Question

What is the pH at the equivalence point in the titration of a 22.6 mL sample of a 0.457 M aqueous hypochlorous acidsolution with a 0.311 M aqueous barium hydroxide solution?

pH =

Answer 1

2HClO(aq) + Ba(OH)2(aq) ----------> Ba(ClO)2(aq) + 2H2O(l)

2 moles           1 mole

HClO                                                 Ba(OH)2

M1   = 0.457M                                     M2 = 0.311M

V1   = 22.6ml                                       V2 =

n1 = 2                                                   n2 =1

             M1V1/n1     =    M2V2/n2

                  V2          =   M1V1n2/n1M2

                                 = 0.457*22.6*1/2*0.311    = 16.6ml

no of moles of Ba(ClO)2 = molarity * volume in L

                                    =       0.457*0.0226    = 0.0103282moles

Total volume of solution = 22.6+16.6   = 39.2ml   = 0.0392L

molarity of Ba(OCl)2   = 0.0103282/0.0392   = 0.2635M

   Ba(OCl)2 (aq) -----------> Ba^2+ (aq) + 2ClO^- (aq)

0.2635M                                                     2*0.2635M

molarity of ClO^-     = 0.527M

                  ClO^- (aq) + H2O -------------> HClO(aq) + OH^- (aq)

I                0.527                                         0                     0

C               -x                                              +x                    +x

E              0.527-x                                        +x                   +x

                      Kb = Kw/Ka

                            = 1*10^-14/3*10^-8   = 3.34*10^-7

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

                3.34*10^-7   = x*x/0.527-x

               3.34*10^-7*(0.527-x)   = x^2

                    x   = 0.00042

            [OH^-]    = x   = 0.00042M

            POH   = -log[OH^-]

                     = -log0.00042

                      = 3.3767

        PH     = 14-POH

                   = 14-3.3767    = 10.6233>>>>>answer

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