Question

At ${25}^{\circ }C,$ consider $1L$ of two solutions as following:
Solution A : Contains $1molC{H}_{3}COONa+0.5molHCl$
Solution B : Contains $1molC{H}_{3}COONa+0.5molC{H}_{3}COOH$
If $K_a\left(C{H}_{3}COOH\right)={10}^{-5}$ then ratio of [$H^{+}$ (aq.)] of solution A to that of solution B is,
(Given: $log1=0andlog2=0.3,log(5\times {10}^{-6}=-5.3$)

• A. $1:2$
• B. $2:1$
• C. $1:4$
• D. $4:1$

Answer 1

The correct option is B $2:1$

(A)
$C{H}_{3}COONa+HCl\rightleftharpoons C{H}_{3}COOH+NaClt=01mole0.5molet=eqm.0.5mole00.5mole0.5mole$
At equilibrium, we have $0.5$ mole of $C{H}_{3}COOH$. So, it is acidic buffer.
$pH=p{K}_a+\log \left[\frac{\text{Anion}}{\text{Acid}}\right]pH=p{K}_a+\log \frac{0.5}{0.5}pH=p{K}_a=5\Rightarrow [H^{+}{]}_A={10}^{-5}$

(B) $C{H}_{3}COOH+C{H}_{3}COONa$
$pH=p{K}_a+\log \frac{1}{0.5}=5+\log \left(2\right)=5.3[H^{+}{]}_B=5\times {10}^{-6}\text{Ratio:}\frac{[H^{+}{]}_A}{[H^{+}{]}_B}=\frac{{10}^{-5}}{5\times {10}^{-6}}=\frac{10}{5}=2:1$