Question

Calculate pH of a solution which is ${10}^{-1}M$ in HCl & ${10}^{-3}M$ in $C{H}_{3}COOH[K_a=2\times {10}^{-5}]$. Also calculate $\left[H^{+}\right]$ From $C{H}_{3}COOH.$

• A. pH=1, $\left[H^{+}\right]=2\times {10}^{-7}$ M
• B. pH=7, $\left[H^{+}\right]=2\times {10}^{-7}$ M
• C. pH=1, $\left[H^{+}\right]=3\times {10}^{-7}$ M
• D. pH=7, $\left[H^{+}\right]=3\times {10}^{-6}$ M

Answer 1

The correct option is A pH=1, $\left[H^{+}\right]=2\times {10}^{-7}$ M

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}t=eqC(1-\alpha )C\alpha +C\alpha$
$H^{+}$ ion can be considered completely from HCl, due to less dissociation of $C{H}_{3}COOH$ (because of common ion effect by $H^{+}$ of $HCl,a<<1$) and its low conc. So, $\left[H^{+}\right]={10}^{-1}M\therefore pH=1.$
From above equilibrium. $2\times {10}^{-5}=\frac{C\alpha \times {10}^{-1}}{C}$
$\alpha =2\times {10}^{-4}$
$\left[H^{+}\right]\text{from}C{H}_{3}COOH=C\alpha ={10}^{-3}\times 2\times {10}^{-4}$
$\left[H^{+}\right]=2\times {10}^{-7}M.$