Question

Calculate the ratio of pH of a solution containing $1$ mole of ${CH}_{3}COONa+1$ mole of $HCl$ per litre and of other solution containing $1$ mole ${CH}_{3}COONa+1$ mole of acetic acid per litre :

• A. $1:1$
• B. $2:1$
• C. $1:2$
• D. $2:3$

Answer 1

Answer: (C)

$1:2$

$C{H}_{3}COONa+HCl⟶C{H}_{3}COOH+NaCl$

Moles before reaction 1 1 0 0

Moles after reaction 0 0 1 1

$\left[C{H}_{3}COOH\right]=\frac{1}{1}=1M$

$\left[H^{+}\right]=c\alpha =c\sqrt{\frac{K_a}{c}}=\sqrt{K_{a}c}=\sqrt{K_a}$ (c=1)

$p{H}_1=\frac{-1}{2}log{K}_a$

For $C{H}_{3}COONa+C{H}_{3}COOH$

1M 1M

$p{H}_2=-\log K_a+\log \frac{1}{1}=-\log K_a$

Hence ratio will be

$\frac{p{H}_1}{p{H}_2}=\frac{1}{2}$