What is the pH of a 0.94 M solution of HBrO that is also 0.59 M in magnesium hypobromite, the salt of it's conjugate base? The Ka for HBrO = 2.8 X 10-9.

Answer 1

.................HBrO ==> H^+ + BrO^-
initial..........0.94.....0......0
change...........-x.......x......x
equilibrium....0.94-x.....x.......x

Mg(BrO)2 ==> Mg^+2 + 2BrO^- (100% ionized)
0.59M........0........0
-0.59.......0.59.....2*0.59
0...........0.59......1.18

Ka = (H^+)(BrO^-)/(HBrO)
Substitute the ICE charts into Ka expression.
(H^+) = x
(BrO^-) = x from HBrO equation + 1.18 from Mg(BrO)2 equation or 1.18+x for (BrO^-).
(HBrO) = 0.94-x.
Solve for (H^+) or x, then convert to pH.
Hint: You can make the equation much simpler to solve if you assume 1.18+x = 1.18 and also assume 0.94-x = 0.94.

Answer 2

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