What is `[H^(+)]` in `mol//L` of a solution that is `0.20 M` in `CH_(3)COONa` and `0.1 M` in `CH_(3)COOH`? `K_(a)` for `CH_(3)COOH` is `1.8xx10^(-5)`?

Options

A. `3.5xx10^(-4)`
B. `1.1xx10^(-5)`
C. `1.8xx10^(-5)`
D. `9.0xx10^(-6)`

Answer

D

Explanation

Key Idea - `CH_(3)COOH` (weak acid) and `CH_(3)COONa` (conjugated salt) form acidic buffer and for acidic buffer,
`pH=pK_(a)+log.(["salt"])/(["acid"])`
and `[H^(+)]=-` antilog pH
`pH=-log K_(a)+log(["salt"])/(["acid"])`
`[because pK_(a)=-log K_(a)]`
`=-log(1.8xx10^(-5))+log.((0.20))/((0.10))`
`=4.74+log2`
`=4.74+0.3010=5.41`
Now, `[H^(+)]=` antilog (-5.045)
`9.0xx10^(-6)mol//L`