A buffer with a pH of 3.98 contains 0.23 M of sodium benzoate and 0.38 M of benzoic acid. What is the concentration of [H3O+] in the solution after the addition of 0.058 mol HCl to a final volume of 1.3 L? Assume that any contribution of HCl to the volume is negligible g

New pH = 3.84

Explanation:

First of all we may think that if the buffer has pH 3.98 and we're adding H⁺, pH's buffer will be lower, as the [H⁺] is been increased.

Let's determine the moles of each compound:

0.23 M . 1.3L = 0.299 moles of NaBz

0.38 M . 1.3L = 0.494 moles of HBz

We add 0.058 of HCl, which is the same as 0.058 moles of H⁻

HCl →  H⁺  +  Cl⁻

As we add the moles of protons, these are going to react to the Bz⁻

In the buffer system we have these dissociations:

NaBz  →  Na⁺  +  Bz⁻

HBz →  H⁺  + Bz⁻

So, as we add protons, we have a new equilibrium:

          Bz⁻      +      H⁺       ⇄   HBz    

In    0.299         0.058          0.494

Eq   0.241               -              0.552

Protons are substracted to benzoate, so the [HBz] is now higher than before. We calculate the new pH, with the Henderson Hasselbach equation

pH = pKa + log (Bz⁻/HBz)

pH = 4.20 + log (0.241 / 0.552) → 3.84