Question

A 25.0 mL sample of a 0.110 M solution of acetic acid is titrated with a 0.138 M solution of NaOH. Calculate the pH of the titration mixture after 10.0, 20.0, and 30.0 mL of base have been added. The Ka for acetic acid is 1.76x10^-5.

Answer 1

The balanced equation for the neutralization reaction is

CH3COOH + NaOH ------ > CH3COONa(aq) + H2O

Moles of acetic acid, CH3COOH taken = MxV = 0.110 mol/L x 0.025 L = 0.00275 mol

Hence moles of NaOH required to achieve the equivalence point = 0.00275 mol NaOH

(a): When 10.0 mL of 0.138M NaOH is added:

Moles of NaOH added = MxV = 0.138 mol/L x 0.010 L = 0.00138 mol

------------- CH3COOH + NaOH ------ > CH3COONa(aq) + H2O

initl.mol: 0.00275 mol, 0.00138 mol, 0 mol

change: - 0.00138 mol, - 0.00138 mol, +0.00138 mol

final mol: 0.00137 mol, 0 mol, ---------- 0.00138 mol

Now this solution will act as buffer solution, whose pH can be calculated as

pH = pKa + log[CH3COONa] / [CH3COOH] = - log(1.76x10-5) + log(0.00138 / 0.00137) = 4.755 (answer)

(b): When 20.0 mL of 0.138M NaOH is added:

Moles of NaOH added = MxV = 0.138 mol/L x 0.020 L = 0.00275 mol

Hence all of the CH3COOH is neutralized to form 0.00275 mol of CH3COONa

Total volume of the solution, Vt = 25.0 mL + 20.0 mL = 0.045 L

Hence [CH3COONa] = 0.00275 mol / 0.045 L = 0.0611 M

Now CH3COONa will undergo hydrolysis and pH can be calculated from salt hydrolysis formulae

pH = (1/2)x(pKw + pKa + log[CH3COONa] )

=> pH = (1/2)x(14 + 4.755 + log0.0611) = 8.77 (answer)

(c): When 30.0 mL of 0.138M NaOH is added:

volume of extra NaOH added = 30.0 - 20.0 = 10.0 mL

Hence moles of extra NaOH added = 0.138 mol/L x 0.010 = 0.00138 mol

Total vlume = 25.0 mL + 30.0 mL = 0.055 L

[NaOH] = [OH-] = 0.00138 / 0.055 L = 0.0251 M

=> pOH = - log[OH-] = - log(0.0251) = 1.60

=> pH = 14 - 1.60 = 12.40 (answer)