!! LONG ANSWER !!
So, you start with a #"0.20000-M"# solution of acetic acid, or #CH_3COOH#. The initial pH of the solution will depend on the concentration of protons in solution. So, using the ICE table method (more here: http://en.wikipedia.org/wiki/RICE_chart), you get
#CH_3COOH_((aq)) rightleftharpoons CH_3COO_((aq))^(-) + H_((aq))^(+)#
I.......0.2.................................0........................0
C.....(-x)................................(+x)......................(+x)
E...(0.2 - x)............................x..........................x
#K_a = ([H^(+)] * [CH_3COO^(-)])/([CH_3COOH]) = (x * x)/(0.2 - x) = 1.75 * 10^(-5)#
Solving for #x# will give you #x = 0.00187#, which is equal to the concentration of protons in solution. Therefore,
#pH_("initial") = -log([H^(+)]) = -log(0.00187) = 2.73#
As you perform the titration, you observe the equivalence point occurs when #"30.41 mL"# of #"0.2000-M"# #NaOH# solution is added.
Since #NaOH# is a strong base, it will react with acetic acid, which is a weak acid, to produce sodium acetate, the salt of its conjugate base, and water.
#CH_3COOH_((aq)) + NaOH_((aq)) -> CH_3COONa_((aq)) + H_2O_((l))#
Notice the #"1:1"# mole ratio you have between acetic acid and #NaOH#. This means that at the equivalence point, the number of moles of acetic acid will be equal to the number of moles of #NaOH#. So,
#n_(NaOH) = C * V = "0.2000 M" * 30.41 * 10^(-3)"L"#
#n_(NaOH) = "0.00610 moles"#
Automatically,
#"0.00610 moles NaOH" * ("1 mole acetic acid")/("1 mole NaOH") = "0.00610 moles acetic acid"#
Since both the initial amount of acetic acid, and the added #NaOH# will be consumed at the equivalence point, the number of moles of #CH_3COONa# produced will be equal to #"0.00610"# (mole ratio again #-># 1 mole of each reactant, 1 mole of each product).
The initial volume of acetic acid you had is
#C = n/V => V = n/C = "0.00610 moles"/("0.2000 M") = "30.5 mL"#
This means that the concentration of #CH_3COONa#, which is equivalent to #CH_3COO^(-)#, will be
#C_(CH_3COO^(-)) = n/V_("total") = "0.00610 moles"/((30.5 + 30.41)*10^(-3)"L") = "0.1 M"#
#CH_3COONa# will hydrolize to produce acetic acid and hydroxide, or #OH^(-)#. Use the ICE table method again
#CH_3COO_((aq))^(-) + H_2O_((l)) rightleftharpoons CH_3COOH_((aq)) + OH_((aq))^(-)#
I.......0.1....................................................0........................0
C.....(-x)..................................................(+x)......................(+x)
E...(0.1 - x)...............................................x..........................x
This time use #K_b#, which is equal to #K_W/K_a = (1.0 * 10^(-14))/(1.75 * 10^(-5))#, or
#K_b = 5.71 * 10^(-10)#
#K_b = ([OH^(-)] * [CH_3COOH])/([CH_3COO^(-)]) = x^2/(0.1 - x)#
Solving for #x# will get you
#x = "0.00000756" = [OH^(-)] = 7.56 * 10^(-6)#. Therefore,
#pOH = -log ([OH^(-)]) = -log(7.56 * 10^(-6)) = 5.12#
Now use the fact that #pH_(solution) = 14 - pOH# to get
#pH_("final") = 14 - 5.12 = 8.88#
