Question

The pH of the solution when $0.2molHCl$ is added to 1 L of solution containing 0.1 M of of $C{H}_{3}CO{O}^{-}$ (acetate ion) is :
$K_a=1.8\times {10}^{-5}$ and Volume= 1 L

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A 1

Given, $n_{HCl}=0.2mol$ $\left[C{H}_{3}CO{O}^{-}\right]=0.1M$
$K_a=1.8\times {10}^{-5}$ and Volume= 1 L
$\left[HCl\right]=\frac{0.2}{1}=0.2M$
$C{H}_{3}CO{O}^{-}+HCl\rightleftharpoons C{H}_{3}COOH+HClInitial:0.10.200Equilibrium:00.10.10.1$
Moles of $C{H}_{3}COOH=0.1$
Since $C{H}_{3}COOH$ is a weak acid and $HCl$ is a strong acid. In solution, both strong acid $HCl$ and weak acid $C{H}_{3}COOH$ are present
$\therefore$ the dissociation of weak acid $\left(C{H}_{3}COOH\right)$ is suppressed by strong acid $\left(HCl\right)$, due to common ion effect.
Hence, we can neglect the $[H^{+}{]}_{ion}$ from $C{H}_{3}COOH$ and consider $[H^{+}{]}_{ion}$ only from $HCl$, that is a strong acid.
$\left[H^{+}\right]=0.1pH=-log\left[H^{+}\right]pH=-log\left(0.1\right)pH=1$