An analytical chemist is titrating of a solution of acetic acid with a 0.3500 M solution of NAOH. The pKa of acetic acid is 4.70. Calculate the pH of the acid solution after the chemist has added 78.4 mL of the NAOH solution to it.

pH = 4.78

Explanation:

There are titrated 65.8mL of a 0.7600M of acetic acid

The reaction of acetic acid, CH₃COOH, with NaOH is:

CH₃COOH + NaOH → CH₃COONa + H₂O

Initial moles of acetic acid are:

0.0658L ₓ (0.7600mol / L) = 0.05000 moles of CH₃COOH

Moles of NaOH added to the acid solution are:

0.0784L ₓ (0.3500mol / L) = 0.02744 moles of NaOH = Moles of CH₃COONa produced -Because NaOH is the limiting reactant-

That means after the reaction moles of CH₃COOH and moles of CH₃COONa are:

CH₃COOH: 0.05000 mol - 0.02744 mol = 0.02256 moles

CH₃COONa: 0.02744 moles

Using H-H equation for acetic acid, pH of the solution is:

pH = 4.70 + log₁₀ [CH₃COONa] / [CH₃COOH]

Replacing:

pH = 4.70 + log₁₀ [0.02744] / [0.02256]

is there multiple choice answers?