Ionic EquilibriumHard
Question
A weak acid HA after treatment with 12 mL of 0.1 M strong base BOH has a pH of 5. At the end point the volume of same base required is 26.6 mL . Calculate Ka of acid is :
Options
A.1.8 × 10-5
B.8.22 × 10-6
C.1.8 × 10-6
D.8.2 × 10-5
Solution
For neutralisation :
Total Meq. of acid = Meq. of base = 26.6 × 0.1 = 2.66.
Now for partial neutralisation of acid.
HA + BOH BA + H2O
Meq. before reaction 2.66 1.2 0 0
Meq. after reaction 1.46 0 1.2 1.2
The resultant mixture acts as a buffer and [HA] and [BA] may be placed in terms of Meq. sincen volume of mixture is constant.
pH = - log Ka + log
or 5 = - logKa + log
s
Ka = 8.22 × 10-6.
Total Meq. of acid = Meq. of base = 26.6 × 0.1 = 2.66.
Now for partial neutralisation of acid.
HA + BOH BA + H2O
Meq. before reaction 2.66 1.2 0 0
Meq. after reaction 1.46 0 1.2 1.2
The resultant mixture acts as a buffer and [HA] and [BA] may be placed in terms of Meq. sincen volume of mixture is constant.
pH = - log Ka + log
or 5 = - logKa + log
sKa = 8.22 × 10-6.
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