Question

A buffer containing $\mathbf{0}\mathbf{.2000}\mathbf{\text{M}}$of acid, $\mathbf{\text{HA}}$, and$\mathbf{0}\mathbf{.1500}\mathbf{\text{M}}$of its conjugate base, ${\mathbf{\text{A}}}^{\mathbf{\text{-}}}$, has apH of 3.35. What is thepH after 0.0015mol of NaoHis added to $\mathbf{0}\mathbf{.5000}\mathbf{\text{L}}$of this solution?

Short answer

The pH of the solution is 3.36.

Step-by-step solution

Addition of a strong base to acidic buffer

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.

Calculation of pKa value

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.

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

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

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

Evaluating the pH value

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