Chemical and Ionic Equilibrium 2 Question 15
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15. $\mathrm{CH}{3} \mathrm{NH}{2}\left(0.1\right.$ mole, $\left.K_{b}=5 \times 10^{-4}\right)$ is added to 0.08 mole of $\mathrm{HCl}$ and the solution is diluted to one litre, resulting hydrogen ion concentration is
======= ####15. $\mathrm{CH}{3} \mathrm{NH}{2}\left(0.1\right.$ mole, $\left.K_{b}=5 \times 10^{-4}\right)$ is added to 0.08 mole of $\mathrm{HCl}$ and the solution is diluted to one litre, resulting hydrogen ion concentration is
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $1.6 \times 10^{-11}$
(b) $8 \times 10^{-11}$
(c) $5 \times 10^{-5}$
(d) $8 \times 10^{-2}$
$(2005,1 \mathrm{M})$
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Solution:
$$ \mathrm{CH}{3} \mathrm{NH}{2}+\mathrm{HCl} \longrightarrow \mathrm{CH}{3} \mathrm{NH}{3}^{+}+\mathrm{Cl}^{-} $$
Initial : $\quad 0.10 \quad 0.08 \quad 0 \quad 0$
$\begin{array}{lllll}\text { Final : } & 0.02 & 0 & 0.08 & 0.08\end{array}$
$\mathrm{pOH}=\mathrm{p} K_{b}+\log \frac{\left[\mathrm{CH}{3} \mathrm{NH}{3}^{+}\right]}{\left[\mathrm{CH}{3} \mathrm{NH}{2}\right]}$
$$ =-\log \left(5 \times 10^{-4}\right)+\log \frac{0.08}{0.02}=3.9 $$
$\mathrm{pH}=14-\mathrm{pOH}=10.1$
$$ \left[\mathrm{H}^{+}\right]=8 \times 10^{-11} $$
