Question

How many milliliters of 1.0 M NaOH should be added to a 100 mL solution containing 12.212g of benzoic acid (F.W. 122.12g ; Ka = 6.28 x 10-5) to get a pH of 4.8?

Answer 1

mass benzoic acid = 12.212 g

moles benzoic acid = (mass benzoic acid) / (molar mass benzoic acid)

moles benzoic acid = (12.212 g) / (122.12 g/mol)

moles benzoic acid = 0.100 mol

K_a = 6.28 x 10^-5

pK_a = -log(K_a)

pK_a = -log(6.28 x 10^-5)

pK_a = 4.202

Let moles NaOH added = x

moles conjugate base formed = moles NaOH added

moles conjugate base formed = x

moles benzoic acid remaining = (initial moles benzoic acid) - (moles conjugate base formed)

moles benzoic acid remaining = 0.1 mol - x

According to Henderson - Hasselbalch equation,

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

pH = pK_a + log(moles conjugate base formed / moles benzoic acid remaining)

4.8 = 4.202 + log[x / (0.1 mol - x)]

log[x / (0.1 mol - x)] = 4.8 - 4.202

log[x / (0.1 mol - x)] = 0.598

x / (0.1 mol - x) = 10^0.598

x / (0.1 mol - x) = 3.96

x / 3.96 = 0.1 mol - x

0.252x = 0.1 mol - x

0.252x + x = 0.1 mol

1.252x = 0.1 mol

x = (0.1 mol) / (1.252)

x = 0.08 mol

volume NaOH = (moles NaOH) / (molarity NaOH)

volume NaOH = (0.08 mol) / (1.0 M)

volume NaOH = 0.08 L

volume NaOH = 80 mL