Question

A buffer solution contains 0.15 moles of both HSO4- anion and SO4^2- anion. An aliquot of 0.05 moles of NaOH is added to the buffer solution. What is the new pH of the buffer solution?

The answers are multiple choice and they are 6.66, 7.04, 6.96, 6.74, or 7.26

Answer 1

given concentrations of HSO₄^- and SO₄^2- = 0.15 moles

The normal reaction between in a solution is given by

0.15 NA x x are the concentrations corresponding molecules in reaction

by Henderson Hasselbach equation pKa of HSO₄^- = 1.92

pH of buffer when 0.00 moles of NaOH added is given by

pH = 1.92

when 0.05 moles of NaOH is added then

0.15 0.05 0.15 NA

if we neutralize acid by base then it gets diluted so the concentration of HSO₄^- becomes 0.15 - 0.05 = 0.10 moles and concentration of SO₄^2- will be increased 0.15 + 0.05 = 0.20 moles therefore

pH = 1.92 - 0.30 = 1.61