Question

(a) Calculate the percent ionization of 0.00840 M hypochlorous acid (Ka = 3e-08). % ionization = % (b) Calculate the percent ionization of 0.00840 M hypochlorous acid in a solution containing 0.0190 M sodium hypochlorite. % ionization = %

Answer 1

a) HClO + H₂O <=> ClO^- + H₃O⁺

K_a= [ClO^-][H₃O⁺]/[HClO]

If [ClO^-] = [H₃O⁺] = x, then [HClO] = 0.00840 M - x

3*10^-8 = x²/(0.00840 M - x)

x² + 3*10^-8 x - 2.52 x 10^-10 = 0

Solving the quadratic equation we get x = [H₃O⁺] = 1.586*10^-5 M

The % of ionization is equal to ratio of ionized acid and the total acid concentration:

(1.586*10^-5 M/0.00840)*100% = 0.1888 % = 0.19%

b) If a solution contains 0.0190 M sodium hypochlorite, [ClO^-] = 0.0190 M

K_a= [ClO^-][H₃O⁺]/[HClO] => [H₃O⁺] = Ka[HClO]/[ClO^-]

We can approximate that [HClO] = 0.00840 M, bescause the ionization will be very suppressed in presence of sodium hypochlorite, therefore

[H₃O⁺] = 3*10^-8 * 0.00840 M/0.0190 M = 1.326*10^-8M

The % of ionization will be:

(1.326*10^-8M/0.00840)*100% = 1.58 x 10^-4 %