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.
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.
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