91Ó°ÊÓ

A Basic buffer solution is formed by the combination of a weak base and the salt of its conjugate acid. When a strong acid is added, then the weak base reacts with it and forms the salt of its conjugate acid.

First, manipulate the Henderson-Hassle Balch equation to find the${\text{pK}}_{\text{a}}$ of the base.

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{B}\right]}{\left[{\mathrm{BH}}^{+}\right]} \\ {\mathrm{pK}}_{\mathrm{a}}=\mathrm{pH}-\log \frac{\left[\mathrm{B}\right]}{\left[{\mathrm{BH}}^{+}\right]} \\ =8.88-\log \frac{[0.40]}{[0.25]} \\ {\mathrm{pK}}_{\mathrm{a}}=8.68 \end{gathered}$$

Solve for the added HCL molarity now.

$$\begin{gathered} \mathrm{M}=\frac{\mathrm{mol}}{\mathrm{L}} \\ =\frac{0.0020\text{mol}}{0.25\text{L}} \\ =0.008\text{M} \end{gathered}$$

In water, HCL will dissociate.

${\text{HCl +H}}_{\text{2}}\text{O}→{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{+ Cl}}^{\text{-}}$

All${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ will then react with$\mathrm{B}$ as a strong acid.

${\text{B +H}}_{\text{3}}{\text{O}}^{\text{+}}→{\text{BH}}^{\text{+}}{\text{+H}}_{\text{2}}\text{O}$

As can be seen, the reaction also yielded ${\text{BH}}^{\text{+}}$. Since all of the${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ has been consumed, the concentration of$\text{B}$ will decrease by $\text{0.008 M}$, while the concentration of ${\text{BH}}^{\text{+}}$will increase by $\text{0.008 M}$.

$$\begin{gathered} \left[\mathrm{B}\right]=0.40-0.008 \\ =0.392\text{M} \\ \left[{\mathrm{BH}}^{+}\right]=0.25+0.008 \\ =0.258\text{M} \end{gathered}$$

Now, using the${\text{pK}}_{\text{a}}$ and the new [${\text{BH}}^{\text{+}}$] and [$\text{B}$] values, calculate the new pH.

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{B}\right]}{\left[{\mathrm{BH}}^{+}\right]} \\ =8.68+\log \frac{0.392}{0.258} \\ \mathrm{pH}=8.86 \end{gathered}$$

Therefore, $\text{pH = 8.86}$.