Question

A buffer with a pH of 4.28 contains 0.31 M of sodium benzoate and 0.26 M of benzoic acid. What is the concentration of [H ] in the solution after the addition of 0.050 mol of HCl to a final volume of 1.3 L? Assume that any contribution of HCl to the volume is negligible.

Answer 1

Equilibrium between sodium benzoate and benzoic acid is:

C₆H₅COOH + H₂O ⇌ C₆H₅COO⁻ + H⁺_(aq)

When you add HCl to your buffer solution, sodium benzoate reacts with the acid, and the aforementioned equilibrium is displaced to the left:

C₆H₅COO⁻ + H⁺_(aq) ⇌ C₆H₅COOH + H₂O

Now, the equilibrium constant of this reaction is:

K=1/Ka=0.16×10⁵ >10³

The reaction is quantitative (total). So, we can rewrite it:

C₆H₅COO⁻ + H⁺_(aq)⟶C₆H₅COOH + H₂O

Now, we have to construct an ICE table: Assuming that we didn't dilute the solution when we added HClHCl (this point is not clear in your question):

[C₆H₅COOH]₀=0.26 M

[C₆H₅COO⁻]₀=0.31M

[H+]₀=(0.050/1.3+10^−4.28)M≈0.0385M

As you can clearly seen [H+]0 is the lowest, and so ion hydronium is the limiting reactant. At the end of the reaction, we have:

[C₆H₅COO^−]eq=(0.31−0.0385)=0.2715M

[C₆H₅COOH]eq=(0.26+0.0385)=0.2985M

[H⁺]eq=ϵ

Substituting all these numbers in the expression of K

K=0.2985/0.2715ϵ=0.16×10⁵

By solving the above equation, we find:

[H⁺]=ϵ=6.8715×10⁻⁵ M