Exercise 5 | Buffers and Titration of Acids and Bases

Henderson-Hasselbalch equation can be used:

pH = pKa + log $\frac{{\left[\mathrm{base}\right]}_0}{{\left[\mathrm{acid}\right]}_0}$

pH = 7.17 + log $\frac{{\left[{{\mathrm{SO}}_3}^{2-}\right]}_0}{{\left[{{\mathrm{HSO}}_3}^{-}\right]}_0}$

We want pH = 7.00 ⇒ $\frac{{\left[{{\mathrm{SO}}_3}^{2-}\right]}_0}{{\left[{{\mathrm{HSO}}_3}^{-}\right]}_0}$ = 10^-0.17

If X is the volume (in mL) of the KHSO₃ used to prepare the buffer solution, 100 – X is the volume of K₂SO₃ used.

[SO₃^2-]₀ = 75.0 x $\frac{100 - \mathrm{X}}{100}$

[HSO₃^-]₀ = 150 x $\frac{\mathrm{X}}{100}$

$\frac{{\left[{{\mathrm{SO}}_3}^{2-}\right]}_0}{{\left[{{\mathrm{HSO}}_3}^{-}\right]}_0}$ = 10^-0.17

⇒ $\frac{75.0 \times \left(100 - \mathrm{X}\right)}{150 \times \mathrm{X}}$ = 10^-0.17

⇒ 176.4 X = 7500

⇒ X = 42.5 mL

We need to mix 42.5 mL of 0.150 M of KHSO₃ and 57.5 mL of 75.0 M of K₂SO₃ to prepare a 100 mL of a pH =7.00 buffer solution.