Question

Do the calculations for the titration of 50.00 mL of a 0.1000 M solution of H2SO3 with a 0.2000 M solution of NaOH. Calculate the pH after the addition of 0.00, 12.50, 25.00, 37.50, 50.00, and 60.00 mL of NaOH.

Ka1(H2SO3)=1.23×10-2; Ka2(HSO3-)=6.60×10-8.

Please show all of your work.Thanks!

Answer 1

(a) No base added

H2SO3 <==> H+ + HSO3-

let x amount has dissociated,

Ka1 = [H+][HSO3-]/[H2SO3]

1.23 x 10^-2 = x^2/0.1

x = [H+] = 0.0351 M

pH = -log[H+] = 1.455

(b) 12.5 ml of 0.2 M NaOH added

moles of acid = 5 x 10^-3 mols

moles of base = 2.5 x 10^-3 mols

remaining moles of acid = 2.5 x 10^-3 mols

this is half equivalence point

pH = pKa1 = -log(1.23 x 10^-2) = 1.91

(c) 25 ml of 0.2 M NaOH added

moles of acid = 0.1 x 0.05 = 5 x 10^-3 mols

moles of base = 0.2 x 0.025 = 5 x 10^-3 mols

This is first equivalence point

pH = pka1 + pKa2/2 = 1.91 + 7.18 = 4.545

(d) 37.50 ml of 0.2 M NaOH

moles of NaOH = 0.2 x 0.0125 = 2.5 x 10^-3 mols

[SO3^2-] = 2.5 x 10^-3/0.0875 = 0.0286 M

[HSO3-] = 0.0286 M

pH = pKa2 + log([SO3^2-]/[HSO3-])

      = 7.18 + log(0.0286/0.0286)

      = 7.18

(e) 50 ml of 0.2 M NaOH

moles of NaOH = 0.2 x 0.025 = 5 x 10^-3 mols

[SO3^-] = 5 x 10^-3/0.1 = 0.05 M

this is second equivalence point,

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

with x amount of salt hydrolyzed,

Kb2 = Kw/Ka2 = 1 x 10^-14/6.6 x 10^-8 = x^2/0.05

x = [OH-] = 8.704 x 10^-5 M

pOH = 4.06

pH = 9.94

(f) 60 ml of 0.2 of NaOH added

excess moles of NaOH = 0.2 x 0.01 = 2 x 10^-3 mols

x = [OH-] = 0.0182 M

pOH = 1.74

pH = 12.26