Question

(a) Calculate the percent ionization of 0.00360 M hypobromous acid (Ka = 2.5e-09).

% ionization = ________%

(b) Calculate the percent ionization of 0.00360 M hypobromous acid in a solution containing 0.0260 M sodium hypobromite.

% ionization = _________%

Answer 1

a)
BrOH ---> BrO- + H+
0.00360 0 0 (initial)
0.00360-x x x (at equilibrium)

Ka = x*x / (0.00360-x)
since Ka is very small, x will be small and can be ignored as compared to0.00360
above expression thus becomes,
Ka = x*x / (0.00360)
2.5*10^-9 = x^2/ (0.00360)
x = 3.0*10^-6 M

% ionization = 3.0*10^-6 * 100/0.00360
= 0.083 %

b)
BrOH ---> BrO- + H+
0.00360 0.0260 0 (initial)
0.00360-x 0.0260+x x (at equilibrium)

Ka = (0.0260+x)*x / (0.00360-x)
since Ka is very small, x will be small and can be ignored as compared to0.00360 and 0.0260
above expression thus becomes,
Ka = (0.0260)*x / (0.00360)
2.5*10^-9 = (0.0260)*x/ (0.00360)
x = 3.46*10^-10 M

% ionization = 3.46*10^-10 * 100/0.00360
= 9.62*10^-6 %