Question

Calculate the pH and pOH of a 0.50 M solution of Acetic Acid. The Ka of HOAc is 1.8 x10-5

[H3O+]/(0.50-0.00095)=1.8x10-5 where did they get the .00095, its said to take that as the second assumption

[H3O+]=9.4x10-4

[H3O+]/(0.50-0.00094)=1.85x10-5

[H3O+]=9.4x10-4 (this is said to be the third assumption)

Please, please help I really dont understand this, please show all steps and be as descriptive as possible.

Answer 1

         CH3COOH(aq) + H2O(l) <-----> CH3COO^-(aq) + H3O^+(aq)

initial      0.5 M                             0 M         0 M

change       x                                                         x              x

equilibrium    0.5-x M                                           x              x

Ka = [CH3COO-][H3O+]/[CH3COOH]

1.8*10^-5 = X^2/(0.5-X)

X = 0.003

So that,

at equilibrium

[H3O+] = X = 0.003 M

pH = -log(H3O+)

   = -log0.003

   = 2.52

pOH = 14 - pH

    = 14 - 2.52

    = 11.48