Calculate the ph of a 1.60 m kbro solution. k a for hypobromous acid, hbro, is 2.0 × 10-9

KBrO is produced by reaction of a strong base (KOH) with a weak acid (HBrO).

KOH + HBrO ==> KBrO + H2O

This salt, when diisoved in H2O, produce an alkaline solution (produced from a STRONG base). Since OH- is produced, this is a Kb problem.

Molarity . . . . . . . . .BrO- + H2O <==> HBrO + OH-

Initial1.6 0 0 0

Final1.6-x x x x

Kb BrO- = Kw / Ka HBrO = (1 x 10^-14) / (2.0 x 10^-9) = 5.0 x 10^-6

Kb = [HBrO][OH-] / [BrO-] = (x)(x) / (1.6 - x) = 5.0 x 10^-6

Since Kb is small compared to 1.6, we can neglect the -x term to simplify the calculation.

x^2 / 1.6 = 5.0 x 10^-6

x^2 = 8.0 x 10^-6

x = 2.8 x 10^-3 = [OH-]

pOH = -log[OH-] = -log (2.0 x 10^-3) = 2.55

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 2.55 = 11.4

h let me think

dynamic equilibrium is the answer.