Question

A 49.2 mL sample of a 0.537 M aqueous acetic acid solution is titrated with a 0.415 M aqueous solution of sodium hydroxide. How many milliliters of sodium hydroxide must be added to reach a pH of 4.502?

?mL​

Answer 1

Volume of NaOH required = 36.4 mL

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pH of the solution can be calculated by the formula, pH = -log[H3O+]

Henderson–Hasselbalch equation for weak acid/base is given by

pH= pKa +log [A-]/[HA], [HA] is the molarity of undissociated weak acid, [A⁻] is the molarity of this acid's conjugate base.

Acetate buffer exists in following equilibrium,

CH3COOH <-==> CH3COO- + H3O+,

Ka = [H3O+][CH3COO-] / [CH3COOH] = 1.8 x 10^-5

pKa = - log(Ka) = -log 1.8 x 10^-5 = 4.745

pH of the buffer required = 4.502

pH= pKa +log [A-]/[HA]

4.502 = 4.745 + log [A-]/[HA]

log [A-]/[HA] = 4.502 - 4.745 = -0.243

[A-]/[HA] = 10^(-0.243) = 0.5715

No. of moles of acetic acid in 49.2 mL, 0.537 M = 0.537 mol/L x 0.0492 L = 0.0264 mol

[A-] = 0.5715 x [HA] = 0.5715 x 0.0264 = 0.0151 mol

Molarity of NaOH = 0.415 M

Volume of 0.415 M NaOH equal to 0.0151 mol = 0.0151 mol / 0.415 Mmol/L = 0.03638 L = 36.38 mL