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 4.77 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 62mug/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 equilibrium shown below. Estimate from the diagram the values of K_1, K, . and K3, indicating what these K values mean.

Explanation / Answer

(b) acetic acid

Ka = 1.8 x 10^-5 = [H+][CH3COO-]/[CH3COOH]

1.8 x 10^-5 = x^2/0.1

x = [H+] = [CH3COO-] = 1.34 x 10^-3 M

[CH3COOH] in solution remained = 0.1 - 1.34 x 10^-3 = 0.0986 M

(c) For the phosphate solution,

at pH 4.3

pH = pKa + log([HPO4^2-]/[H2PO4-])

fraction ([HPO4^2-]/[H2PO4-]) = inv.log(4.3 - 7.2) = 0.00125

(d) total phosphate = 62 ug/L

4.3 = 7.2 + log([HPO4^2-]/[H2PO4-])

[HPO4^2-] = 0.00125[H2PO4-]

[H2PO4-] + [HPO4^2-] = 62 ug/L

[H2PO4-] + 0.00125[H2PO4-] = 62 ug/L

[H2PO4-] = 62/1.00125 = 61.92 ug/L

[HPO4^2-] = 62 - 61.92 = 0.08 ug/L

(e) From the plot,

K1 = inv.log(-2.2) = 6.31 x 10^-3

K2 = inv/log(-7.2) = 6.31 x 10^-8

K3 = inv.log(-12.5) = 3.16 x 10^-13

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.