Q19.28P - Chemistry: Molecular Nature Of Matter And Change

Question

A buffer that contains $\mathbf{\text{0.40 M}}$ base, B, and $\mathbf{\text{0.25 M}}$ of its conjugate acid, ${\mathbf{\text{BH}}}^{\mathbf{\text{ + }}}$, has a pH of $\mathbf{\text{8.88}}$. What is the pH after $\mathbf{\text{0.0020}}$ mol of HCL is added to $\mathbf{\text{0.25 L}}$ of this solution?

Step-by-Step Solution

Answer

The pH of the solution is 8.86.

Step 1: Buffer solution

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.

Step 2: Explanation

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.

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

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

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

Step 3: 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 $\text{0.008 M}$, 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, $\text{pH = 8.86}$.