Question

A titration of 200.0 mL of 1.00 M H2A was done with 1.18 M NaOH. For the diprotic acid H2A, ka1 = 2.5 x 10^-5 and ka2 = 3.1 x 10^-9. Calculate the pH after 100.0 mL of 1.18 M NaOH have been added

Answer 1

1. pH of the 1,00 M diprotic acid solution is:

$$pH = \frac{1}{2} \left(- \lg K_{a1} - \lg C_{a}^{0}\right) = \frac{1}{2} \left(- \lg 2.5 \cdot 10^{-5} - \lg 1.00\right) = 4.60$$

2. Acid concentration after 100.00 mL of NaOH have been added:

$$C_{a} = \frac{C_{a}^{0} \cdot V_{a} - C_{t} \cdot V_{t}}{V_{a} + V_{t}} = \frac{1,00 \cdot 200,00 - 1,18 \cdot 100,00}{200,00 + 100,00} = 0,273\ \text{M}$$

3. Salt concentration is:

$$C_{s} = \frac{C_{t} \cdot V_{t}}{V_{a} + V_{t}} = \frac{1,18 \cdot 100,00}{200,00 + 100,00} = 0,393\ \text{M}$$

5. pH of the acid solution after 15.00 mL of titrant have been added is:

$$pH = - \lg K_{a2} - \lg \frac{0,273}{0,393} = 8.65$$

Answer: 8,65.