Question

What is $[H+]$ in mol per litre of the solution that is $0.02M$ in ${CH}_{3}COONa$ and $0.01M$ in ${CH}_{3}COOH$ ? $(K_{a}for{CH}_{3}COOH=1.8\times {10}^{-5})$

• A. $3.5\times {10}^{-4}$
• B. $1.1\times {10}^{-5}$
• C. $1.8\times {10}^{-4}$
• D. $9.0\times {10}^{-6}$

Answer 1

The correct option is D $9.0\times {10}^{-6}$

$9.0\times {10}^{-6}$

Given:-

$K_a=1.8\times {10}^{-5}$

$\therefore p{K}_a=-log{K}_a$

$\Rightarrow p{K}_a=-log(1.8\times {10}^{-5})$

$\Rightarrow p{K}_a=4.74$

As we know that,

$pH=p{K}_a+\log \frac{\left[salt\right]}{\left[acid\right]}$

$\Rightarrow pH=4.74+\log \frac{0.02}{0.01}=4.74+log2=4.74+0.3010=5.041$

Now,

$pH=-log\left[H^{+}\right]$

$\left[H^{+}\right]=antilog(-pH)$

$\Rightarrow \left[H^{+}\right]=antilog(-5.041)=9\times {10}^{-6}mol{L}^{-1}$