Question

Find the pH of a 0.183 M aqueous solution of hypobromous acid (HOBr), for which Ka = 2.06 × 10–9.

Answer 1

HOBr dissociates as:

HOBr -----> H+ + OBr-

0.183 0 0

0.183-x x x

Ka = [H+][OBr-]/[HOBr]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((2.06*10^-9)*0.183) = 1.942*10^-5

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

so, x = 1.942*10^-5 M

So, [H+] = x = 1.942*10^-5 M

use:

pH = -log [H+]

= -log (1.942*10^-5)

= 4.7118

Answer: 4.71