Weak buffer solution titrated with a strong base

Question

A 50.0 mL sample of buffer solution contains 0.120 M acetic acid and 0.150 M sodium acetate.

If 5.55 mL of 0.092 M NaOH is added to this solution, identify the resulting number of moles of acetic acid, sodium acetate, and NaOH.

Answer 1

A solution containing acetic acid (CH3COOH) and sodium acetate (CH₃COONa) represents a buffer solution because you have a weak acid and its conjugate base. When base (OH^-) is added, it will react with the acid to produce the conjugate base + H₂O, as seen below:

CH₃COOH + OH^- ==> CH₃COO^- + H₂O.

Initial moles CH₃COOH = 50.0 ml x 1 L / 1000 ml x 0.120 mol / L = 6.00x10^-3 moles

Initial moles CH₃COO- = 50.0 ml x 1 L / 1000 ml x 0.150 mol / L = 7.50x10^-3 moles

moles OH- added = 5.55 ml x 1 L / 1000 ml x 0.092 mol / L = 5.11x10^-4 moles

CH₃COOH + OH^- ==> CH₃COO^- + H₂O

6.00x10^-3...........5.11x10^-4....7.50x10^-3.............Initial

-5.11x10^-4.........-5.11x10^-4...+5.11x10^-4............Change

5.49x10^-3................0...........8.01x10^-3.............Equilibrium

Thus, at equilibrium, we should have the following number of moles of each species:

moles of acetic acid (CH3COOH) = 5.49x10^-3 moles

moles of sodium acetate (CH3COONa) = 8.01x10^-3 moles

moles of sodium hydroxide (NaOH) = 0 moles

Answer 2

The usual approach is to assume that the OH^- reacts completely to the acid and and produces the conjugate base (that is, if the OH^- is covered by the amount of acid) and then use the K_a expression and the changes necessary to get to equilibrium to get all the final concentrations:

In this particular example, the NaOH is in solution, so their will be dilution of concentration before we start:

M_HAc = (50/55.55) (.12 M) = .108 M M_Ac = .135 M (same dil) M_OH = (5.55/55.55) * (.092 M) = .009192 M

Letting the OH^- neutralize the acid:

M_HAc = .108 - .009192 = .09881 M M_Ac = .135 + .009192 = .1442 M_OH is about 0

Now we can calculate the equilibrium for some extent of reaction, x for the rxn HAc → H⁺ + Ac^-

M_HAc,eq = .09881 -x M_Ac,eq = .1442+x M_H+ = x approximately

Now Plug equilibrium concentrations in terms of x into K_A expression for Acetic Acid:

1.75 x 10^-5 = (.1442+x)x/(.09881-x)

You can solve the quadratic for x or assume that x<<.09881 which allows explicit solution.

After you solve for x, you can plug x into the expressions for the equilibrium concentrations.

For some reason the question asks for moles, which you obtain by multiplying by .05555 liters.

This could also be done with moles instead of molarity, remembering to divide by .05555 liters when you have to plug into the K expression.