A student is asked to determine the value of K a for hypochlorous acid by titrat
ID: 499087 • Letter: A
Question
A student is asked to determine the value of Ka for hypochlorous acid by titration with barium hydroxide.
The student begins titrating a 47.9 mL sample of a 0.543 M aqueous solution of hypochlorous acid with a 0.267 M aqueous barium hydroxide solution, but runs out of standardized base before reaching the equivalence point. The student's last observation is that when 17.0 milliliters of barium hydroxide have been added, the pH is 7.162.
What is Ka for hypochlorous acid based on the student's data?
Explanation / Answer
mol of HClO present initially = M*V
= 0.543 M *47.9 mL
= 26 mmol
mol of Ba(OH)2 added = M*V
= 0.267 M * 17.0 mL
=4.54 mmol
This is equal to 4.54*2 = 9.08 mmol of OH-
HClO and OH- will react to form ClO-
mol of ClO- formed = 9.08 mmol
mol of HClO remaining = 26 - 9.08 = 16.9 mmol
use:
pH = pKa + log {[ClO-]/[HClO]}
7.162 = pKa + log (9.08/16.9)
pKa = 7.432
now use:
pKa = -log Ka
7.432 = -log Ka
Ka = 3.70*10^-8
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