Question

A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (pKa1) and 7.172 (pKa2).

A) Calculate pH at the first equivalence point.

B) Calculate pH at the second equivalence point.

Answer 1

a)

in the 1st equivalence point:

pH can be calcualted between pKA1 and pKA2, since it is exactly in between

i.e. half points will give the middle point

pH = 1/2*(pKa1 + pKa2)

pH = 1/2*(1.857+7.172) = 4.51

b)

in 2nd equivalence point:

SO3-2 is present, which ionizes to form

SO3-2 + H2O <-> HSO3- + OH-

Kb = [HSO3-][OH-]/[SO3-2]

Kb = Kw/KA = (10^-14)/(10^-7.172) = 1.485*10^-7

[HSO3-] = x= [OH-]

[SO3-2] = mmol of SO3-2 / Vtotal

mmol of H2SO3 = mmol of SO3-2 = MV = 0.112*60 = 6.72

V of NaOh required =2*mmol/M = 2*6.72/0.112 = 120 mL

Total V in equivlanec epoints = 60+120 = 180 mL

[SO3-2] = 6.72/120 = 0.056 M

now...

substitute

Kb = [HSO3-][OH-]/[SO3-2]

1.485*10^-7 = x*x/(0.056 -x)

x = [OH-] = 9.11*10^-5

pOH = -log(OH) = -log( 9.11*10^-5) = 4.04

pH = 14-pOH = 14-4.04 = 9.96 approx