Question

What is the pH of a 1.13M solution of ascorbic acid? Ka1 = 8.0 x 10-5 Ka2 = 1.6 x 10-12

Answer 1

Setting up an ICE table by assuming an arbitrary Dibasic acid H₂A that dissociates by a fraction $\alpha$ at equilibrium:

                                                H₂A      $\rightarrow$      HA^-           +      H⁺

Initial concentration:              1.13 M                         0                            0

equilibrium concentration: 1.13(1 - $\alpha$) 1.13$\alpha$ 1.13$\alpha$

The K_a1 value is equated to determine $\alpha$_:

K_{a1 =} [HA^-] * [H⁺]/[H₂A]

8.0 x 10^-5 = 1.13$\alpha$ * 1.13$\alpha$ / {1.13 * (1 - $\alpha$)}

Since $\alpha$ << 1, ( 1 - $\alpha$) = 1

Solving, we get: $\alpha$ = 8.41 * 10^-3

Since $\alpha$ is already so negligible and K_a1 is nearly 10⁶ times K_a2, the dissociation of HA^- can be neglected.

[H⁺] = 1.13 * $\alpha$ = 1.13 * 8.41 * 10^-3 = 9.50 * 10^-3

pH = -log([H⁺]) = - log (9.50 * 10^-3) = 2.022