Discuss three factors that influence the distribution of chemical species in an
ID: 490046 • Letter: D
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 1.8x10^-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_3 PO_4 doubleheadarrow H^+ + H_2 PO_4, K_1 = ? H_2 PO_4 doubleheadarrow H + HPO_4, K_2 = ? HPO_4 doubleheadarrow H^+ + PO_4, K_3 = ? -4 CH3C00W * o. o 4- * H3P04 - + H2PO4, K, = ? H2P(V ~ H + HP(V K. - ? HPO42' - H* + PO41. K, - ? y 77Explanation / Answer
Factors are : polarity of the chemical species
Density and temperature
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CH3COOH <==> CH3COO- + H+
pH = 4.7 So, [H+] = 10^-pH = 10^-4.7 = 1.99*10^-5 M
Ka = [H+][CH3COO-]/[CH3COOH]
If [H+] = [CH3COO-] = 1.99*10^-5 M
[Ch3COOH] = (0.1-1.99*10^-5) = 0.099 M
The solution is composed of 0.099 M CH3COOH, 1.99*10^-5 M OH- and 1.99*10^-5 M CH3COO-.
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Fractional value :
[H3PO4/H2PO4-] = 0.007
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(d) decimal fraction of [H3PO4] = 0.007/1+0.007 = 6.95*10^-3
concentration of H3PO4 = 62 ug/L * 6.95*10^-3 = 0.43 ug/L
decimal fraction of H2PO4^- = 1/1.007 =0.99
concentration of H2PO4^- = 62ug/L *0.99 = 61.57 ug/L
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