The weak acid, HA, is 0.1 % ionized in a 0.2 M solution. (a) What is the equilibrium constant for the dissociation of the acid (Ka)? (b) What is the pH of the solution
.............HA ==> H^+ + A^-
I............0.2..........0...........0
.......0.001*0.2 which is 2E-4 M
C.........-2E-4........+2E-4....+2E-4
E.......0.2-2E-4.......2E-4......2E-4
Plug the E line into the Ka expression and solve for Ka.
The pH = -log(H^+). You have H^+ from above. Solve for pH. Post your work if you get stuck.