What is the $\mathrm{pH}$ of a $0.24\ \mathrm M$ solution of benzylammonium chloride ($\ce{C6H5CH2NH3Cl}$)?
$K_\mathrm b$ for benzylamine is $2.2\times10^{-5}$
Your answer must be within $\pm 0.4\ \%$

I was able to determine that $[\ce{H3O^+}]=-1.8\times10^{-5}\ \mathrm M$ using $K_\mathrm b$. Then I used the formula $\mathrm{pH}=-\log[\ce{H3O^+}]$ to determine that the $\mathrm{pH}$ is $0.0000414465$. However, this answer keeps being marked as incorrect.

What am I doing wrong?

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    $\begingroup$ Welcome on chemistry.se! How did you come to the negative proton concentration? $\endgroup$ – pH13 - Yet another Philipp Oct 10 '16 at 21:19
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    $\begingroup$ The problem is idiotic. It calls for an answer good to +/- 0.4%, but the $K_b$ is only good to +/- 2.3% and the molarity is only good to +/- 2.1%. Absurd. The problem should have specified 0.240 M and $K_b = 2.20 \times 10^{-5}$ $\endgroup$ – MaxW Oct 11 '16 at 6:14

The $K_\mathrm{b}$ value of benzylamine, the conjugated base of benzylmmoniumchloride, is given. The acid constant of benzylammoniumchloride can be inferred via the identity $$ \mathrm{p}K_a + \mathrm{p}K_b = 14. $$ Thus, we have $$ \mathrm{p}K_\text{b,benzylamin} = 4.65$$ $$ \mathrm{p}K_\text{a,benzylammoniumchloride} = 9.35 $$ $$ \Rightarrow K_\text{a,benzylammoniumchloride} = 10^{-9.35}$$ As furthermore $$ K_\text{a,benzylammoniumchloride} = \frac{[\mathrm{H}^\oplus] \cdot [\mathrm{PhCH_2NH_2}]}{[\mathrm{PhCH_2NH_3^\oplus]}} = \frac{[\mathrm{H}^\oplus]^2}{[\mathrm{PhCH_2NH_3^\oplus]}}$$ and we see that $$ K_\text{a,benzylammoniumchloride} = \frac{[\mathrm{H}^\oplus]^2}{[\mathrm{PhCH_2NH_3^\oplus]}}$$ This leaves us with an equation with only one unknown variable and is easy to solve as $[\mathrm{PhCH_2NH_3^\oplus}] = 0.24~\mathrm{\frac{mol}{L}}$ is given: $$ [\mathrm{H^\oplus}] = \sqrt{[\mathrm{PhCH_2NH_3^\oplus]}\cdot K_\text{a,benzylammoniumchloride}} \approx 1.0354 \cdot 10^{-5} ~\mathrm{mol/L} $$ And as the pH is the negative of the logarithm with base ten we have: $p\mathrm{H} \approx 4.985$

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    $\begingroup$ By the way, a pH of about 0 should strike you as odd for such a weak acid. A pH of about 0 you would only get with strong acids like $\mathrm{H_2SO_4}$ or HCl. $\endgroup$ – logical x 2 Oct 10 '16 at 21:51

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