Q19.29P - Chemistry: Molecular Nature Of Matter And Change

Question

A buffer that contains $\mathbf{0}\mathbf{.}\mathbf{110} \mathbf{M} \mathbf{HY}$ and ${\mathbf{\text{0.220 M Y}}}^{\mathbf{\text{ - }}}$ has a pH of $\mathbf{\text{8.77}}$. What is the pH after $\mathbf{\text{0.0015}}$ mol of ${\mathbf{\text{Ba(OH)}}}_{\mathbf{\text{2}}}$ is added to $\mathbf{\text{0.350 L}}$ of this solution?

Step-by-Step Solution

Answer

The pH of the solution is 8.82.

Step 1: Acidic buffer stability

When a strong base is added to an acidic buffer solution, then the weak acid reacts with the strong base and forms its conjugate base and thus the pH remains unchanged.

Step 2: Explanation

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

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[{\mathrm{Y}}^{-}\right]}{\left[\mathrm{HY}\right]} \\ {\mathrm{pK}}_{\mathrm{a}}=\mathrm{pH}-\log \frac{\left[{\mathrm{Y}}^{-}\right]}{\left[\mathrm{HY}\right]} \\ =8.77-\log \frac{[0.220]}{[0.110]} \\ {\mathrm{pK}}_{\mathrm{a}}=8.47 \end{gathered}$$

Solve for the added ${\text{Ba(OH)}}_{\text{2}}$ molarity now.

$$\begin{gathered} \mathrm{M}=\frac{\mathrm{mol}}{\mathrm{L}} \\ =\frac{0.0015 \text{mol}}{0.350\text{ L}} \\ =0.0043\text{ M} \end{gathered}$$

In water, ${\text{Ba(OH)}}_{\text{2}}$ will dissociate.

${\text{Ba(OH)}}_{\text{2}}\to {\text{Ba}}^{\text{2 + }}{\text{ + 2OH}}^{\text{ - }}$

All ${\text{OH}}^{\text{ - }}$ will then react with $\text{HA}$ as a strong base.

${\text{HY + OH}}^{\text{ - }}\to {\text{H}}_{\text{2}}{\text{O + Y}}^{\text{ - }}$

Step 3: Evaluating the pH

As can be seen, the reaction also yielded ${\text{Y}}^{\text{ - }}$. Since all of the data-custom-editor="chemistry" ${\text{OH}}^{\text{ - }}$ has been consumed, the concentration of $\text{HY}$ will decrease by $2\times 0.0043\text{ M}$, while the concentration of ${\text{Y}}^{\text{ - }}$ will increase by $2\times 0.0043\text{ M}$. Since one mole of ${\text{Ba(OH)}}_{\text{2}}$ produces two moles of ${\text{OH}}^{\text{ - }}$, we multiply the concentration by two.

$$\begin{gathered} [\mathrm{HY}]\; =\; 0.110\; -\; 0.0086 \\ =\; 0.1014\mathrm{M} \\ \left[{\mathrm{Y}}^{ - }\right] =\; 0.220\; +\; 0.0086 \\ =\; 0.2286\mathrm{M} \end{gathered}$$

Now, using the ${\text{pK}}_{\text{a}}$ and the new [$\text{HY}$] and [${\text{Y}}^{\text{ - }}$] values, calculate the new pH.

$$\begin{gathered} {\text{pH = pK}}_{\text{a}}\text{ + log}\frac{{\text{[Y}}^{\text{ - }}\text{]}}{\left[\text{HY}\right]} \\ \text{ = 8.47 + log}\frac{\text{0.2286}}{\text{0.1014}} \\ \text{pH = 8.82} \end{gathered}$$

Therefore, $\text{pH = 8.82}$.