[moondaughter]

I don't know what I did wrong in this question.

Calculate the pH of a solution prepared by mixing 20.00ml of the aspirin solution in question #4 with 10.00ml of 0.200M NaOH. BEFORE using ICE box to solve for the pH of this mixture, think about whether the pH should be less than 7, equal to 7 or greater than 7 simply by comparing the moles of aspirin and the NaOH.

(#4: Aspirin is a weak acid with a Ka of 3.0X10-5. Calculate the pH of a solution by dissolving 0.65g of aspirin in water and diluting it to 50.0ml. You may use the symbol RCOOH to represent aspirin when writing chemical reaction. Molecular weight of aspirin = 180.0g/mol)

At past the end point, there is a solution of conjugate base and OH- ions.

I calculated the molarity of the aspirin solution to be (0.65 g/180 g/mol)/.050 L
The number of moles in 0.020 L is ((0.65 g/180 g/mol)/.050 L )(.020L)
The number of moles of OH- is (.010L * .200M)
The amount of leftover OH- ions is

(.010L * .200M) - ((0.65 g/180 g/mol)/.050 L )(.020L)

Divided by 0.030, which is the total amount of solution, [OH-] = 0.018518518

14 - log 0.018518518 = 12.27 but the answer my professor gave us is 12.31.

What did I do wrong?

[AWK]

Seems your calculations are OK