A buffer that contains $\mathbf{0}\mathbf{.}\mathbf{110} \mathbf{M} \mathbf{HY}$ and${\mathbf{\text{0.220 MY}}}^{\mathbf{\text{-}}}$has a pHof$\mathbf{\text{8.77}}$. 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?

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.

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.

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

As can be seen, the reaction also yielded ${\text{Y}}^{\text{-}}$. Since all of thedata-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, $\text{pH = 8.82}$.