Finding pH question--?

Question

Calculate the pH of a 0.35 M solution of KBrO (Ka of HBrO = 2.3 x 10^-9)

Please show how you get the answer. Thanks!!! :)

THE CORRECT ANSWER IS: 4.55 but I don't know how the person got that... thanks!! :)

Answer 1

Since HBrO is a very weak acid (Ka is very small), nearly all of it will exist as HBrO rather than BrO- and H+. That means nearly all of the BrO- ions are going to react.

If x mol/L of BrO- ions react to form HBrO, we know x >> 1.0×10^-7, so the concentration of OH- ions will be essentially x and the concentration of H+ ions will be 1.0×10^-14 / x. Also, the concentration of HBrO will be x and the concentration of BrO- will be 0.35 - x.

So, Ka = [H+][BrO-] / [HBrO]
=> 2.3×10^-9 = [(1.0×10^-14 / x).(0.35 - x)] / x
=> 2.3×10^-9 x^2 = (1.0×10^-14) (0.35 - x)
=> 2.3×10^5 x^2 + x - 0.35 = 0
=> x = (-1 + √(1 - 2.3×10^5 (-0.35))) / 4.6×10^-5
= 6.15×10^-4.

So [H+] = 10^-14 / x = 1.63×10^-11
and pH = -log[H+] = 10.8 to 1 d.p.

Answer 2

KBrO is a strong electrolyte

KBrO >> K+ + BrO-

The equilibrium is

BrO- + H2O <> HBrO + OH-
initial concentration
0.35

at equilibrium
0.35-x . . . . . . . . . . x . . . . . .x

This is the hydrolysis reaction and its equilibrium constant

Kh = Kw / Ka = 4.35 x 10^-6

4.35 x 10^-6 = (x)(x) / 0.35 - x

x = [OH-] = 0.00123

pOH = 2.91

pH = 11.09


pH can not be 4.55 because KBrO derives from a strong base( KOH )and a weak acid (HBrO) and in this case pH > 7