Question

Calculate the percent ionization of hypobromous acid (HBrO) in solutions of each of the following concentrations (Ka = 2.5e-09.)

(a) 0.142 M

= %

(b) 0.448 M

= %

(c) 0.625 M

= %

Answer 1

Ka = 2.5 x 10^-9

a) 0.142 M HBrO

[H+] = sqrt(Ka x C)

        = sqrt (2.5 x 10^-9 x 0.142)

        = 1.88 x 10^-5 M

% ionization = ([H+] / C ) x 100

                    = (1.88 x 10^-5 / 0.142 ) x 100

% ionization = 0.0133 %

(b)

[H+] = sqrt(Ka x C)

        = sqrt (2.5 x 10^-9 x 0.448)

        = 3.35 x 10^-5 M

% ionization = ([H+] / C ) x 100

                    = (3.35 x 10^-5 / 0.448 ) x 100

% ionization = 7.47 x 10^-3 %

(c)

[H+] = sqrt(Ka x C)

        = sqrt (2.5 x 10^-9 x 0.625)

        = 3.95 x 10^-5 M

% ionization = ([H+] / C ) x 100

                    = (3.95 x 10^-5 / 0.625 ) x 100

% ionization = 6.32 x 10^-3 %