Question

A solution was made up of $0.01M$ in chloroacetic acid $Cl{CH}_{2}COOH$ and also $0.002M$ in sodium chloracetate $Cl{CH}_{2}COONa$. What is $\left[H^{+}\right]$ in the solution? $K_a=1.5\times {10}^{-3}$.

• A. $\left[H^{+}\right]=1.25\times {10}^{-4}$
• B. $\left[H^{+}\right]=2.5\times {10}^{-3}$
• C. $\left[H^{+}\right]=5\times {10}^{-4}$
• D. $\left[H^{+}\right]=7.5\times {10}^{-3}$

Answer 1

The correct option is D $\left[H^{+}\right]=7.5\times {10}^{-3}$

Its a buffer solution and pH of buffer solution is

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

Substituting,

$pH=3-log1.5+log\frac{0.002}{0.01}=3-log1.5+log2-1=2.12$

$\left[H^{+}\right]={10}^{-pH}=7.5\times {10}^{-3}$