Chemical And Ionic Equilibrium Result Question 15

15. $\mathrm{CH}_3 \mathrm{NH}_2\left(0.1 \mathrm{~mole}, K_b=5 \times 10^{-4}\right)$ is added to $0.08$ mole of HCl and the solution is diluted to one litre, resulting hydrogen ion concentration is

(2005,1 M)

(a) $1.6 \times 10^{-11}$

(b) $8 \times 10^{-11}$

(c) $5 \times 10^{-5}$

(d) $8 \times 10^{-2}$

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Answer:

Correct Answer: 15. ( b )

Solution:

$\begin{aligned} \mathrm{pOH} & =\mathrm{p} K_b+\log \frac{\left[\mathrm{CH}_3 \mathrm{NH}_3^{+}\right]}{\left[\mathrm{CH}_3 \mathrm{NH}_2\right]} \\ & =-\log \left(5 \times 10^{-4}\right)+\log \frac{0.08}{0.02}=3.9 \\ \mathrm{pH} & =14-\mathrm{pOH}=10.1 \\ {\left[\mathrm{H}^{+}\right] } & =8 \times 10^{-11}\end{aligned}$

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