Question

25.0 CM3 OF 0.100M acetic acid is titrated with 0.900M of NaOH. Ka for acetic acid =1.85*10^-5. calculate
1)the intial pH of the solution.
2) the pH of the mixture at 1cm3 before the equivalence point.
3) the pH of the mixture at the equivalence point
4) the pH of the mixture at 1 cm3 after the equivallence point.

Answer 1

Are you sure that's 0.900 M NaOH? and not 0.0900 M? I'll assume 0.900 is correct. And I'll write acetic acid, HC2H3O2, (or CH3COOH) as HAc.
Millimols HAc initially = 25 x 0.1 = 2.5.
millimols NaOH added = mL x 0.900

The first thing to do is to determine the equivalence point.
mLa x Ma = mLb x Mb
25 x 0.1 = mLb x 0.9
mL base = (25 x 0.1)/0.9 = 2.77which makes me almost certain there is a typo in the 0.9.

a. For initial pH.
..........HAc ==> H^+ + Ac^-
initial.. 0.1......0.....0
change.....-x......x......x
equil.....0.1-x....x......x
Plug this into Ka expression, solve for x = (H^+) and convert to pH.

c. at the equivalence point. What's in the solution? That's acetate ion and water. That's it. So the pH is determined by the hydrolysis of the acetate ion. (Ac^-) at the equivalence point is mmols/mL (thats 2.5 mmols for Ac^- and mL is 25 to start + volume NaOH added). That's 25 + 2.78 from the above or 27.78 mL = about 0.09
............Ac^- + HOH --> HAc + OH^-
initial.....0.09............0......0
change.....--x..............x.....x
equil.....0.09-x............x.....x

Kb for Ac^- = (Kw/Ka for HAc) = (HAc)(OH^-)/(Ac^-)
Substitute from the ICE chart for the hydrolysis into the Kb expression and solve for x = OH^-, then convert to pH.

b. For all point BEFORE the equivalence point.
mols HAc - mols NaHO = mols Ac formed
pH = pKa + log (Ac^-)/(HAc)

d. For all points AFTER the E.P., it's excess NaOH.
pOH = (excess NaOH) and change to pH.