Calculate the pH after the addition of 50 mL of KOH in the titration of 50.0 mL of 0.250 M HClO(aq) with 0.250 M KOH(aq). The ionization constant for HClO is 4 x 10^-8.
I'm confused about what to do after I find the Kb value....the work I've done on it so far is in the comments

Question

Calculate the pH after the addition of 50 mL of KOH in the titration of 50.0 mL of 0.250 M HClO(aq) with 0.250 M KOH(aq). The ionization constant for HClO is 4 x 10^-8.
I'm confused about what to do after I find the Kb value....the work I've done on it so far is in the comments

anonymous · Answer

(50mL) (.25M )= 12.5 mmol ClO

[ClO]  =  (12.5 mmol)/(50mL + 50 mL) = .125 M

Kb = (1 x 10^-14)/(4 x 10^-8)

Kb = 2.5 x 10^-7

What do I do next?