If HOCl has a Ka=3.2*10^-12, what is the pH of 0.1M KOCl?

Answer 1

........OCl^- + HOH --> HOCl + OH^-
initial..0.1............0......0
change...-x..............x.....x
equil...0.1-x...........x.......x

Kb = (Kw/Ka) = (HOCl)(OH^-)/(OCl^-)

Substitute equilibrium numbers from the ICE chart into the Kb expression and solve for OH^-, then convert to pH.