Question

Consider the titration of a 12.0 mL solution containing 0.200 mol/L acetic acid (CH3COOH, Ka = 1.8 x 10-5 ), with a solution of 0.250 mol/L KOH. (12 points total) i) Determine the pH of the solution when 4.80 mL of KOH has been added.

ii) Determine the pH of the solution when 14.00 mL of KOH has been added.

Answer 1

(i) initial concentration of acetic acid = 0.200 mol/L

volume of acetic acid solution = 12.0 mL

initial moles of acetic acid = (initial concentration of acetic acid) * (volume of acetic acid solution)

initial moles of acetic acid = (0.200 mol/L) * (12.0 mL)

initial moles of acetic acid = 2.40 mmol

Similarly, moles of KOH added = 1.20 mmol

moles of conjugate base (acetate) formed = moles of KOH added

moles of conjugate base (acetate) formed = 1.20 mmol

moles of acetic acid remaining = (initial moles of acetic acid) - (moles of acetate formed)

moles of acetic acid remaining = (2.40 mmol) - (1.20 mmol)

moles of acetic acid remaining = 1.20 mmol

K_a = 1.8 x 10^-5

pK_a = -log(K_a)

pK_a = -log(1.8 x 10^-5)

pK_a = 4.74

According to Henderson - Hasselbalch equation,

pH = pK_a + log([conjugate base] / [weak acid])

pH = pK_a + log(moles of acetate formed / moles of acetic acid remaining)

pH = 4.74 + log(1.20 mmol / 1.20 mmol)

pH = 4.74 + log(1)

pH = 4.74 + 0

pH = 4.74

(ii) initial moles of acetic acid = 2.40 mmol

moles of KOH added = (concentration of KOH) * (volume of KOH)

moles of KOH added = (0.250 mol/L) * (14.00 mL)

moles of KOH added = 3.50 mmol

excess moles of OH^- = (moles of KOH added) - (initial moles of acetic acid)

excess moles of OH^- = (3.50 mmol) - (2.40 mmol)

excess moles of OH^- = 1.10 mmol

Total volume = (volume acetic acid) + (volume KOH)

Total volume = (12.0 mL) + (14.0 mL)

Total volume = 26.0 mL

[OH^-] = (excess moles of OH^-) / (total volume)

[OH^-] = (1.10 mmol) / (26.0 mL)

[OH^-] = 0.0423 M

pOH = -log[OH^-]

pOH = -log(0.0423 M)

pOH = 1.37

pH = 14 - pOH

pH = 14 - 1.37

pH = 12.63