Question

The weak acid, HA, is 2.4% dissociated in a 0.22M Solution. Calculate (a) the Ka, (b) the pH of the solution, and (c) the amount of 0.1M KOH required to neutralize 550 ml of the weak acid solution.

Answer 1

a) Ka = C*X^2

     X = degree of dissociation = 2.4/100 = 0.024

     C = concentration of HA = 0.22 M

   Ka = 0.22*(0.024)^2 = 1.27*10^-4

b) pH of weak acid = 1/2(pka-logC)

   pka = -logka = -log(1.27*10^-4) = 3.9

                   = 1/2(3.9-log0.22)

          = 2.28

c) no of mol of weak acid = V*M

   = 0.55*0.22

   = 0.121 mol

no of mol of KOH required = 0.121 mol

volume of KOH required = n/M = 0.121/0.1 = 1.21 L