Here is where the sodium hydroxide comes in. Sodium hydroxide,
Sodium acetate will exist as sodium cations,
The balanced chemical eqution for this neutralization rection looks like this - I'll use the net ionic equation
#"CH"_3"COOH"_text((aq]) + "OH"_text((aq])^(-) -> "CH"_3"COO"_text((aq])^(-) + "H"_2"O"_text((l])#
Notice that you have a
Now, you don't want to completely neutralize the acid, since that would only leave with its conjugate base in solution.
Instead, you want to convert about half of the moles of acetic acid to moles of acetate anions by adding sodium hydroxide.
This means that every mole of the former will react with one mole of sodium hydroxide and produce one mole of acetate anions.
You're working with a
This means that if you take a volume
#n_"acetic" = V * 0.100#
This means that you will need a total of
#n_"NaOH" = 1/2 * n_"acetic" = 1/2 * V * 0.100 = 0.050 * V#
moles of sodium hydroxide.
The volume of the sodium hydroxide solution must be
#V = (0.050 * V" moles")/"0.200 M" = 0.25 * V = 1/4V#
So, if you mix a volume
Now the solution will be buffer, since it will have equal concentrations of acetic acid and of acetate anions.