Question

Calculate the pH of a 0.033 M solution of sodium ascorbate, Na2C6H6O6. For ascorbic acid, H2C6H6O6, Ka1 = 7.9 × 10-5 and Ka2 = 1.6 × 10-12.

Kb = 1E-14 / 7.9E-5 = 1.266E-10

I 0.033 0 0
C -x +x +x
E 0.033-x x x

x^2/-0.033 = 1.266E-10
x = 2.044E-6
pOH= -log(2.044E-6)= 5.69
pH= 14 - 5.69 = 8.31

Answer 1

I think Kb for the ascorbate ion is kw/k2 instead of kw/k1. Everything else (the procedure that is) looks ok to me.
You can prove this is you wish by writing the k1 and k2 expression as well as Keq for the hyrolysis equation for ascorbate ion.

Answer 2

Is this the right answer?

Kb = 1E-14 / 1.6E-12 = 6.26E-3
x^2/-0.033 = 6.26E-3
x = 1.436E-2
pOH= -log(1.436E-2)= 1.84
pH= 14 - 1.84 = 12.16

Answer 3

You have assumed 0.033-x = 0.033 and I'm not sure you can do that. If we compare the answer of 0.0144M for OH^- (rounded) is with the assumption but 0.033-0.0144 appears it isn't = 0.033. So I think you need to go through the quadratic equation. A quick guess is that will change the answer by about 25% or so.

Answer 4

x^1 - 6.25E-3 + 2.0625E-4 = 0

x = {-(-6.25E-3) +/- sqrt[(6,25E-3)^2 - (4*1*2.0625E-4)]}/(2*1)
x = -1.089E-2 (Not x) or 1.714E-2
pOH = -log(1.714E-2 = 1.76
pH = 14-1.76 = 12.24

Answer 5

I got x^2 + 6.25E-3 - 2.06E-4 = 0 and when I solved that I came up with 0.0116M (vs 0.0144 before) for x. That gave me 1.93 for pOH and 12.06 for pH.