Ascorbic acid is vitamin C, C6H8O6,(176g/mol) is a diprotic acid where ka1=6.8*10^-5 and ka2=2.7*10^-12.?

What is the pH of the solution that contains 10mg of acid per mL of solution?
C6H8O6+H2O=C6H7O6+H3O

Answer:
Molar mass = 176.13 g/mol

10mg = 5.7*10^-5 moles

(5.7*10^-5 moles)/.001L = .057M

Now make a table;

______C6H8O6+H2O=C6H7O6+H3O
I________.057__________0_____0
C________-X__________+X_____+X
E_____.057 - X__________X______X

Ka1 = [H+][C6H7O8]/[C6H8O6]

Plug in values

6.8*10^-5 = [x][x]/[.057]

We assume .057-x to be the same as .057 by itself since x is probably very small compared to .057

6.8*10^-5 = [x^2]/[.057]

Solve for x

x = .0020 M
so
[H] = .0020M

We are only going to base the results on the first Ka since the second Ka is so much smaller. The results from the second Ka are negligible compared to the first.

pH = -log[H]
ph = 2.7


Hope that helps.
Vit C + H2O -----> VitCminusH + H3O+


10mg/mL * 1000 mL/L * 1g/1000 mg * 1 mole/176 gram = 0.057 moles/L

Vit C + H2O----->VitCminusH+H3O+
0.057 who cares 0 0
-x and each of the products would be +x

Ka1 = 6.8 x 10^-5 = [H3O+][VitC minus H]/[VitC]

6.8 x 10^-5 = [x][x]/[0.057-x]

Assume x is small

6.8 x 10^-5 = x2/[0.057]

x = 1.97 x 10^-3

Check assumption: 1.97 x 10^-3/0.057 * 100 = 3.5% - using the 5% rule, this is acceptable!

So [H3O^+] = 1.97 x 10^-3 M
and [VitC minus H] = 1.97 x 10^-3 M also

Here's where Ka2 comes in

VitC minus H + H2O -----> VitC minus 2 H's + H3O+
1.97 x 10^-3 (from above) None 1.97 x 10^-3
-x ignore water and the products are each +x

Ka 2 = 2.7 x 10^-12 = [H3O^+][VitC - 2Hs]/[VitC-1H]

2.7 x 10^-12 = [1.97 x 10^-3 +x][x]/[1.97 x 10^-3 -x]

assume x is small

2.7 x 10^-12 = [1.97 x 10^-3][x]/[1.97 x 10^-3]

So x = 2.7 x 10^-12

[H3O^+] = 1.97 x 10^-3 + 2.7 x 10^-12 which therefore equals (sig figs!!) 1.97 x 10^-3 M

This happens with di and tri protic acids. The production of the H^+ ion from the first dissociation FAAARRRR outweighs the production of H^+ ions from the second or third dissociations (for the weaker acids)

pH = -log[H3O^+1]

pH = -log (1.97 x 10^-3)

pH = 2.71

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