Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.230 M HClO (aq) with 0.230 M KOH (aq). The ionization constant (Ka) for HClO is 4.00 x 10^-8.

Question

(b) after addition of 25.0 mL of KOH
(d) after addition of 50.0 mL of KOH

Answer 1

You must learn to recognize where you are on the titration curve.
1. Start by calculating mL KOH to reach the equivalence point.
50.0 x 0.230 = mL x 0.230
Obviously it will be 50.0 mL KOH and that will answer part. That is determined by the hydrolysis of the salt, KClO.
.........ClO^- + HOH ==> HClO + OH^-
initial..0.150M............0.....0
change....-x...............x.....x
equil....0-.150-x...........x.....x

Kb for ClO^- = (Kw/Ka for HClO) = (x)(x)/(0.150-x)
Solve for x = (OH^-) and convert to pH.

For part a.
........HClO + KOH ==> KClO + H2O
....50*0.230 + 25.0*0.230 .....
mmol....11.5.. 5.75......0.....0
change..-5.75..-5.75....5.75...5.75
equil..4.25......0.......5.75..5.75
Ka = (H^+)(ClO^-)/(HClO)
Plug in x for (H^+), get HClO and ClO^- from the ICE chart, solve for x and convert to pH.
Alternatively, you can use the Henderson-Hasselbalch equation.