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 base0acid0

pH = 7.17 + log SO32-0HSO3-0

 

We want pH = 7.00 ⇒ SO32-0HSO3-0 = 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 100 - X100

[HSO3-]0 = 150 x X100

 

SO32-0HSO3-0 = 10-0.17

⇒ 75.0 × 100 - X150 × 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.