Question

Calculate the percent ionization of arsenous acid (H₃AsO₃) in solutions of each of the following concentrations (K_a = 5.1e-10.)

(a) 0.211 M

%



(b) 0.497 M

%



(c) 0.886 M

%

Answer 1

H₃AsO₃(aq) -----------> H₂AsO₃^-(aq) + H⁺(aq)

1) Initailly [H₃AsO₃] = 0.,211 M & [H⁺] = [H₂AsO₃^-] = 0 M

Let at eqb., [H₃AsO₃] = (0.,211 - x) M & [H⁺] = [H₂AsO₃^-] = x M

Thus, K_a = x²/(0.211 - x)

or, 5.1*10^-10 = x²/(0.211 - x)

or, x² + (5.1*10^-10)x - 1.0761*10^-10 = 0

or, x = 1.037*10^-5 M

thus, % ionization = (x/0.211)*100 = 4.91*10^-3 %

2) Initailly [H₃AsO₃] = 0.,497 M & [H⁺] = [H₂AsO₃^-] = 0 M

Let at eqb., [H₃AsO₃] = (0.,497 - x) M & [H⁺] = [H₂AsO₃^-] = x M

Thus, K_a = x²/(0.497 - x)

or, 5.1*10^-10 = x²/(0.497 - x)

or, x² + (5.1*10^-10)x - 2.5347*10^-10 = 0

or, x = 1.592*10^-5 M

thus, % ionization = (x/0.211)*100 = 3.203*10^-3 %

3) Initailly [H₃AsO₃] = 0.,886 M & [H⁺] = [H₂AsO₃^-] = 0 M

Let at eqb., [H₃AsO₃] = (0.,886 - x) M & [H⁺] = [H₂AsO₃^-] = x M

Thus, K_a = x²/(0.886 - x)

or, 5.1*10^-10 = x²/(0.886 - x)

or, x² + (5.1*10^-10)x - 4.5186*10^-10 = 0

or, x = 2.13*10^-5 M

thus, % ionization = (x/0.866)*100 = 2.455*10^-3 %