Question

Ascorbic acid or Vitamin C, H₂C₆H₆O₆, is a weak diprotic acid. What are the concentrations of H⁺, H₂C₆H₆O_6, HC₆H₆O₆^- and C₆H₆O₆^2- in a 0.10 M aqueous solution of ascorbic acid? For ascorbic acid Ka1=7.9x10^-5 and Ka2= 1.6x10^-12.

Answer 1

The given diprotic acid H2C6H6O6 dissociates in two steps as-

step 1 : H2C6H6O6 + H2O ------------> HC6H6O6- + H3O+ Ka1 = 7.9 x 10-5  

No we know the acid dissociation constant is the ratio between the product of concentration of product to the concentration of acid taken . i.e for the above step

Ka1 = [HC6H6O6-] * [H3O+] / [H2C6H6O6] where al the concentrations are at equilibrium.

Now to find these concentrations, we have to form the ICE table as-

Reaction	H2C6H6O6 + H2O ------------>	HC6H6O6- +	H3O+
Initial	0.10 M	0	0
Change	-x	+x	+x
Equilibrium	0.10 - x	x	x

So putting the values-

Ka1 = [HC6H6O6-] * [H3O+] / [H2C6H6O6]

(7.9 * 10-5) = [x] * [x] / [0.10 - x]

(7.9 * 10-5) * [0.10 - x] = x2  

(0.79 * 10-5) - (7.9 * 10-5)x = x2  

x2 + (7.9 * 10-5)x - (0.79 * 10-5) = 0

Solving this-

x = 0.00085

So putting the value, in this step

[H2C6H6O6] = 0.10 - x = 0.10 - 0.00085 = 0.09915‬ M

[HC6H6O6-] = x = 0.00085 M

[H3O+] = x = 0.00085 M

Again

step 2 : HC6H6O6- + H2O ------------> C6H6O6-2 + H3O+ Ka2 = 1.6 x 10-12  

No we know the acid dissociation constant is the ratio between the product of concentration of product to the concentration of acid taken . i.e for the above step

Ka2 = [C6H6O6-2] * [H3O+] / [HC6H6O6-] where al the concentrations are at equilibrium.

Now to find these concentrations, we have to form the ICE table as-

Reaction	HC6H6O6- + H2O ------------>	C6H6O6-2 +	H3O+
Initial	0.00085	0	0.00085
Change	-x	+x	+x
Equilibrium	0.00085 - x	x	0.00085 + x

So putting the values-

Ka1 = [C6H6O6-2] * [H3O+] / [HC6H6O6-]

(1.6 x 10-12) = [x] * [0.00085 + x] / [0.00085 - x]

(1.6 x 10-12) * [0.00085 - x] = 0.00085x + x2  

(1.36 * 10-15) - (1.6 x 10-12)x = 0.00085x + x2  

x2 + 0.00085x + (1.6 x 10-12)x - (1.36 * 10-15) = 0

x2 + 0.00085x - (1.36 * 10-15) = 0 (since value of (1.6 x 10-12) is very negligible it can be taken as zero)

Solving this-

x = 1.6 * 10-12  

So putting the value, in this step

[H3O+] = 0.00085 + x = 0.00085 + 1.6 * 10-12 = 0.00085 M

(since value of (1.6 x 10-12) is very negligible it can be taken as zero)

[C6H6O6-2] = x = 1.6 * 10-12 M

[HC6H6O6-] = 0.00085 - x = 0.00085 - 1.6 * 10-12 = 0.00085 M

So finally we have-

[H3O+] = 0.00085 M

[HC6H6O6-] = 0.00085 M

[C6H6O6-2] = 1.6 * 10-12 M

[H2C6H6O6] = 0.09915‬ M