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.

The answer is d.potassium