Question

1. Consider a 0.575 M solution ascorbic acid (H.C.H.O.) also known as Vitimin C (pKa = 4.10 and where pKaz = 11.80.) a. First write out the two acid dissociation equilibrium reactions that occur in this solution (4pts) b. What is the pH, [HC.H.06 ) and [CH.02- ] (8pts) 2. What is the approximate pH of a solution of 0.620M NaC.H.O.? Here you can refer to the equilibrium constants in question 2, assume this compound is fully soluble. (4pts)

Answer 1

a) The dissociations that occur are:

H2C6H6O6 = H + + HC6H6O6-

HC6H6O6- = H + + C6H6O6-2

b) Ka is calculated:

Ka1 = 10 ^ -pKa1 = 10 ^ -4.1 = 7.9x10 ^ -5

Ka2 = 10 ^ -11.8 = 1.6x10 ^ -12

You have the expression of Ka1:

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

7.9x10 ^ -5 = X ^ 2 / 0.575

It clears X = 0.007 M

PH is calculated:

pH = - log 0.007 = 2.15

You have the second dissociation:

Ka2 = [C6H6O6-2] * [H +] / [HC6H6O6-]

1.6x10 ^ -12 = [C6H6O6-2] * 0.007 / 0.007

It clears [C6H6O6-2] = 1.6x10 ^ -12 M

2) Kb2 is calculated:

Kb2 = Kw / Ka2 = 10 ^ -14 / 1.6x10 ^ -12 = 6.25x10 ^ -3

The expression of Kb2 has:

6.25x10 ^ -3 = X ^ 2 / 0.62

It clears X = 0.062 M

POH and pH are calculated:

pOH = - log 0.062 = 1.21

pH = 14 - 1.21 = 12.79

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