Find the pH of a 0.000100 M solution of HOBr.
Given: Ka = 2.0 * 10-9
Using a RICE table, Ka = [OBr-][H3O+] / [HOBr]
Let [OBr-] = [H3O+] = x
I got the 2nd degree polynomial: 0 = x2 + 2.0 * 10-9 x - 2.0 * 10-13
Solving for x, I get: x = 4.462 * 10-7 = [H3O+]
Then I find that the pH = 6.350.
The answer supposedly has a pH of 6.262. What did I do wrong?