How would you go about solving this pH problem? 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?
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:
$p\mathrm{H} \approx 4.985$