Question

$C{H}_{3}N{H}_2$ (0.1 mole $K_b$ = 5 $\times {10}^{-4})$ is added to 0.08 mole of HCl and the solution is diluted to one litre, resulting hydrogen ion concentration is

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

Answer 1

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

$C{H}_{3}N{H}_2+HCl\to C{H}_{3}N{H}_3^{+}+C{l}^{-}$

Initial:	0.10	0.08	0	0
Final:	0.02	0	0.08	0.08

$pOH=p{K}_b+log\frac{\left[C{H}_{3}N{H}_3^{+}\right]}{C{H}_{3}N{H}_2}$
$=-log(5\times {10}^{-4})+log\frac{0.08}{0.02}=3.9$
pH = 14 - pOH = 10.1
$\left[H^{+}\right]=8\times {10}^{-11}$