Question

Calculate $\left[H^{+}\right]$ from a $C{H}_{3}COOH$ in a solution which is ${10}^{-1}M$ in $HCl$ and ${10}^{-3}M$ in $C{H}_{3}COOH;[K_a=2\times {10}^{-5}]$.

• A. $2\times {10}^{-4}$
• B. $4\times {10}^{-7}$
• C. $2\times {10}^{-7}$
• D. None of the above

Answer 1

The correct option is C $2\times {10}^{-7}$

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