(10 points please help!) Find the pH at the equivalence point? Acid/Base titration Question

50 mL of 0.01 M NaC2H3O2 ( Kb=5.71x10-10) is titrated with 1 M HCl.

Show work please! :) thanks

C2H3O2- + HCl ---> C2H3O2H + Cl-
First we need to see how much HCl must be added to neutralize the C2H3O2-.
50 mL * 0.1 M = 0.5 mmol C2H3O2-
Volume HCl added = 0.5 mmol / 1 M = 0.5 mL
..................................C2H3O… . . . . . . HCl . . . . . . . .C2H3O2H
Initial Amount.............. 0.5 mmol ........ 0.5 mmol ....... 0 mmol
Change..................... -0.5 mmol......... -0.5 mmol........ +0.5 mmol
Final Amount............. 0 mmol............ 0 mmol ............ 0.5 mmol
Final Concentration......0 M................. 0 M...............0.5 mmol / 50.5 mL = 9.901*10^-3 M

So now all of the C2H3O2- has been neutralized by HCL, but there is still C2H3O2H (acetic acid) in solution. We need to calculate the [H+] when this solution reaches equilibrium.
C2H3O2H <---> C2H3O2- + H+
Ka = [H+][C2H3O2-] / [C2H3O2H] = x^2 / [9.901*10^-3 M]
Ka = Kw / Kb = 1*10^-14 / 5.71*10^-10 = 1.75*10^-5
1.75*10^-5 = x^2 / [9.901*10^-3]
x^2 = 1.73*10^-7
x = [H+] = 4.16*10^-4 M
pH = -log[H+] = -log[4.16*10^-4]
pH = 3.38

Thus, the pH at the equivalence point is 3.38