Question

Calculate the percent ionization of hypobromous acid (HBrO) in solutions of each of the following concentrations (Ka = 2.5e-09.) (a) 0.263 M (b) 0.477 M (c) 0.868 M

Answer 1

HBrO ßà H+ + BRO-

C            0        0   (initial)

C-x           x       x (at equilibrium)

Ka = x*x / (C-x)

since Ka is very small, x will be small and can be ignored as compared to C

Above expression will thus become,

Ka = x*x / C

x = sqrt (Ka*C)

percent ionization = x*100/C

                   =(100/C) sqrt (Ka*C)

a)

percent ionization = (100/0.263)*sqrt(2.5*10^-9 * 0.263)

                   = 0.00975 %

b)

percent ionization = (100/0.477)*sqrt(2.5*10^-9 * 0.477)

                   = 0.00724 %

c)

percent ionization = (100/0.868)*sqrt(2.5*10^-9 * 0.868)

                   = 0.00537 %