Calculate the pH for this case in the titration of 500 mL of

Calculate the pH for this case in the titration of 50.0 mL of 0.230 M HClO(aq) with 0.230 M KOH(aq). The ionization constant for HClO = 4.0*10^-8

ONLY ANSWER THIS:

(d) after addition of 50.0 mL of KOH

I get 9.73 and this is wrong, not sure why.

no of moles of HClO = molarity * volume in L

= 0.23*0.05

= 0.0115 moles

no of moles of KOH = molarity * volume in L

= 0.23*0.05

= 0.0115moles

HClO + KOH ------------------> KClO + H2O

I 0.0115 0.0115 0

C -0.0115 -0.0115 0.0115

E 0 0 0.0115

conc of ClO^- = no of moles/total volume

= 0.0115/0.1 = 0.115 M

ClO^- + H2O ----------------> HClO + OH^-

I 0.115 0 0

C -x +x +x

E 0.115-x +x +x

Kb = Kw/Ka

= 1*10^-14/4*10^-8

= 2.5*10^-7

Kb = [HClO][OH^-]/[ClO^-]

2.5*10^-7 = x*x/(0.115-x)

2.5*10^-7 *(0.115-x) = x^2

x = 0.00017

[OH^-] = x = 0.00017M

POH = -log[OH^-]

= -log0.00017

= 3.7695

PH = 14-POH

= 14-3.7695 = 10.23 >>>>answer