Calculate the acetate ion concentration in a solution prepared by dissolving 2.80×10-3 mol of HCl(g) in 1.00 L of 5.00×10-1 M aqueous acetic acid (Ka = 1.80×10-5).

Answer 1

I'll call acetic acid HAc. That ionizes as
...................HAc ==> H^+ + Ac^-
I.................0.5 M........0...........0
C.................-x.............x............x
E.............0.5-x.............x............x

Also, the HCl is a strong acid and ionizes completely as
...............HCl ==>H^+ + Cl^-
I.........0.00280 M..0.........0
C.....-0.00280...0.00283..0.00280
E..............0......0.00280...0.00280

The H^+ from the HCl acts as a common ion to the HAc, it forces the ionization equilibrium of HAc to the left and that reduces the H^+ and the Ac^- (from the HAc)
Ka = (H^+)(Ac^-)/(HAc)
For H you substitute x from HAc and 0.00280 from the HCl (as 0.00280 + x), substitute x for the (Ac^-) and for HAc you substitute 0.5 -x. Then solve for x, evaluate 0.00280 + x and convert that total H^+ to pH.