Asked by Francesca

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.150 M HClO(aq) with 0.150 M KOH(aq). The ionization constant for HClO is 4.0x10^-8.

pH before the addition of any KOH?
pH after the addition of 25 mL of KOH?

Please show the steps to solving this problem.

Answers

Answered by DrBob222
HClO + KOH ==> KClO + H2O

At the beginning you have pure HClO.
.......HClO ==> H^+ + ClO^-
I.....0.150......0......0
C........-x......x......x
E...0.150-x......x......x

Substitute the E line into the Ka expression and solve for x = H^+ and convert to pH.

After reaction:
HClO + KOH ==> KClO + H2O

mols HClO initially = M x L = ?
mols KOH added = M x L = ?
How much of the HClO is used? how much is left?
How much of the KOH has been used? How much is left.
You have prepared a buffer solution at this point and the pH can be determined by using te Henderson-Hasselbalch equation. Substitute and solve for pH. Post your work if you get stuck.
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