Question

Determine the percent ionization of the following solutions of phenol at 25 degrees Celsius.

a) 0.681 M

b) 0.250 M

c) 1.68 x 10^-6 M (enter answer in scientific notation)

Answer 1

C₆H₅OH(aq) <---------> C₆H₅O^-(aq) + H⁺(aq) ; K_a = {[C₆H₅O^-]*[H⁺]}/[C₆H₅OH] = 1.6*!0^-10

1) Initially, [C₆H₅OH] = 0.681 M ; [C₆H₅O^-] = [H⁺] = 0 M

Let at eqb., [C₆H₅OH] = (0.681-x) M ; [C₆H₅O^-] = [H⁺] = x M

Thus, 1.6*10^-10 = x²/(0.681-x)

or, x² + (1.6*10^-10)x - (1.1*10^-10) = 0

or, x = 1.049*10^-5 M

Thus, % ionization = (x/initial concentration of phenol) = (1.049*10^-5/0.681)*100 = (1.54*!0^-3)%

2) Initially, [C₆H₅OH] = 0.25 M ; [C₆H₅O^-] = [H⁺] = 0 M

Let at eqb., [C₆H₅OH] = (0.25-x) M ; [C₆H₅O^-] = [H⁺] = x M

Thus, 1.6*10^-10 = x²/(0.25-x)

or, x² + (1.6*10^-10)x - (0.4*10^-10) = 0

or, x = 6.324*10^-6 M

Thus, % ionization = (x/initial concentration of phenol) = (6.324*10^-6/0.25)*100 = (2.53*!0^-3)%

3) Initially, [C₆H₅OH] = 1.68*10^-6 M ; [C₆H₅O^-] = [H⁺] = 0 M

Let at eqb., [C₆H₅OH] = (1.68*10^-6-x) M ; [C₆H₅O^-] = [H⁺] = x M

Thus, 1.6*10^-10 = x²/(1.68*10^-6-x)

or, x² + (1.6*10^-10)x - (2.688*10^-16) = 0

or, x = 1.265*10^-5 M

Thus, % ionization = (x/initial concentration of phenol) = (1.632*10^-8/1.68*10^-6)*100 = (0.9714%