A solution is made initially with 0.550 M HIO₃ (Kc = 0.170). Once the equilibrium below is established, what is the equilibrium concentration of H⁺ ions? HIO₃ (aq) ⇌ H+ (aq) + IO₃⁻ (aq)

Answer 1

HIO3(aq) <==> H+(aq) + IO3-(aq)

Kc = [H+][IO3-] / [HIO3]

Set up ICE table:

HIO3(aq) <==> H+(aq) + IO3-(aq)

0.550...............0...............0.........Initial

-x...................+x...............+x.......Change

0.550 - x.........x.................x........Equilibrium

Plug these values into the Kc expression:

Kc = [x][x] / 0.550 - x and solve for x

0.170 = x² / 0.550 - x

x² = 0.0935 - 0.170x

x² + 0.170x - 0.0935 = 0

x = 0.232 M = [H⁺] ions