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 the solution diluted to one litre. The $H^{+}$ ion concentration in the solution will be:

• 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 $8\times {10}^{-11}$

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

here, $\left[O{H}^{-}\right]=kb\frac{\left[C{H}_{3}N{H}_2\right]}{\left[C{H}_{3}N{H}_3\right]}$

$\Rightarrow 5\times {10}^{-4}\times \frac{0.02}{0.08}=1.25\times {10}^{-4}$

$\because H^{+}=\frac{kw}{\left[O{H}^{-}\right]}=\frac{{10}^{-14}\times 4}{5\times {10}^{-}4}=8\times {10}^{-11}$