Question

A 52.0 mL volume of 0.25 M HBr is titrated with 0.50 M KOH. Calculate the pH after addition of 26.0 mL of KOH at 25 ?C.

Answer 1

B

mmol of acid = MV = 52*0.25 = 13mmol of acid

mmol of base = MV = 0.5*26 = 13 mmol of base

then

pH = 7

since strong acid/base titration

C)

HB+ is formed

the next equilibrium is formed, the conjugate acid and water

HB+ + H2O <-> H3O+ + B

Ka by definition since it is an acid:

Ka = [H3O+][B]/[HB+]

Ka can be calcualted as follows:

Ka = Kw/Kb = (10^-14)/(1.8*10^-5) = 5.55*10^-10

and;

assume x = [H3O+]

[H3O+] = x= [B] due to equilibrium

[B] = M-x

[NH4+] = (M1V1)/(50+50) = 0.1

[B] = M-x = 0.1-x

Ka = [H3O+][B]/[HB+]

5.55*10^-10 = x*x/(0.1 -x)

solve for x with quadratic equation

x = H3O+ = 7.45*10^-6 M

[H3O+] = 7.45*10^-6 M

pH = -log(H3O+) = -log(7.45*10^-6) = 5.13

pH = 5.13