Calculate the ph of a 0.20 m solution of iodic acid (hio3, ka = 0.17).

Iodic acid partially dissociates into H + and IO3-

Assuming that x is the concentration of H + at equilibrium, and sine the equation says the same amount of IO3 - will be released as that of H+, its concentration is also X. The formation of H + and IO3 - results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x

Ka = [H+] [IO3-] / [HIO3];

Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20;

At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x;

so 0.17 = x² / (0.20 - x);

Solving for x using the quadratic formula:

x = [H+] = 0.063 M or pH = - log [H+] = 1.2.