A solution is prepared by dissolving 0.56 g of benzoic acid (C6H5CO2H, Ka? 6.4? 10?5) in enough water to make 1.0 L of solution. Calculate [C6H5CO2H], [C6H5CO2?], [H?], [OH?], and the pH of this solution.

[H⁺] = 6.083x10⁻⁴ M, [C₆H₅OO⁻] = 6.083x10⁻⁴ M, [C₆H₅OOH] = 3.98x10⁻³M, pH = 3.22

Explanation:

dа ta: we have 0.56 gr of benzoic acid, disolved in 1Lt of water. Kₐ = 6.4x10⁻⁵

M (molar mass) of BA (Benzoic Acid) = 122 g/mol

Then, the inicial concentration is 0.56/122 = 4.59x10⁻³ M

We should consider the equation once it reaches the equilibrium:

C₆H₅COOH ⇄ C₆H₅COO⁻ + H⁺

C - x x x

And, for the Kₐ:

Kₐ = [H⁺][C₆H₅COO⁻]/[C₆H₅COOH] = x² / (C-x), where C = 4.59x10⁻³

Then: x² + Kₐx - KₐC = 0

x² + 6.4x10⁻⁵ - 2.9x10⁻⁷ = 0

Resolving this cuadratic equation (remember to use Baskara equation), we obtain:

x = 6.083x10⁻⁴ M

Then: [H⁺] = [C₆H₅COO⁻] = 6.083x10⁻⁴ M

[C₆H₅COOH] = C - x = 3.98x10⁻³ M

pH = - Log [H⁺] = 3.22