posted by Emma .
Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.140 M HClO(aq) with 0.140 M KOH(aq).HClO is a weak acid with a Ka of 4.0× 10–8. It reacts with strong base to produce ClO–.
(e) after addition of 60.0 mL of KOH
HClO + KOH ==> KClO + H2O
mols HClO = M x L = 0.140 x 0.0500 = about 0.007
mols KOH = 0.0600 x 0.14 = 0.0084
This is past the equivalence point.
mols excess KOH = 0.0084-0.007 = 0.0014
M (OH^-) = 0.0014/0.110 L = ? and convert to pOH, then to pH.