A student titrates 25.0 mL of a 0.100 M solution of acetic acid with a 0.100 solution of sodium hydroxide. Calculate the pH at the equivalence point.

Answer 1

OH- + HC2H3O2 --> H20 + C2H3O2-
0.0025 mol 0.0025 mol - 0
-0.0025 -0.0025 - +0.0025
0 0 - 0.0025
0.0025 mol/0.050 L = 0.050 M

C2H3O2- + H20 <--> HC2H3O2 + OH-
0.050 M - 0 0
-x - +x +x
0.050-x - x x

Kb = 10^(-14)/(1.8x10^(-5) = 5.56x10^(-8)
= x^2/(0.050-x)
x = 5.27x10^(-6)
pOH = -log(x)
= 5.28
pH = 14 - 5.28 = 8.72

Is that correct? The answer key says that the answer is 2.87, but that doesn't seem logical to me either.

Answer 2

I think 8.72 is the correct answer.