Question

A 5.55 −g sample of a weak acid with Ka=1.3×10−4 was combined with 5.20 mL of 6.20 M NaOH and the resulting solution was diluted to 750 mL. The measured pH of the solution was 4.25. What is the molar mass of the weak acid?

Answer 1

pKa = - log Ka = - log (1.3 * 10^-4) = 3.886

5.20 mL of 6.20 M NaOH = 0.00520 L * 6.20 mole / L = 0.03224 mole.

let there are x mole weak acid.

mole of salt = (x - 0.03224) mole.

resulting solution is buffer solution.

Using Henderson equation,

pH = pKa + log [salt] / [acid]

or

4.25 = 3.886 + log [0.03224 / (x - 0.03224)]

or

x = 0.04619 mole.

thus

mole of weak acid = 0.04619 mole.

molar mass of weak acid = mass / mole = 5.55 g / 0.04619 mole = 120.16 g / mole