Calculate the pH of a 0.04 M solution of ascorbic acid (K_a1= 7.9 $\times$10^-⁵;K_a2is 1.6 $\times$10^-¹²). A)11.2 B)2.8 C)5.5 D)8.5 E)11.8

Question

Quizplus · Accepted Answer

Ascorbic acid is a weak acid and its ionization reaction is:
H2Ascorbic acid ⇔ H+ + HAscorbate-
The dissociation constant of ascorbic acid (KS1U1B1a1) is given as 7.9 F1F1F1S1F1F1F10 10S1U1P1-S1S1P0S1U1P15S1S1P0. 
So, [H+] = √(KS1U1B1a1 x C) = √(7.9 F1F1F1S1F1F1F10 10S1U1P1-S1S1P0S1U1P15S1S1P0 x 0.04) = 7.2 x 10^-3 M 
pH = -log[H+] = -log(7.2 x 10^-3) ≈ 2.8

Calculate the PH of a 0 ×\times× 10-5;Ka2 is 1 ×\times× 10-12)

Calculate the PH of a 0 $\times$ 10^-⁵;K_a2is 1 $\times$ 10^-¹²)