Question

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.

Answer 1

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