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.

Rxn => Addition of 0.0092M OH⁻ will react with 0.0092M H⁺ shifting buffer . equilibrium to the right decreasing [HOAc] and increasing [OAc⁻] by . 0.0092M each.

Therfore ...

Given: HOAc ⇄ [H⁺] + [OAc⁻]

C(i) 0.12M 0M 0.15M

ΔC - 0.0092M +x +0.0092M

C(f) 0.1108M x 0.1592M => New Concentrations . after adding 0.0092M . NaOH

Substituting new acid and ion concentrations into Ka expression ...

Ka = [H⁺][OAc⁻]/[HOAc] = (x)(0.1592M)/(0.1108M) = 1.75 x 10⁻⁵M

=> x = [H⁺](new) = (1.75 x 10⁻⁵M*)(0.1108M)/(0.1592M) = 1.22 x 10⁻⁵M in H⁺ ions

*units of Ka are Molar

FYI => Adding a strong base to a buffer solution will shift pH to more basic.

Adding a strong acid to a buffer solution will shift pH to more acidic.

=> (such is a good way to check that your buffer calculations are correct.)

NOTE => Question asks for moles of HOAc, Na⁺OAc⁻ & NaOH after adding base. Giving answers in terms of Molarity (moles/Liter) is same as moles. Therefore ...

[HOAc] = 0.1108M

[NaOAc] = 0.1592M

[NaOH] = ∅M (from rxn of H⁺ + OH⁻ => H₂O, all NaOH was consumed in acid/base reaction. Remaining are only Na⁺ as a spectator ion and OH⁻ as a function of the new concentration of H⁺ => [OH⁻] = Kw/[H⁺] = 1 x 10⁻¹⁴/1.22 x 10⁻⁵ = 8.2 x 10⁻¹⁰M.

Hope this helps. :-)