Be sure to answer all parts. What is the [H3O+] and the pH of a benzoic acid-benzoate buffer that consists of 0.24 M C6H5COOH and 0.35 M C6H5COONa? (Ka of benzoic acid = 6.3 * 10-5) Be sure to report your answer to the correct number of significant figures.

[H3O+] = 4.32 * 10^-5 M

pH = 4.36

Explanation:

Step 1: Data given

Molarity of C6H5COOH = 0.24 M

Molarity of C6H5COONa = 0.35 M

(Ka of benzoic acid = 6.3 * 10^-5

Step 2: Calculate [H3O+]

Benzoic acid = weak acid which, in aqueous solution dissociates as follows:

C6H5COOH (aq) C6H5COO - (aq) + H + (aq)

Ka = [C6H5COO-][H+] / [C6H5COOH] = 6.3 * 10^-5

Because benzoic acid is a weak acid and so only slightly dissociated in aqueous solution, we can assume that all of the C6H5COOH comes from the benzoic acid, and all of the C6H5COO - comes from the ionic sodium benzoate in the buffer mixture.

Ka = 6.3 * 10^-5 = 0.35 x [H + (aq) ] / 0.24

[H+] = [H3O+] = 6.3 * 10^-5 * 0.24 / 0.35

[H3O+] = 4.32 * 10^-5 M

Step 3: Calculate pH

pH = - log [H3O+] = - log (4.32 * 10^-5) = 4.36