What is the solubility of benzoic acid C6H5COOH in water (in g dm^-3) if its solution has pH=2,9 and Ka=6,46?

What is the solubility of benzoic acid C6H5COOH in water (in g dm^-3) if its solution has pH=2,9 and Ka=6,46?

C6H5COOH + H2O → H3O+1 + C6H5COOH-1

According to the balanced equation above, 1 mole of C6H5COOH molecules will produce 1 mole of H3O+1 ions and 1 mole of C6H5COOH-1 ions.

pH = -1 * log [H3O+1]

[H3O+1] = 10^(-2.9) = 1.259 * 10^-3 moles per liter

Since the solution contains 1.259 * 10^-3 moles of H3O+1 ions, it must also contain 1.259 * 10^-3 moles of C6H5COOH-1 ions per liter of solution, (x – 1.259 * 10^-3) moles of C6H5COOH molecules that did not dissociate.

Ka = 6.46 * 10^-5

Ka = [H3O+1] * [C6H5COOH-1] ÷ [C6H5COOH]

6.46 * 10^-5 = (1.259 * 10^-3) * (1.259 * 10^-3) ÷ (x – 1.259 * 10^-3)

Multiply both sides by (x – 1.259 * 10^-3)

(6.46 * 10^-5) * (x – 1.259 * 10^-3) = (1.259 * 10^-3)^2

(6.46 * 10^-5 * x) – (8.13 * 10^-8) = 1.585 * 10^-6

6.46 * 10^-5 * x = (8.13 * 10^-8) + (1.585 * 10^-6)

x = 2.58 * 10^-2 moles of C6H5COOH molecules per liter of solution!

Molar mass of C6H5COOH = 72 + 5 + 12 + 32 + 1 = 122 grams

Mass of C6H5COOH = 2.58 * 10^-2 * 122 = 3.1476 grams

Solubility = 3.15 grams per liter of solution!!