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
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
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.
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.
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
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.
To calculate the pH of the 0.033 M solution of sodium ascorbate, Na2C6H6O6, we need to consider the basicity of the solution. Sodium ascorbate is a salt formed by the reaction of ascorbic acid, H2C6H6O6, with sodium hydroxide, NaOH.
Since ascorbic acid is a weak acid, it can partially dissociate in water. The dissociation of ascorbic acid occurs in two steps, resulting in two equilibrium constants, Ka1 and Ka2.
Now, we need to find the Kb value of the ascorbate ion, C6H6O6^-2, which is the conjugate base of ascorbic acid. We can find Kb using the equation Kb = Kw/Ka, where Kw is the ion product of water and is equal to 1.0 × 10^-14 at room temperature.
Kw is equal to Ka1 * Ka2, so we have Kw = (7.9 × 10^-5) * (1.6 × 10^-12) = 1.266 × 10^-16.
Therefore, Kb = 1.0 × 10^-14 / 1.266 × 10^-16 = 1.266 × 10^-10.
To calculate the concentration of hydroxide ions, OH^-, in the solution, we need to set up an equilibrium expression for the reaction between the ascorbate ion and water. Let the concentration of OH^- be x.
H2O + C6H6O6^-2 ⇌ H2C6H6O6 + OH^-
To facilitate calculations, we assume that the concentration of ascorbate ion, C6H6O6^-2, is the same as the concentration of sodium ascorbate, 0.033 M.
Using an ICE table, we have:
Initial: 0.033 M 0 M 0 M
Change: -x +x +x
Equilibrium: (0.033 - x) M x M x M
Now, we can set up the Kb equilibrium expression:
Kb = [OH^-][H2C6H6O6] / [C6H6O6^-2]
Substituting known values and simplifying:
1.266 × 10^-10 = x * (0.033 - x) / (0.033 - x)
Since Kb is very small compared to the initial concentration of sodium ascorbate, we can assume that x is negligible in comparison to 0.033. Therefore, we can approximate (0.033 - x) as 0.033.
1.266 × 10^-10 ≈ x * 0.033 / 0.033
1.266 × 10^-10 ≈ x
Solving for x, we find:
x = 1.266 × 10^-10
This represents the concentration of hydroxide ions, OH^-, in the solution. To calculate the pOH, we take the negative logarithm of the concentration of hydroxide ions:
pOH = -log(1.266 × 10^-10) ≈ 5.69
Finally, to find the pH of the solution, we subtract the pOH from 14:
pH = 14 - 5.69 ≈ 8.31
Therefore, the pH of the 0.033 M solution of sodium ascorbate is approximately 8.31.