Question

Please show all steps.

If you titrate 25.00 mL of a 0.09797 M solution of HBrO, having Ka = 2.8 x 10^-9, with 0.05000 M Ba(OH)₂ solution:

a) Calculate the number of mL of Ba(OH)₂ solution needed to reach the equivalence point.

b) Calculate the pH inititally, before any Ba(OH)₂ solution is added.

c) Calculate the pH at halfway.

d) Calculate the pH at the equivalence point.

Answer 1

A) nO of mol of HBrO = 25*0.09797/1000 = 0.00245 mol

   2HBrO + Ba(OH)2 ---> BaBrO + H2O

2 mol HBrO = 1 mol Ba(OH)2

nO of mol of Ba(OH)2 = 0.00245/2 = 0.001225 mol

volume of Ba(OH)2 = n/v = 0.001225/0.05 = 0.0245 L

                      = 24.5 ml

b) at half equivalence point , pH = pka

   pka = -log(2.8*10^-9) = 8.55

   pH = 8.55

c) at the equivalence point

   concentration of salt formed = 0.001225/(24.5+25)*1000 = 0.0247 M

pH of salt = 7+1/2(pka+logC)

              = 7+1/2(8.55+log0.0247)

              = 10.47