Calculate the [H3O+] of 0.35 M solution of benzoic acid, HC7H3O2

4.70x10⁻³ = [H₃O⁺]

Explanation:

The Ka of benzoic acid is 6.3x10⁻⁵

The aqueous equilibrium of benzoic acid with its conjugate base could be written as:

HC₇H₃O₂(aq) + H₂O(l) ⇄ H₃O⁺(aq) + C₇H₃O₂(aq)

Where Ka is defined as:

Ka = 6.3x10⁻⁵ = [C₇H₃O₂] [H₃O⁺] / [HC₇H₃O₂]

As both H⁺ and C₇H₃O₂ are obtained from the same equilibrium:

[C₇H₃O₂] = [H₃O⁺]

And as [HC₇H₃O₂] = 0.35M we can write:

6.3x10⁻⁵ = [H₃O⁺] [H₃O⁺] / [0.35M]

2.205x10⁻⁵ = [H₃O⁺]²

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