91Ó°ÊÓ

When a strong base is added to an acidic buffer solution, then the weak acid present in the buffer reacts with the strong base and forms the conjugate base of the weak acid.

To begin, manipulate the Henderson-Hessl Balch equation to determine the acid's ${\text{pK}}_{\text{a}}$.

$$\begin{gathered} {\text{pH = pK}}_{\text{a}}\text{+ log}\frac{\left[{\text{A}}^{\text{-}}\right]}{\text{[HA]}} \\ {\text{pK}}_{\text{a}}\text{= pH - log}\frac{\left[{\text{A}}^{\text{-}}\right]}{\text{[HA]}} \\ \text{= 3.35 - log}\frac{\text{[0.1500]}}{\text{[0.2000]}} \\ {\text{pK}}_{\text{a}}\text{= 3.47} \end{gathered}$$

Solve for the added NaoH molarity now.

$$\begin{gathered} \mathrm{M}=\frac{\mathrm{mol}}{\mathrm{L}} \\ =\frac{0.0015\text{mol}}{0.5000\text{L}} \\ =0.003\text{M} \end{gathered}$$

In water, Naoh will dissociate.

$\text{NaOH}→{\text{Na}}^{\text{+}}{\text{+ OH}}^{\text{-}}$

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

${\text{HA + OH}}^{\text{-}}→{\text{H}}_{\text{2}}{\text{O +A}}^{\text{-}}$

As can be seen, the reaction also yielded ${\text{A}}^{\text{-}}$. Since all of the${\text{OH}}^{\text{-}}$ has been consumed, the concentration of$\text{HA}$ will decrease by role="math" localid="1663292293657" $\text{0.003 M}$, while the concentration of${\text{A}}^{\text{-}}$ will increase by $\text{0.003 M}$.

$$\begin{gathered} \text{[HA]}=\; 0.2000\; -\; 0.003 \\ =\; 0.1970\text{M} \\ \left[{\text{A}}^{\text{-}}\right]=\; 0.1500\; +\; 0.003 \\ =\; 0.1530\text{M} \end{gathered}$$

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

$$\begin{gathered} {\text{pH = pK}}_{\text{a}}\text{+ log}\frac{\left[{\text{A}}^{\text{-}}\right]}{\text{[HA]}} \\ =\; 3.47\; +\mathrm{lo}g\frac{0.1530}{0.1970} \\ \text{pH}=\; 3.36 \end{gathered}$$

Therefore, $\text{pH}=\; 3.36$.