Question

In: Chemistry

Calculate the percent ionization of arsenous acid (H3AsO3) in solutions of each of the following concentrations...

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



(a) 0.211 M

%



(b) 0.497 M

%



(c) 0.886 M

%

Solutions

Expert Solution

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, Ka = 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, Ka = 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, Ka = 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 %


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