What Is the pH of a Solution with Double Dissociation of Ascorbic Acid?

[relativitydude]

Hello again,

Yes, it's me with another disassociation problem.

If you have .0213M of ascorbic acid, what is the pH of a solution containing the second disassociated form?

Ok,

HC6H7O6 <--> H + C6H7O6

8.0e-5 = x^2 / (.0213 - x)
x = 1.265e-3M

C6H7O6 <---> H + C6H6O6
.02M - x (1.265-3 + x) (1.265-3 + x)

so,

1.6e-12 = (1.265e-3+x)^2/(.02-x)

Now I apparently set up the second part wrong as I am getting imaginary molarities and pHs. My chemistry teacher said I was not supose to square the rpoducts like that.

I do not understand why

 

[GCT]

C6H7O6 <---> H + C6H6O6
.02M - x (1.265-3 + x) (1.265-3 + x)

The initial concentration of the conjugate anion is 0. Also the initial concentration of the acid for the second Ka should be the same as the hydronium concentration.

 

[ravenprp]

you are squaring the products in the second part of your equation. The dissociation of ascorbic acid is a simple equilibrium reaction where the acid (HC6H7O6) dissociates into a proton (H+) and the conjugate base (C6H6O6). The equilibrium constant (Ka) for this reaction can be expressed as [H+][C6H6O6]/[HC6H7O6]. Therefore, to find the concentration of H+ in the solution, you can use the Ka value and the initial concentration of the acid (HC6H7O6). The pH can then be calculated using the H+ concentration. I hope this helps clarify the problem for you.