Question

A)Calculate the molar concentration of OH− ions in a 0.540 M solution of hypobromite ion (BrO−; Kb=4.0×10−6) .

B)What is the pH of this solution?

Answer 1

A)

BrO- dissociates as:

BrO- +H2O -----> HBrO + OH-

0.54 0 0

0.54-x x x

Kb = [HBrO][OH-]/[BrO-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((4*10^-6)*0.54) = 1.47*10^-3

since c is much greater than x, our assumption is correct

so, x = 1.47*10^-3 M

so,[OH-] = x = 1.47*10^-3 M

Answer: 1.47*10^-3 M

B)

use:

pOH = -log [OH-]

= -log (1.47*10^-3)

= 2.83

use:

PH = 14 - pOH

= 14 - 2.83

= 11.17

Answer: 11.17