Question

Solve this:

$$\begin{gathered} \mathbf{32}\mathbf{.}0.1moleofC{H}_{3}N{H}_2\left(K_b=5\times {10}^{-4}\right)ismixedwith0.08moleofHClanddilutedtoonelitre.The\left[H^{+}\right]insolutionis: \\ A)8\times {10}^{-2}MB)8\times {10}^{-11}M \\ C)1.6\times {10}^{-11}MD)8\times {10}^{-5}M \end{gathered}$$

Answer 1

Dear Student,

$$\begin{gathered} C{H}_{3}N{H}_2+HCl=C{H}_{3}N{H_3}^{+}+C{l}^{-} \\ Initial0.10.08- \\ AtEq.0.02-0.08 \\ \left[O{H}^{-}\right]=\frac{Kb\left[C{H}_{3}N{H}_2\right]}{\left[C{H}_{3}N{H_3}^{+}\right]} \\ =\frac{5\times {10}^{-4}\times 0.02}{0.08} \\ =1.25\times {10}^{-4} \\ \left[H^{+}\right]=\frac{Kw}{\left[O{H}^{-}\right]} \\ =\frac{{10}^{-14}\times 4}{1.25\times {10}^{-4}} \\ =8\times {10}^{-11} \end{gathered}$$

Regards,