Ascorbic acid, H2C6H6O6, is a diprotic acide with a K1= 7.9 E-5 and a K2=1.6E-12. In a .005 M aq solution of ascorbic acid which of the following species is present in the lowest concentration.
1. H2O
2. H3O+
3. H2C6H6O6
4. HC6H6O6-
5. C6H6O6-2
C6H6O6-2
To determine which species is present in the lowest concentration in a .005 M aqueous solution of ascorbic acid, we need to consider the acid dissociation reactions.
First dissociation reaction:
H2C6H6O6 (ascorbic acid) ⇌ H+ + HC6H6O6-
Second dissociation reaction:
HC6H6O6- ⇌ H+ + C6H6O6-2
Taking into account the given equilibrium constants (K1 = 7.9E-5 and K2 = 1.6E-12), we can compare the concentrations of the species based on their equilibrium expressions.
For the first dissociation reaction:
K1 = [H+][HC6H6O6-] / [H2C6H6O6]
For the second dissociation reaction:
K2 = [H+][C6H6O6-2] / [HC6H6O6-]
In a solution of ascorbic acid, we start with a concentration of 0.005 M. Since both K1 and K2 values are relatively small, the concentrations of H+ and the dissociated species will be much smaller.
To determine the lowest concentration species, we need to compare the equilibrium expressions for each species.
1. H2O - Water is not involved in any of the dissociation reactions.
2. H3O+ - This is not directly involved in the dissociation reactions, but it can be formed in small amounts through the autoionization of water (H2O ⇌ H+ + OH-).
3. H2C6H6O6 - This is the initial concentration of ascorbic acid and will be present in the highest concentration.
4. HC6H6O6- - This is the dissociated form of ascorbic acid (first dissociation), and its concentration will be lower than that of H2C6H6O6.
5. C6H6O6-2 - This is the dissociated form of HC6H6O6- (second dissociation), and its concentration will be the lowest among the given species.
Therefore, the species present in the lowest concentration in a .005 M aqueous solution of ascorbic acid is C6H6O6-2 (option 5).
To determine which species is present in the lowest concentration in a .005 M aqueous solution of ascorbic acid, we need to consider the ionization of ascorbic acid and the equilibrium concentrations of its various species.
Ascorbic acid, H2C6H6O6, is a diprotic acid, meaning it can donate two protons (H+ ions). It ionizes in water according to the following reactions:
1. H2C6H6O6 ⇌ H+ + HC6H6O6- (Reaction 1)
2. HC6H6O6- ⇌ H+ + C6H6O6-2 (Reaction 2)
The equilibrium constants for these reactions are given as K1 = 7.9E-5 for Reaction 1 and K2 = 1.6E-12 for Reaction 2.
To find the concentration of each species at equilibrium, we need to apply the equilibrium expressions:
K1 = [H+][HC6H6O6-] / [H2C6H6O6]
K2 = [H+][C6H6O6-2] / [HC6H6O6-]
Given the concentration of ascorbic acid as .005 M, we can assume that initially, before any reaction has occurred, [H2C6H6O6] = .005 M.
We can then solve for the equilibrium concentrations of the species using the equilibrium expressions and the previously mentioned concentration of ascorbic acid:
For Reaction 1:
K1 = [H+][HC6H6O6-] / [.005]
7.9E-5 = [H+][HC6H6O6-] / .005
Since [H+] will be much smaller than .005 (due to the given equilibrium constants), [HC6H6O6-] will be much smaller than [H2C6H6O6]. Therefore, [HC6H6O6-] is present in the lowest concentration among the given species.
The correct answer is 4. HC6H6O6-.