Determine the volume of strontium hydroxide that needs to be added to create a buffer.

Question

Determine the volume (in mL) of 0.230 M strontium hydroxide (Sr(OH)₂) that must be added to 83.9 mL of 0.0920 M sodium hydrogen sulfite (NaHSO₃) to yield a pH of 6.81.

pK_a of HSO₃ is 7.19 and assume the 5% approximation is valid.

Answer 1

HSO₃^- is a weak acid (pKa = 7.19

SO₃^2- is the conjugate base

Together, these make a buffer.

Initial moles HSO₃^- = 83.9 ml x 1 l / 1000 ml x 0.0920 mol / L = 0.007719 moles

Using the Henderson Hasselbalch equation, we can determine the moles of SO₃^2- needed for pH of 6.81

pH = pKa + log [SO₃^2-] / [HSO₃^-]

6.81 = 7.19 + log R where R = ratio of SO₃^2- / HSO₃^-

-0.38 = log R

R = 0.4169 = SO₃^2- / HSO₃^-

x / 0.007719 - x = 0.4169

x = [SO₃^2-] = 0.00227 moles

In order to obtain 0.00227 moles of SO₃^2-, how many moles of Sr(OH)₂ are needed?

2HSO₃^- + 2OH^- ==> 2SO₃^2- + 2H2O net ionic

0.00227 mols SO₃^2- x 1 mol Sr(OH)2 / 2 mol SO₃^2- = 0.00114 mols Sr(OH)₂ needed

Volume Sr(OH)₂ = 0.00114 mols x 1 L / 0.230 mol = 0.00494 L = 4.94 mls Sr(OH)₂