What is the ph of a solution of 0.750 m kh2po4, potassium dihydrogen phosphate?

Ka1 = 7.5 x 10^-3 and Ka2 = 6.2 x 10^-8.

Since there is a very large difference between Ka2 and Ka3, we only need to be concerned how much H3O+ is produced by the 2nd ionization step, corresponding to Ka2.

Ka2 = [H3O+][HPO4 2-] / [H2PO4-] = (x)(x) / (0.750 - x) = 6.2 x 10^-8

Because Ka2 is so small, the -x term after 0.750 will be negligible, so we can delete it to simplify the math.

x^2 / 0.750 = 6.2 x 10^-8
x^2 = (0.750)(6.2 x 10^-8) = 4.7 x 10^-8
x = 2.2 x 10^-4 M = [H3O+]

pH = -log [H3O+] = -log (2.2 x 10^-4) = 3.67

The answer is 65.3312