Question

1. During a titration experiment, a student titrated 25.00 mL of a 0.100 M sodium hydroxide solution with 5.10 mL of a 0.250 M sulphuric acid solution. What is the pH of the resulting solution? [8]

Answer 1

pH = 2.80

-----------------------------------

Since NaOH is strong base an H2SO4 is strong acid, neutralization reaction takes place

H2SO4 + 2 NaOH ==> Na2SO4 +2 H2O

1 mole of H2SO4 reacts with 2 moles of NaOH get neutralized.

No. of moles of H2SO4 in 5.10 mL, 0.250 M solution = 0.250 x 5.10/1000 = 0.001275 mol

No. of moles of NaOH in 25.00 mL, 0.100 M solution = 0.100 x 25.00/1000 = 0.0025 mol

0.0025 mol of NaOH reacts with 0.0025/2 = 0.00125 mol of H2SO4.

No. of moles of H2SO4 added = 0.001275 mol

No. of moles of H2SO4 added in excess = 0.001275 - 0.00125 = 2.5 x 10^-5 mol.

Total volume of solution = 25.00 mL + 5.10 mL = 30.10 mL

Molarity of H2SO4 added in excess = 2.5 x 10^-5 mol in 30.10 mL = 2.5 x 10^-5 mol / 0.03010 L = 8.3056 x 10^-4 M

Even though 1 mole of H2SO4 contains 2 mole of H+ ions,

H2SO4(aq) + H2O(l) ==> H3O^+(aq) + HSO4^-(aq), Ka1 = 1

HSO4^-(aq) + H2O(l) <==> H3O^+(aq) + SO4^2-(aq), Ka2 = 0.012

Concentration of H+ from Ka1 = 8.3056 x 10^-4 M (complete ionisation)

pH of solution if we only consider Ka1 = -log (8.3056 x 10^-4) = 3.08

Concentration of H+ from Ka2,

Ka2 = 0.012 = [H3O^+][SO4^2-]/[HSO4^-]

Suppose dissociation of x moles of HSO4^- takes place, then

[H3O^+] = 8.3056 x 10^-4 + x

[SO4^2-] = x

[HSO4^-] = 8.3056 x 10^-4 - x

0.012 = (8.3056 x 10^-4 + x)x/(8.3056 x 10^-4 - x), solving x,

x= 7.3472 x 10^-4

[H3O^+] = 8.3056 x 10^-4 + 7.3472 x 10^-4 = 0.001565 M

pH = -log[H+] = -log (0.001565) = 2.805