Question

We place 0.133 mol of a weak acid, HA, in enough water to produce 1.00 L of solution. The final pH of the solution is 1.23 . Calculate the ionization contant, K_a, of HA.

Answer 1

first find molarity of weak acid HA from given information

no of moles of HA = 0.133 moles , volume of HA solution = 1 liter

molarity = no of moles/volume in liter , substitute values to find molarity of HA

molarity of weak acid HA = 0.133/1 = 0.133 M

given pH of solution = 1.23 , we know that [H⁺] = 10^-pH , substitute value [H⁺] = 10^-1.23 = 0.05888 M

HA acid dissociates as

HA +H₂O ⇌ A_(aq)^- + H₃O⁺

K_a = [A_(aq)^-][H₃O⁺] / [HA]

HA is monoprotic acid therefore [H₃O⁺] = [A^-] = 0.05888 M

Substitute the values in equation  

K_a = (0.05888 x 0.05888)/ 0.133 = 260.66574 x 10^-4

ans ionization constant Ka of weak acid HA is 260.6657 x 10^-4