10. What is the pH of a solution that consists of 0.5 M H2C6H6O6 (ascorbic acid) and
0.75 M NaHC6H6O6 (sodium ascorbate)? Ka₁ = 6.8 x 10-5, Ka2 = 2.8 × 10-12
A. 11.4
B. 3.99
C. 4.34
D. 4.57
E. 11.7



Answer :

To find the pH of the solution containing 0.5 M ascorbic acid (H2C6H6O6) and 0.75 M sodium ascorbate (NaHC6H6O6), we need to consider the acidic dissociation of ascorbic acid in two steps. 1. Write down the dissociation reactions: H2C6H6O6 ⇌ H⁺ + HC6H6O6⁻ HC6H6O6⁻ ⇌ H⁺ + C6H6O6²⁻ 2. Set up the initial concentrations: [H2C6H6O6] = 0.5 M [HC6H6O6⁻] = 0 M [C6H6O6²⁻] = 0 M 3. Calculate the equilibrium concentrations using the dissociation reactions and Ka values: Ka₁ = [H⁺][HC6H6O6⁻] / [H2C6H6O6] = 6.8 x 10⁻⁵ Ka₂ = [H⁺][C6H6O6²⁻] / [HC6H6O6⁻] = 2.8 x 10⁻¹² 4. Solve for [H⁺] using the equilibrium expressions and ICE (Initial, Change, Equilibrium) table: For the first dissociation: [H⁺] = sqrt(Ka₁ * [H2C6H6O6]) = sqrt(6.8 x 10⁻⁵ * 0.5) = 0.0049 M For the second dissociation: [H⁺] = sqrt(Ka₂ * [HC6H6O6⁻]) = sqrt(2.8 x 10⁻¹² * 0.0049) = 6.13 x 10⁻⁶ M 5. Calculate the pH using the relation pH = -log[H⁺]: pH = -log(6.13 x 10⁻⁶) ≈ 5.21 Therefore, the correct answer is not listed among the options provided.

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