Question

For the titration of 40.0 mL of 0.200 M acetic acid with 0.150 M sodium hydroxide, determine the pH when:

(a) 40.0 mL of base has been added.

(b) 53.3 mL of base has been added.

(c) 66.6 mL of base has been added.

Answer 1

Given

Va = 40 ml = 0.04 L

Ma = 0.200 M = 0.2 mol/L

Mb = 0.15 M = 0.15 mol/L

Ka of acetic acid = 1.75 * 10^-5

CH3COOH (aq) + H2O(l) ⇌ H3O+ (aq) + CH3COO− (aq)

Ka = [H3O+] [CH3COO-] / [CH3COOH] = 1.75 * 10^-5 = x * x / (0.2 -x)

x = 1.86 * 10^-3 M = 0.00186 M

so when no base is added pH = 2.73

when 40 ml = 0.04 L base added

pH = pKa + log [CH3COO-] / [CH3COOH]

[CH3COO-] = No. of moles of NaOH added / total volume = 0.04 L * 0.15 mol/L / (0.04 + 0.04) L = 0.075 mol/L

[CH3COOH] = Intial moles of CH3COOH - moles of NaOH added = 0.04 * 0.2 - 0.04 * 0.15 / 0.04 + 0.04

[CH3COOH] = 0.025 mol/L

pH = - log (1.75 * 10^-5) + log ( 0.075 /0.025) = 5.23 Answer (a)

when 53.3 = 0.0533 ml of base is added

[CH3COO-] = No. of moles of NaOH added / total volume = 0.053 L * 0.15 mol/L / (0.053 + 0.04) L = 0.0855 mol/L

[CH3COOH] = Intial moles of CH3COOH - moles of NaOH added = 0.04 * 0.2 - 0.053 * 0.15 / 0.04 + 0.053

[CH3COOH] = 5.376 * 10^-4 mol/L

pH = - log (1.75 * 10^-5) + log ( 0.0855 /5.376 * 10^-4 ) = 6.95 Answer (b)