What is the difference in `pH` for `1//3` and `2//3` stages of neutralization of `0.1 M CH_(3)COOH` with `0.1 M NaOH`?

Options

A. `2 log 3`
B. `2 log (1//4)`
C. `2 log (2//3)`
D. `2 log 2`

Answer

D

Explanation

`CH_(3)COOH+OH^(-) rarrCH_(3)COO^(-)+H_(2)O`
Difference in `pH` between `(1)/(3)` and `(2)/(3)` stages of neutralisation
` =[pK_(a)+log``((2//3)/(1//3))]-pK_(a)+log``((1//3)/(2//3))=2 log 2`