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J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 16.34

Calculate the concentration of Cu2+ in parts per billion (ppb) in a saturated solution of copper(II) arsenate, Cu3(AsO4)2(aq). (Hint: Recall that 1 ppb signifies 1 g Cu2+ per 109 g solution.)




Answer:

First, treat the solubility equilibrium in units of molarity and then change the concentration to ppb.
Cu3(AsO4)2(s)
3 Cu2+(aq)
+
2 AsO43–(aq)
Ksp = [Cu2+]e3[AsO43–]e2 = 7.6×10–36
Initial
0
Change
+ 3x
+ 2x
Equilibrium
3x
2x

7.6×10–36 = [3x]3[2x]2 = 108x5

x = 3.7×10–8 M = molar solubility of copper(II) arsenate.

The concentration of Cu2+ ions = 3x = 3(3.7×10–8) = 1.1×10–7 M.

Assuming the density of the dilute solution is about the same as the density of water, 1 g/mL = 103 g/L, then to find ppb requires 106 L of solution:

[Cu2+] = (1.1×10–7 mol/L)(63.5 g/mol)(106 L solution per 109 g solution) = 7.0 ppb