UPSEE Solved Model Paper 7

Given
Volume of Base. BOH = 40 mL
Normality of HCl solution = 0.1 N
Volume of HCl added V1 = 5 mL
Volume of HCl added V2 = 20 mL
pH1 = 10.04
pH2 = 9.14
W e have to find the dissociation constant of base, Kb
Let the initial millimoles of base be a
Reaction can be shown as:
Case I
On adding 5 mL acid:

pH = 10.04
Thus. pOH = 14 - 10.04
= 3.96
pOH is related with Kb as :
pOH = - log Kb + log

[Salt]

[Base]

pOH = - log Kb + log

[BCl]

[BOH]

... (ii)
where, [BCI] = Concentration of salt
[BOH] = Concentration of base
On substitution all the values in equation (j), we get
3.95 = - log Kb + log

0.5

(a−0.5)

... (ii)
Case II
On adding 20 mL add:

pH - 9.14
Poh = 14 - 9.14
= 4.86
On substituting all the values in equation (I), we get
4.86 = - log Kb + log (

2

a−2

) ... (iii)
Subtracting equation (iii) from (ii) , we get
3.96 = - log Kb + log (

0.5

a−0.5

)
4.86 = - log Kb + log (

2

a−2

)
─────────────────────
- 0.9 = log (

0.5

a−0.5

) - log (

2

a−2

)
As, log a - log b = log

a

b

Thus, - 0.9 = log [

0.5 (a−0.5)

2 (a−2)

]
- 0.9 = log [

1 2(2a−1)

2×2 (a−2)

]
- 0.9 = log [

(a−2)

2(2a−1)

]

(a−2)

2(2a−1)

= 10−(0.9)

(a−2)

2(2a−1)

= 0.126
(a - 2) = 0.126 (4a - 2)
(a - 2) = 0.504a - 0.252
a - 0.504a = 2 - 0.252
0.496 a = 1.748
1.748
a =

10748

0.496

a = 3.52
On submitting the values of 'a' in equation (ii) we get
3.96 = - log Kb + log [

0.5

3.52−0.5

]
3.96 = - log Kb + log (

0.5

3.02

)
3.96 = - log Kb + log 0.165
3.96 = - log Kb + (- 0.78)
3.96 + 0.78 = - log Kb
- log Kb = 4.74
Kb = 10−(4.74)
Kb = 0.000018197
Kb = 1.8 × 10−5
Hence, option 'C' is correct.