What is the pH of a 1.13M solution of ascorbic acid? Ka1 = 8.0 x 10-5 Ka2 = 1.6 x 10-12
Setting up an ICE table by assuming an arbitrary Dibasic acid H2A that dissociates by a fraction at equilibrium:
H2A HA- + H+
Initial concentration: 1.13 M 0 0
equilibrium concentration: 1.13(1 - ) 1.13 1.13
The Ka1 value is equated to determine :
Ka1 = [HA-] * [H+]/[H2A]
8.0 x 10-5 = 1.13 * 1.13 / {1.13 * (1 - )}
Since << 1, ( 1 - ) = 1
Solving, we get: = 8.41 * 10-3
Since is already so negligible and Ka1 is nearly 106 times Ka2, the dissociation of HA- can be neglected.
[H+] = 1.13 * = 1.13 * 8.41 * 10-3 = 9.50 * 10-3
pH = -log([H+]) = - log (9.50 * 10-3) = 2.022
What is the pH of a 1.13M solution of ascorbic acid? Ka1 = 8.0 x 10-5...
Calculate the pH of a 0.05 M solution of ascorbic acid ( H2C6H6O6 = H2ASC) Ka1 = 1 x 10-5 Ka2= 5 x 10-12 H2ASC (aq) + H2O (l) <=> H3O+ (aq) + HASC- (aq)
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