Titration help?

Question

2. If the pH of a 1.50 M weak acid solution is 4.00.

a. What is the % dissociation of the acid.

b. What is the Ka of the acid.

i forget all of this stuff. math and explanation please! thanks!

Answer 1

pH = 4.00 so [H+] = 1x10^-4

(pH = -log[H+]; [H+] = 10^-pH)

0.0001 M / 1.50 M = 0.0067% dissociation (0.000067)

HA <--> H^+1 + A^-1

Ka = [H+][A-] / [HA]

Ka = 0.0001 x 0.0001 / (1.50 - 0.0001) = 0.0001^2/1.50

Ka = 6.67x10^-9

.