Calculate the pH of a solution prepared by dissolving 0.150 mol of benzoic acid and 0.300 mol of sodium benzoate in water sufficient to yield 1.00 L of solution. The K_a of benzoic acid is 6.30 × 10^-5. A)2.516 B)3.892 C)4.502 D)10.158 E)4.195

Question

Quizplus · Accepted Answer

First, we need to determine whether the solution is a buffer or not. Since the solution contains both benzoic acid and its conjugate base (sodium benzoate), it can be considered a buffer.

Next, we need to determine the pH of the buffer. We can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where pKa is the dissociation constant of benzoic acid, [A-] is the concentration of benzoate ions, and [HA] is the concentration of benzoic acid.

The pKa of benzoic acid is given as 6.30 × 10-5.

The concentration of benzoic acid is 0.150 mol/1.00 L = 0.150 M.

The concentration of benzoate ions is 0.300 mol /1.00 L = 0.300 M.

Plugging in these values, we get:

pH = 6.30 × 10-5 + log(0.300/0.150) = 4.502

Therefore, the pH of the solution is 4.502. The correct answer choice is C.

Calculate the PH of a Solution Prepared by Dissolving 0