Question

We place 0.565 mol of a weak acid, HA, and 11.6 g of NaOH in enough water to produce 1.00 L of solution. The final pH of this solution is 4.88 . Calculate the ionization constant, K_a, of HA.

Answer 1

Mass of NaOH = 11.6 g, Molar mass of NaOH = 40 g/mol

Number of moles of NaOH = mass of NaOH in gram / Molar mass of NaOH = 11.6 g / 40 g/mol = 0.29 mol

Reaction :- HA(aq) + NaOH(aq) ----> NaA(aq) + H2O (l)

From reaction, 1.0 mol of HA reacts with 1.0 mol of NaOH produces 1.0 mol of NaA then 0.29 mol of NaOH will react with 0.29 mol of HA will produce 0.29 mol of NaA.

Number of moles of HA remaining = (0.565 - 0.29) mol = 0.275 mol

From Henderson equation

pH = pKa + log {[Conjugate base] / [Acid]}

4.88 = pKa + log {[NaA] / [HA]}

4.88 = pKa + log ( 0.29 mol / 0.275 mol)

4.88 = pKa + log 1.05 = pKa + 0.02

pKa= 4.88 - 0.02 = 4.86

We know Ka = 10^-pKa = 10^-4.86 = 1.38 * 10^-5

Ka = 1.38 * 10^-5