# How do you calculate the pH at the equivalence point for the titration of .190M methylamine with .190M HCl? The Kb of methylamine is 5.0x10^-4.

Jun 1, 2016

$p H = 5.86$

#### Explanation:

The net ionic equation for the titration in question is the following:

$C {H}_{3} N {H}_{2} + {H}^{+} \to C {H}_{3} N {H}_{3}^{+}$

This exercise will be solved suing two kinds of problems: Stoichiometry problem and equilibrium problem .

Stoichiometry Problem :
At the equivalence point, the number of mole of the acid added is equal to the number o fmole of base present.

Since the concentrations of base and acid are equal, the concentration of the conjugate acid $C {H}_{3} N {H}_{3}^{+}$ can be determined as follows:

Since equal volumes of the acid and base should be mixed, and since they are additive, the concentration of $C {H}_{3} N {H}_{3}^{+}$ will be half the initial concentration of $C {H}_{3} N {H}_{2}$.

Thus, $\left[C {H}_{3} N {H}_{3}^{+}\right] = 0.095 M$

Equilibrium Problem :
The conjugate acid that will be the major species at the equivalence point, will be the only significant source of ${H}^{+}$ in the solution and therefore, to find the pH of the solution we should find the $\left[{H}^{+}\right]$ from the dissociation of $C {H}_{3} N {H}_{3}^{+}$:

$\text{ " " " " " " " " } C {H}_{3} N {H}_{3}^{+} r i g h t \le f t h a r p \infty n s C {H}_{3} N {H}_{2} + {H}^{+}$
$I n i t i a l : \text{ " " " " "0.095M" " " " "0M" " " " } 0 M$
$\text{Change": " " " " " "-xM" " " " "+xM" " } + x M$
$\text{Equilibrium": (0.095-x)M" " " "xM" " } x M$

${K}_{a} = \frac{\left[C {H}_{3} N {H}_{2}\right] \left[{H}^{+}\right]}{\left[C {H}_{3} N {H}_{3}^{+}\right]}$

Note that ${K}_{w} = {K}_{a} \times {K}_{b}$

$\implies {K}_{a} = \frac{{K}_{w}}{{K}_{b}} = \frac{1.0 \times {10}^{- 14}}{5.0 \times {10}^{- 4}} = 2.0 \times {10}^{- 11}$

$\implies {K}_{a} = \frac{\left[C {H}_{3} N {H}_{2}\right] \left[{H}^{+}\right]}{\left[C {H}_{3} N {H}_{3}^{+}\right]} = \frac{x \cdot x}{0.095 - x} = \frac{{x}^{2}}{0.095 - x} = 2.0 \times {10}^{- 11}$

Solve for $x = 1.38 \times {10}^{- 6} M = \left[{H}^{+}\right]$

Therefore, the pH of the solution is $p H = - \log \left[{H}^{+}\right]$

$\implies p H = - \log \left(1.38 \times {10}^{- 6}\right) = 5.86$

Here is a video that explains in details the titration of a weak acid by a strong base:
Acid - Base Equilibria | Weak Acid - Strong Base Titration.