1.5 × 10–15 [Mol. wt of FeSO4 = 152 gm/mol]
Ans :
Fe(OH)2 Fe2+(aq) + 2OH–(aq)
(1 : 2 type)
S
S
S
= [Fe2+] [OH–]2
K sp
∴ [Fe2+] =
2
OH– 
∴ Ksp
=
1.5 × 10 −15
[0.02 ]2
=
1.5 × 10 −15
4.0 × 10 −4
∴ [Fe2+]
= 3.75 × 10–12
∴ [FeSO4]
= 3.75 × 10–12
∴ [Fe2+]
= [FeSO4]
If
[Fe2+]
( ∵ FeSO4 → Fe2+(aq) + SO2–4(aq) )
= [FeSO4] = 3.75 × 10–12 M, then IP = Ksp, hence pptn will not occur.
If [Fe2+] = [FeSO4] is slightly more than 3.75 × 10–12 M, then IP > Ksp and pptn of
Fe(OH)2 will occur.
Wt in gm
1000
S
=
×
∴
V in ml
Mol. wt
Wt in gm
1000
∴3.75 × 10–12 =
×
500
152
3.75 × 10 –12 × 500 × 152
1000
∴ Wt in gm = 2.85 × 10–10 gm
∴ Wt in gm =
If we add FeSO4 slightly more than 2.85 × 10–10 gm, then IP > Ksp and pptn of
Fe(OH)2 will occur.
116.
In a sample of water, the concn. of F– is 3.0 × 10–5 M. How much solid CaCl2 has to
be added to precipitate F– ? Ksp of CaF2 = 1.7 × 10–10 and Mol. wt of CaCl2 is 111
gm/mol.
117.
Calculate minimum amount of NaOH required to precipitate Mg(OH)2 from 500
ml 0.1 M MgCl2 Solution. (Ksp of Mg(OH)2 is 2.8 × 10–16 & mol. wt of NaOH is
40 g/mol).
26
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118.
What is the maximum concentration of equimolar solutions of ferrous sulphate
and sodium sulphide so that when mixed in equal volumes, there is no
precipitation of iron sulphide ? (For iron sulphide, Ksp = 6.3×10–18)
119.
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 these
solutions precipitation will take place ?
Type XIII
120.
:
Common Ion Effect on Solubility of Sparingly soluble
salts
Find out the maximum amount in terms of gm of Zn(OH)2 that can be dissolved
in 2 lit of 0.02 M NaOH solution. (Ksp for Zn(OH)2 is 4.5 × 10–17). (Mol. wt. of
Zn(OH)2 is 99 gm/mol)
Ans : Solubility in water :
Zn(OH)2
S
S
2 OH
(1 : 2 type)
2S
4S3
∴ Ksp
=
∴ S
=
3
Ksp
4
=
3
4.5 × 10 -17
4
=
3
1.125 × 10 -17
=
3
11.125 × 10 -18
=
3
11.125 × 10 -6
∴ S
–
(aq)
Zn2+(aq) +
∴ S
= 2.2324 × 10–6 mole/lit.
Solubility in 0.02 M NaOH solution :
NaOH
→
Na+(aq)
+
–
OH
(aq)
0.02 M
0.02 M
Suppose solubility of Zn(OH)2 in 0.02 M NaOH solution is ‘S’ mole/lit.
Therefore,
Zn2+(aq) +
2 OH
S
S
2 S + 0.02
[Zn2+]
[OH–]2
∴ Ksp
=
∴ Ksp
= [S] [2S + 0.02]2
But
∴ Ksp
27
–
(aq)
Zn(OH)2(s) 2S < < 0.02
(∵ NaOH is strong electrolyte, while Zn (OH)2 is
sparing soluble salt)
= [S] [0.02]2
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4.5 × 10 −17
4 × 10–4
∴ Solubility in gm/lit
∴ S
=
∴ Solubility in gm/2 lit
OR
S
=
1.125 × 10–13 =
= 1.125 × 10–13 mole/lit
= 1.125 × 10–13 × 99
= 1.1137 × 10–11 gm/lit
= 2 × 1.1137 × 10–11
= 2.2275 × 10–11 gm/2 lit
Wt in gm
1000
×
V in ml
Mol. wt
Wt in gm
1000
×
2000
99
121.
Calculate the maximum amount of AgCl (in terms of gm) that can be dissolved in
2 lit solution of 0.02 M AgNO3 (Ksp of AgCl = 2.8 × 10–10 and Mol. wt = 143.5
g/mol).
122.
The value of solubility product Ksp of sparingly soluble salt Mg(OH)2 is
1.8 × 10–11 M3 at 298 K. If a solution of 0.1 M NaOH is added to it, what will be
the concentration of Mg(OH)2 ? Discuss result. (Text. Book Illus. – 21 Page 122)
123.
The ionization constant of benzoic acid is 6.46×10–5 and Ksp for silver benzoate is
2.5×10–13. How many times is silver benzoate more soluble in a buffer of pH 3.19
compared to its solubility in pure water ?
The best teachers will not give you something
to drink, they will make you thirsty.
They will put you on a path to seek answers.
28
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