What is the PH at 25°C of the solution obtained by dissolving a 5.0g tablet of aspirin in 0.50L of water? The tablet contains 0.325g of the acid HC9H7O4. The acid is monoprotic and Ka=3.3*10^-4 at 25°C.
HC9H7O4 = HA and molar mass = 180
moles HA = grams/molar mass = 0.325/180 = 0.00181 and
molarity = 0.00181 moles/0.5 L = 0.00361 M
..................HA ==> H^+ + A^-
I...............0.00361.....0.......0
C..................-x...........x........x
E.........0.00361-x........x.........x
Ka = (H^+)(A^-)/(HA) = 3.3E-4
Plug the E values into the ka expression and solve for x = (H^+)
Then pH = -log(H^+)
Post your work if you get stuck. This is quite a weak acid AND it is a very dilute solution; therefore, the quadratic MUST be employed. In addition, I have not included the small H^+ from the ionization of water but I think that can be ignored; i.e., it won't affect the pH that much.