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 380 mL of the sodium hydroxide solution has been added? Ka for chlorous acid is 1.1 x 10^-2. (hint: your answer should contain four significant figures)

Answer 1

Given:

M(HClO2) = 0.4 M

V(HClO2) = 75 mL

M(NaOH) = 0.125 M

V(NaOH) = 380 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 * 380 mL = 47.5 mmol

We have:

mol(HClO2) = 30 mmol

mol(NaOH) = 47.5 mmol

30 mmol of both will react

excess NaOH remaining = 17.5 mmol

Volume of Solution = 75 + 380 = 455 mL

[OH-] = 17.5 mmol/455 mL = 0.0385 M

use:

pOH = -log [OH-]

= -log (3.846*10^-2)

= 1.415

use:

PH = 14 - pOH

= 14 - 1.415

= 12.585

Answer: 12.59