We’re being asked **to calculate the pH** of a **0.16 M CH _{3}NH_{3}Cl** using an ICE chart.

CH_{3}NH_{3}^{+} is a conjugate acid of the base CH_{3}NH_{2}. We need to use the K_{b} for CH_{3}NH_{2} to find the K_{a} of its conjugate acid, CH_{3}NH_{3}^{+}. Let us just omit Cl^{- }since it is neutral.

We can find on books and the internet that the K_{b} of CH_{3}NH_{2} is 4.4x10^{-4}. Therefore, the Ka of CH_{3}NH_{3}^{+} is:

$\overline{){{\mathbf{K}}}_{{\mathbf{w}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathbf{K}}}_{{\mathbf{a}}}{{\mathbf{K}}}_{{\mathbf{b}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{{\mathbf{a}}}\mathbf{}\mathbf{=}\frac{\mathbf{}{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{a}}\mathbf{}\mathbf{=}\frac{\mathbf{}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{4}\mathbf{.}\mathbf{4}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}$

**K _{a} = 2.27x10**

Since CH_{3}NH_{3}^{+} has a very low K_{a} value, it’s a weak acid. Remember that ** weak acids** partially dissociate in water and that

CH_{3}NH_{3}^{+}(aq) + H_{2}O(l) ⇌ H_{3}O^{+}(aq) + CH_{3}NH_{2}(aq); **K _{a} = 2.27×10^{–11}**

From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table.

The ** K_{a} expression** for CH

$\overline{){{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}\frac{\left[{H}_{3}{O}^{+}\right]\left[{\mathrm{CH}}_{3}{\mathrm{NH}}_{2}\right]}{\left[{\mathrm{CH}}_{3}{{\mathrm{NH}}_{3}}^{+}\right]}}$

*Liquids are ignored in the K _{a} expression.*

Note that each concentration is raised by the stoichiometric coefficient: [CH_{3}NH_{3}^{+}], [H_{3}O^{+}] and [CH_{3}NH_{2}] are raised to 1.

Solve an equilibrium problem (using an ICE table) to calculate the pH of each of the following solutions.

0.16 M CH_{3} NH_{3} Cl