Question

0.1 mole of $C{H}_{3}N{H}_2(K_b=5\times {10}^{-4})$ is mixed with 0.08 mole of HCl and diluted to one litre. What will be the $H^{+}$ concentration in the solution?
log 5 =0.6989
log 2= 0.3010
${10}^{-10.09}=8.128\times {10}^{-11}$

• A. $8\times {10}^{-2}M$
• B. $8\times {10}^{-11}M$
• C. $1.6\times {10}^{-11}M$
• D. $8\times {10}^{-5}M$

Answer 1

The correct option is B $8\times {10}^{-11}M$

All HCl reacts with aniline to form a salt. Being a salt of weak base and strong acid, it hydrolyses to give a basic solution with pH more than 7
$\begin{array}{cccccc}& C{H}_{3}N{H}_2& +& HCl& \to & C{H}_{3}N{H}_3^{+}C{l}^{-}\\ \text{Initial moles}& 0.1& & 0.08& & 0\\ \text{Moles after mixing}& 0.02& & 0& & 0.08\end{array}$
As it is a basic buffer solution.
$pOH=p{K}_b+log\frac{\text{salt}}{\text{base}}$
$pOH=p{K}_b+log\frac{0.08}{0.02}$
$pOH=-log5\times {10}^{-4}+log4$
$pOH=3.30+0.602=3.902$
$pH=14-3.902=10.09$
$pH=-log\left[H^{+}\right]$
$10.09=-log\left[H^{+}\right]$
$\left[H^{+}\right]=8.128\times {10}^{-11}\approx 8\times {10}^{-11}M$