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.
