Be sure to answer all parts What is the original molarity of

Be sure to answer all parts What is the original molarity of a solution of a weak acid whose K, is 3.5 x 10 S and whose pH is 5.20 at 25° ×10 M (Enter your answer in scientifie notation.)

Solution

use:
pH = -log [H+]
5.20 = -log [H+]
[H+] = 6.31*10^-6 M
HA dissociates as:

HA          ----->     H+   + A-
c                 0         0
c-x               x         x

[H+] = x = 6.31*10^-6 M

Ka = [H+][A-]/[HA]
Ka = x*x/(c-x)
3.5*10^-5 = (6.31*10^-6)*(6.31*10^-6) / (c - 6.31*10^-6)
c - 6.31*10^-6 = 1.14*10^-6
c = 7.45*10^-6 M

Answer: 7.45*10^-6 M