Question

The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 °c is 4.48. what is the value of ka for hbro?

Answer 1

hbro dissociate as follows
HBro---> H+ + BrO-
Ka= (H+)(BrO-) / HBro
PH = -log (H+)
therefore (H+) = 10^-4.48= 3.31 x10^-5
ka is therefore= ( 3.31 x 10^-5)^2/0.55=1.99 x10^-9