Acids and Bases Chem Help?

HA ⇌ H{+} + A{-}

Let "z" be the concentration (in mol/L) of H{+} ions at equilibrium, [H=].

Then [A-] is also z.

[HA] = 0.28 - z

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

4.0×10^−6 = z × z / (0.28 - z)

Solve for z algebraically, discarding the negative solution:

z = 0.00105630 = 0.0011 mol/L H{+}

I agree that it looks like some sort of rounding trouble.  Rounding the results from a quadratic formula is tricky business.  What I did above is probably not exactly correct.

I could make an argument for only "0.001 M H{+}.

pka = -log(4.0*10^(-6)) = 5.38

pH = 1/2(pka-logC)

   = 1/2(5.38-log0.28) = 2.97

[H+]=10^(-pH)

       = 10^(-2.97) = 0.01072 M

You calculated the pH.

The [H^+] is 0.0010583 M (to use a few extra digits)

You then did this extra step:

pH = -log 0.0010583 = 2.975