Acid, Base, and Buffer Titration Problem

I missed this question on an exam and I was wondering if anyone can explain to me how to do it.

Homework Statement

The pKa of acetic acid, HC2H3O2, is 4.76. A buffer solution was made using an unspecified amount of acetic acid and 0.300 moles of NaC2H3O2 in enough water to make 2 Liters of solution. Its pH was measured as 4.40 on a meter. How many moles of HC2H3O2 were used?

Homework Equations

K= [products]/[reactions]
pH= pKa + log (A/HA)

The Attempt at a Solution

Ka= 10^(-4.76) = 1.74 x 10 ^(-5)
Kb = 5.75 x 10^(-10)
pH =4.4 [H]=3.98 x 10^(-5)
pOH= 9.6 [OH]= 2.51 x 10^(-10)

moles of OH= 2.51 x 10^(-10) x 2= 5.02 x 10^(-10) moles

Base_________+____H2O -->___Acetic Acid_+_____OH +______Na
.3 moles______________________x_________________0
-5.024 x 10^(-10)__________5.02 x 10^(-10)_______5.02 x 10^(-10)
.2999 moles_______________x+ 5.02 x 10^(-10)______5.02 x 10^(-10)

pH=pKa + log(A/HA)
4.4=4.76+ log (.3/(x+5.02 x 10^(-10)))
=10^-(-.36)= 10^[log (.3/(x+5.02 x 10^(-10)))]
2.29= .3/(x+5.02 x 10^(-10)))

(2.29)(x+5.02 x 10^(-10))=.3
x=.131 moles

.3 + .131= 0.43 moles