Question

4. Please calculate the resulting pH when 25 mL of 0.10 M acetic acid is titrated with 10 mL of 0.10 M NaOH (K. acetic acid = 1.8 x 10-5): CH,COOH (aq) + NaOH(aq) → CH3COONa (aq) + H20 (1)

Answer 1

Answer

pH = 4.56

Explanation

pKa = -logKa

pKa of CH3COOH = - log( 1.8×10^-5) = 4.74

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

1:1molar reaction

Initial moles of CH3COOH = ( 0.10mol/1000ml) × 25ml = 0.0025mol

moles of NaOH added = ( 0.10mol/1000ml) × 10ml = 0.0010mol

0.0010moles of NaOH react with 0.0010moles of CH3COOH

After reaction

number of moles of CH3COOH = 0.0025mol - 0.0010mol = 0.0015mol

number of moles of CH3COO^- = 0.0010mol

Total volume = 25ml + 10ml = 35ml

[CH3COOH] = (0.0015mol/35ml)×1000ml = 0.04286M

[ CH3COO^-] = ( 0.0010mol/35ml) × 1000ml = 0.02857M

Henderson - Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

pH = 4.74 + log( 0.02857M /0.04286M)

pH = 4.74 - 0.18

pH = 4.56