Question

Question:Consider the titration of 100.0 mL of 0.100 M H₃A(K_a1 = 5.0 x 10^-4, K_a2= 1.0 x 10^-8, K_a3 = 1.0 X 10^-11) with 0.0500 M NaOH.
a. Calculate the pH after 100.0 mL of 0.0500 M NaOHhas been added.
b. What total volume of 0.0500 M NaOH is required toreach a pH of 8.67?

Short answer

The pH after addition and total volume can be found.

Step-by-step solution

Calculating the pH after 100.0 mL of 0.0500 M NaOH has been added

The mole number of H₃A can be

$\frac{0.1 \mathrm{mol}}{1000 \mathrm{mL}}\times 100 \mathrm{mL}=0.01 \mathrm{mol}.$

The NaOH for spending of the first equivalence point is

$\frac{1000 \mathrm{mL}}{0.05 \mathrm{mol}}\times 0.01 \mathrm{mol}=200 \mathrm{mL}.$

At this point, $\left[{\mathrm{H}}_3\mathrm{A}\right]=\left[{\mathrm{H}}_2{\mathrm{A}}^{-}\right]$

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}1}+\log \frac{\left[{\mathrm{H}}_2{\mathrm{A}}^{-}\right]}{\left[{\mathrm{H}}_3\mathrm{A}\right]} \\ \mathrm{pH}=-\log (5\times {10}^{-4}) \\ \mathrm{pH}=3.3 \end{gathered}$$

Calculating the pH at the second equivalence point

The pH can be calculated as:

$$\begin{gathered} \mathrm{pH}=\frac{{\mathrm{pK}}_{\mathrm{a}2}+{\mathrm{pK}}_{\mathrm{a}3}}{2} \\ \mathrm{pH}=\frac{-\log (1\times {10}^{-8}))+(-\log (1\times {10}^{-11}))}{2} \\ \mathrm{pH}=9.5 \end{gathered}$$

${\mathrm{H}}_2{\mathrm{A}}^{-}+{\mathrm{OH}}^{-}\to {\mathrm{HA}}^{2-}+{\mathrm{H}}_2\mathrm{O}$

The mole number of H₂A^- is 0.01 mol.

Step 3: Calculating a total volume of 0.0500 M NaOH is required to reach a pH of 8.67

The pH can be expressed after reaching 8.67:

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}2}+\log \frac{\left[{\mathrm{HA}}^{2-}\right]}{\left[{\mathrm{H}}_2{\mathrm{A}}^{-}\right]} \\ 8.67=8+\log \frac{\left[\mathrm{x}/\mathrm{V}\right]}{\left[(0.01-\mathrm{x})/\mathrm{V}\right]} \\ 0.67=\log \frac{\mathrm{x}}{0.01-\mathrm{x}} \\ \mathrm{x}=0.008\mathrm{mol} \end{gathered}$$

The number of moles of NaOH to achieve 8.67 as 0.01 mol+0.008 mol=0.018 mol. The total volume of NaOH needed can be

$\frac{1000\mathrm{mL}}{0.05 \mathrm{mol}}\times 0.018 \mathrm{mol}=360 \mathrm{mL}.$