Question

A buffer with a pH of 4.06 contains 0.13 M of sodium benzoate and 0.18 M of benzoic acid. What is the concentration of [H ] in the solution after the addition of 0.052 mol of HCl to a final volume of 1.3 L? Assume that any contribution of HCl to the volume is negligible.

Answer 1

Initial concentration of benzoate= 0,13M

Initial moles of benzoate=MxL=0,13Mx1,3L=0,169mol of benzoate

Initial Concentration of benzoic acid=0,18M

Initial moles of acid=MxL=0,18Mx1,3L=0,234mol of benzoic acid

molesHCl add=0,052mol

HCl is neutralized for benzoate, changing concentration of benzoate and of acid, according following reaction:

H⁺ + benzoate <=====> benzoic acid

moles of benzoate and benzoic acid in the change:

moles benzoate= (initial moles - moles HCl)=(0,169-0,052)mol=0,117mol

Benzoate Concentration after add HCl= 0,177mol/1,3L=0,09M

moles benzoic acid= (initial moles + moles HCl)=(0,234+0,052)mol=0,286mol

Benzoic Acid Concentration after add HCl=0,286mol/1,3L=0,22M

With Concentrations of benzoate and benzoic acid we will calculate concentration of H after add HCl, but first should calculate Ka value using initial data of buffer solution

Calculatin Ka using Henderson Hasselbach Equation:

with ka value will calculate new concentration of H

Benzoic Acid <=======> H+ + Benzoate

Solving for H y using concentration calculated after add HCl

After addition of 0,052 mol HCl concentration of H is 1,54.10^-4M