Question

Calculate the ratio of pH of a solution containing 1 mol of $C{H}_{3}COONa$ + 1 mol of $HCl$ per litre to another solution containing 1 mol of $C{H}_{3}COONa$ + 1 mol of acetic acid per litre.

• A. 1:1
• B. 2:1
• C. 1:2
• D. 2:3

Answer 1

The correct option is C 1:2

$C{H}_{3}COONa+HCl\to C{H}_{3}COOH+NaCl$
1 1 0 0
$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$
So, $\left[H^{+}\right]=\sqrt{K_a.c}$ =$\sqrt{K_a}$ (Since c = 1)
= $p{H}_1=-\frac{1}{2}log{K}_a$ = $\frac{1}{2}p{K}_a$
For IInd solution, its a mixture of acetic acid and its salt
using formula
$p{H}_2=p{K}_a+log\frac{\left[salt\right]}{\left[acid\right]}$
Concentration of the acid and the salt are given, using that we get $p{H}_2=p{K}_a$
On calculating their ratio we get,
$\frac{p{H}_1}{p{H}_2}=\frac{1}{2}$