Question

Calculate the equilibrium concentration of carbonate in a 0.067 M solution of carbonic acid. Do not enter units as part of your answer. Use “E” for scientific notation.

Ascorbic acid (Vitamin C, CHO) is a weak diprotic acid, with K = 6.8x10 and K = 2.7x10. What is the pH of a solution that contains 10.0 mg of vitamin C per 1.0 mL of solution?

Answer 1

Part 1)

                                   H₂CO₃               H⁺      +     HCO₃^-

Initial……………....… 0.067……….....…..0…………. 0

Change……………… -x………........……+x………….+x

Equilibrium……...…(0.067-x)……......……x………….. x

Ka1 = x2/(0.067-x)

= x2/(0.067)         [x <<< 0.067]

We have, Ka1 = 4.3 x 10^-7    

Therefore,

4.3 x 10^-7 =   x2/(0.067)

x = [H⁺] = [HCO₃^-] =1.7 x 10^-4 M

                                     HCO₃^-                           H⁺        +         CO₃^2-

Initial……………..… (1.7 x 10^-4)…….........…(1.7 x 10^-4)…….……. 0

Change……………...….–x……………...........……+x……………….+x

Equilibrium…..........((1.7 x 10^-4)-x)….......…((1.7 x 10^-4)+x).............x

Ka2 = [(1.7 x 10^-4))x x]/[( (1.7 x 10^-4)-x)] = x         [x <<< (1.7 x 10^-4)]

We have, Ka2 = 5.6 x 10^-11    

Therefore,

x = [CO₃^2-] =5.6 x 10^-11 M = 5.6E-11

Equilibrium concentration of carbonate = 5.6E-11

Part2)

Ka1 = 6.8 x 10^-5

Ka2 = 2.7 x 10^-12

Molar mass of ascorbic acid =176.12 g/mol

Moles of ascorbic acid = (0.01 g)/ (176.12 g/mol) = 5.679 x 10^-5 moles

Molarity of ascorbic acid = Moles/Volume in Liter = (5.679 x 10^-5)/ 0.001 = 5.679 x 10^-2 M

Here, we can see that Ka1>>>Ka2. Therefore, we need not consider Ka2

                                   Ascorbic acid                           H⁺     +   Ascorbate

Initial…………….……(5.679 x 10^-2)……………......…….0……………….0

Change……………………….-x……………..……………+x……….……..+x

Equilibrium…………….(5.679 x 10^-2 -x)……....…………x…….………….x

Ka1 = x²/(5.679 x 10^-2 -x) = x2/(5.679 x 10^-2)           [x<<< 5.679 x 10^-2]

6.8 x 10^-5 = x²/(5.679 x 10^-2)          

x = [H⁺] = 1.965 x 10^-3

pH = -log[H⁺] = -log(1.965 x 10^-3) = 2.7

pH = 2.7