What is the pH of a 0.24 M benzoic acid solution with a Ka of 6.5 x 10^-5?

To calculate the pH of a 0.24 M benzoic acid solution, we start with the dissociation reaction of benzoic acid (C₆H₅COOH):

C₆H₅COOH <=> C₆H₅COO^– + H⁺

The acid dissociation constant (Ka) is given as 6.5 x 10^-5.

We can set up an expression for Ka:

Ka = &frac{{[C₆H₅COO^{–+6H₅COOH]}}}

Let x be the concentration of H⁺ ions that dissociate from benzoic acid. Therefore:

• [C₆H₅COO^–] = x
• [H⁺] = x
• [C₆H₅COOH] = 0.24 – x

Plugging these into the expression for Ka gives us:

6.5 x 10^-5 = &frac{{x * x}}{{0.24 – x}}

Assuming x is small compared to 0.24, we can simplify this to:

6.5 x 10^-5 = &frac{{x²}}{{0.24}}

Now, solving for x:

x² = 6.5 x 10^-5 * 0.24

x² = 1.56 x 10^-5

x = &radic{{1.56 x 10^-5}}

x ≈ 0.00395 M

This value of x corresponds to the concentration of H⁺ ions. Now, we can find the pH:

pH = -log[H⁺]

pH = -log(0.00395) ≈ 2.40

Therefore, the pH of a 0.24 M benzoic acid solution is approximately 2.40.