Discuss three factors that influence the distribution of chemical species in an

Question

Discuss three factors that influence the distribution of chemical species in an aquatic system. In each case give an example or illustration to support your discussion. Acetic acid (CH_3COOH) has a dissociation constant of 1.8 times 10^-5. What would be the composition of a 0.10M acetic acid in a solution of pH 4.7? Below is a distribution diagram for phosphoric acid in an aqueous medium, showing the fractions of the various phosphate species as a function of pH. What would be the fractional values of the phosphate species in a lake of pH 4.3? If the total phosphate content of the lake is 62 mu g/L, what would be the concentrations of the phosphate species at pH 4.3? The distribution of phosphate in aqueous media is governed by the equilibria shown below. Estimate from the diagram the values of K_1, K_2, and K_3, indicating what these K values mean. H_3PO_4 harr H^+ + H_2PO_4^-, K_1 = ? H_2PO_4^- harr H^+ + HPO_4^2-, K_2 = ? HPO_4^2- harr H^+ + PO^4^3-, K_3 = ?

Explanation / Answer

4.(b): Given the pH of the solution, pH = 4.7

=> - log[H3O+] = 4.7

=> [H3O+] = 10-4.7 = 2.00x10-5 M

The balanced equation for the dissociation of CH3COOH is

-------------- CH3COOH(aq) + H2O <-----> CH3COO-(aq) + H3O+ ; Ka = 1.8x10-5  

Initl con(M): 0.10, ------------------------------ 0 ------------------ 2.00x10-5 M  

change(M): -y ---------------------------------- +y, ----------------- 2.00x10-5 M

eqm.con(M): (0.10 - y), ----------------------- y ------------------- 2.00x10-5 M

Ka = 1.8x10-5 = [CH3COO-(aq)]x[H3O+] / [CH3COOH]

=> 1.8x10-5 = y x 2.00x10-5 M / (0.10 - y)

=> 3.8x10-5y - 1.8x10-6 = 0

=> y = 1.8x10-6/ 3.8x10-5= 0.04737 M

Hence concentration of CH3COOH(aq), [CH3COOH(aq)] = 0.10 - y = 0.10 - 0.04737 M = 0.0526 M (answer) concentration of CH3COO-(aq), [CH3COO-(aq)] = y = 0.04737 M (answer)

[H3O+] = 10-4.7 = 2.00x10-5 M (answer)

(c):

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