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.
