# Exercise 5 | Buffers and Titration of Acids and Bases

How could you prepare a 100 mL of a pH =7.00 buffer solution from a solution of 0.150 M of KHSO3 and a solution of 75.0 M of K2SO3?

Data: Ka (HSO3-/SO32-) = 6.76 x 10-8

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.

[SO32-]0 = 75.0 x

[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

⇒  = 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.