Ascorbic acid (H2C6H6O6; H2Asc for this problem), known as vitamin C, is a diprotic acid (Ka1= 1.0x10–5 and Ka2= 5x10–12) found in citrus fruit. Calculate [HAsc–], [Asc2–], and the pH of 0.050 M H2Asc.

To calculate [HAsc−], [Asc2−], and the pH of 0.050 M H2Asc, we must understand that it's a diprotic acid, meaning it can donate two protons (H+ ions). The first dissociation constant (Ka1) is much larger than the second (Ka2), so we'll assume that the first dissociation is the one that primarily contributes to the acid behavior of the solution.

Here is a step-by-step calculation:

Step 1: Write the dissociation reactions. The dissociation of H2Asc happens in two steps: 1. H2Asc → H+ + HAsc− (with Ka1) 2. HAsc− → H+ + Asc2− (with Ka2)

Step 2: Set up an ICE table for the first reaction. Let's assume x is the amount of H2Asc that dissociates:   [H2Asc]     [H+]       [HAsc−] I   0.050 M      0        0 C  - x          +x         +x E  0.050 M - x  x          x

Step 3: Write the expression for the first dissociation constant. Ka1 = [H+][HAsc−] / [H2Asc]

Step 4: Substitute the equilibrium concentrations into Ka1. Ka1 = x^2 / (0.050 - x)

Equilibrium concentration approximation: Since Ka1 is relatively small, we can assume that x will be much smaller than 0.050 M and so, to simplify the calculation, we can approximate 0.050 M - x ≈ 0.050 M.

This approximation gives us: Ka1 = x^2 / 0.050 M 1.0 x 10^−5 M = x^2 / 0.050 M x^2 = (1.0 x 10^−5 M) * 0.050 M x^2 = 5.0 x 10^−7 M^2 x = √(5.0 x 10^−7 M^2) x ≈ 2.24 x 10^−4 M

Therefore, [H+] ≈ [HAsc−] ≈ 2.24 x 10^−4 M at equilibrium.

Step 5: Calculate pH using the [H+]. pH = -log[H+] pH = -log(2.24 x 10^−4) pH ≈ 3.65

Step 6: Calculate [Asc2−] ignoring the second dissociation. Since the second dissociation constant is much smaller than the first, the concentration of [Asc2−] produced from the second dissociation will be negligible compared to the concentration of [HAsc−]. Thus, [Asc2−] will be very small and close to zero.

Final concentrations (approximations): [HAsc−] ≈ 2.24 x 10^−4 M [Asc2−] ≈ 0 M, as the second dissociation is negligible. pH ≈ 3.65