Question

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?

Short answer

The pH of the solution is 9.49

Step-by-step solution

Addition if strong acid to the acidic buffer 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.

Explanation

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.

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

Evaluating the pH

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, $\text{pH = 9.49}$.