Question

A 75 mL solution of .400 M chlorous acid is titrated with a .125 M solution of sodium hydroxide. What is the pH of the solution after 40 mL of the sodium hydroxide solution has been added? Ka for chlorous acid is 1.1 x 10-2. (hint: your answer should contain three significant figures)

Answer 1

Given:

M(HClO2) = 0.4 M

V(HClO2) = 75 mL

M(NaOH) = 0.125 M

V(NaOH) = 40 mL

mol(HClO2) = M(HClO2) * V(HClO2)

mol(HClO2) = 0.4 M * 75 mL = 30 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.125 M * 40 mL = 5 mmol

We have:

mol(HClO2) = 30 mmol

mol(NaOH) = 5 mmol

5 mmol of both will react

excess HClO2 remaining = 25 mmol

Volume of Solution = 75 + 40 = 115 mL

[HClO2] = 25 mmol/115 mL = 0.2174M

[ClO2-] = 5/115 = 0.0435M

They form acidic buffer

acid is HClO2

conjugate base is ClO2-

Ka = 1.1*10^-2

pKa = - log (Ka)

= - log(1.1*10^-2)

= 1.959

use:

pH = pKa + log {[conjugate base]/[acid]}

= 1.959+ log {4.348*10^-2/0.2174}

= 1.26

Answer: 1.26