Question

A 0.251 L solution of 1.57 M sulfurous acid (Ka1 = 1.5e-2 and Ka2 = 1.0e-7) is titrated with 1.85 M NaOH. What will the pH of the solution be when 0.3600 L of the NaOH has been added?

Answer 1

mol of acid = MV = 1.57 * 0.251 = 0.39407

mol of H+ = 2mol of acid = 2*0.39407 = 0.78814 mol of H+

now...

mol of base = MV = 0.36*1.85 = 0.666 mol of OH-

therefore,

mmol of H+ = 0.78814 -0.666 = 0.12214 mol of acid...

therefore...

mmol of HS- left = 0.12214

mmol of S-2 formed = (0.39407-0.12214) = 0.27193

this is a buffer

since HS- is acting as an acid and S-2 as a conjugate base

therefore

pH = pKa2 + log(S-2/HS-)

pKa2 = -log(Ka2) = -log(10*10^-7)= 7

pH = 7 + log(S-2/HS-)

we calculated --> S-2 = 0.27193; HS- = 0.12214

pH = 7 + log(S-2/HS-)

pH = 7 + log(0.27193/0.12214)

pH = 7.3475 approx