What is the pH when 25.00 mL of 0.20 M CH3COOH has been titrated with 40.00 mL of 0.10 M NaOH?

You are dealing with a neutralization reaction that takes place between acetic acid, #"CH"_3"COOH"#, a weak acid, and sodium hydroxide, #"NaOH"#, a strong base.

Now, the pH of the resulting solution will depend on whether or not the neutralization is complete or not.

If the neutralization is not complete, more specifically if the acid is not completely neutralized, you will have a buffer solution that will contain acetic acid and its conjugate base, the acetate anion..

It's important to note that at complete neutralization, the pH of the solution will not be equal to #7#. Even if you neutralize the weak acid completely, the solution will be left with its conjugate base, which is why you can expect its pH to be greater than #7#.

So, the balanced chemical equation for this reaction is - I'll show you the ionic equation

#"CH"_3"COOH"_text((aq]) + "OH"_text((aq])^(-) -> "CH"_3"COO"_text((aq])^(-) + "H"_2"O"_text((l])#

Notice that #1# mole of acetic acid will react with #1# mole of sodium hydroxide, shown here as hydroxide anions, #"OH"^(-)#, to produce #1# mole of acetate anions, #"CH"_3"COO"^(-)#.

Use the molarities and volumes of the two solutions to determine how many moles of each you're adding

#color(blue)(c = n/V implies n = c * V)#

#n_"acetic" = "0.20 M" * 25.00 * 10^(-3)"L" = "0.0050 moles CH"_3"COOH"#

and

#n_"hydroxide" = "0.10 M" * 40.00 * 10^(-3)"L" = "0.0040 moles OH"^(-)#

Since you have fewer moles of hydroxide anions, the added base will be completely consumed by the reaction.

As a result, the number of moles of acetic acid that remain in solution will be

#n_"acetic remaining" = 0.0050 - 0.0040 = "0.0010 moles"#

The reaction will also produce #0.0040# moles of acetate anions.

This means that you're now dealing with a buffer. Use the Henderson-Hasselbalch equation to find its pH

#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))#

Use the total volume of the solution to find the new concentrations of the acid and of its conjugate base

#V_"total" = V_"acetic" + V_"hydroxide"#

#V_"total" = "25.00 mL" + "40.00 mL" = "65.00 mL"#

The concentrations will thus be

#["CH"_3"COOH"] = "0.0010 moles"/(65.00 * 10^(-3)"L") = "0.015385 M"#

and

#["CH"_3"COO"^(-)] = "0.0040 moles"/(65 * 10^(-3)"L") = "0.061538 M"#

The #pK_a# of acetic acid is equal to #4.75# #-># see here

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

The pH of the solution will thus be

#"pH" = 4.75 + log( (0.061538color(red)(cancel(color(black)("M"))))/(0.015385color(red)(cancel(color(black)("M")))))#

#"pH " = color(green)(" 5.35")#

You can find other examples here

http://socratic.org/questions/200-ml-of-a-50-0-mm-buffer-at-ph-3-95-is-prepared-from-50-0-mm-ch3cooh-and-50-0-

http://socratic.org/questions/a-buffer-was-prepared-by-adding-50ml-0-1-m-naoh-to-100ml-0-1m-ch3cooh-what-is-th

http://socratic.org/questions/calculate-the-ph-of-buffer-prepared-by-mixing-100ml-of-0-2m-ch3cooh-with-100ml-o