Calculate the ph of a 0.800 m kbro solution. ka 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-

Initial0.800 0 0 0

Final0.800-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) / (0.800 - x) = 5.0 x 10^-6

Since Kb is small compared to 0.800, we can neglect the -x term to simplify the calculation.

x^2 / 0.800 = 5.0 x 10^-6

x^2 = 4.0 x 10^-6

x = 2 x 10^-3 = [OH-]

pOH = -log[OH-] = -log (2.0 x 10^-3) = 2.69

pH + pOH = 14.00

pH = 14.00 - pOH = 14.00 - 2.69 = 11.3

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

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