A buffer that contains$\mathbf{\text{1.05 M B}}$and${\mathbf{\text{0.750 M BH}}}^{\mathbf{\text{+}}}$has a pH of $\mathbf{\text{9.50}}$. What is the pHafter$\mathbf{\text{0.0050}}$mol of HCLis added to$\mathbf{\text{0.500 L}}$of this solution?

When a strong acid is added to an acidic buffer solution, the conjugate base of the weak acid reacts with it and forms the weak acid of the buffer solution and maintains constant pH.

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

$$\begin{gathered} {\text{pH = pK}}_{\text{a}}\text{+ log}\frac{\text{[B]}}{\left[{\text{BH}}^{\text{+}}\right]} \\ {\text{pK}}_{\text{a}}\text{= pH - log}\frac{\text{[B]}}{\left[{\text{BH}}^{\text{+}}\right]} \\ \text{= 9.50 - log}\frac{\text{[1.05]}}{\text{[0.750]}} \\ {\text{pK}}_{\text{a}}\text{= 9.35} \end{gathered}$$

Solve for the added HCL molarity now.

$$\begin{gathered} \mathrm{M}=\frac{\mathrm{mol}}{\mathrm{L}} \\ =\frac{0.0050\text{mol}}{0.500\text{L}} \\ =0.01\text{M} \end{gathered}$$

In water, HCL will dissociate.

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

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

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

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 $\text{0.01 M}$, while the concentration of ${\text{BH}}^{\text{+}}$will increase by $\text{0.01 M}$.

$$\begin{gathered} \text{[B] = 1.05 - 0.01} \\ \text{= 1.04 M} \\ \left[{\text{BH}}^{\text{+}}\right]\text{= 0.750 + 0.01} \\ \text{= 0.760 M} \end{gathered}$$

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

$$\begin{gathered} {\text{pH = pK}}_{\text{a}}\text{+ log}\frac{\text{[B]}}{\left[{\text{BH}}^{\text{+}}\right]} \\ \text{= 9.35 + log}\frac{\text{1.04}}{\text{0.760}} \\ \text{pH = 9.49} \end{gathered}$$

Therefore, $\text{pH = 9.49}$.