6.5 Precipitation and dissolution  (Page 6/17)

In the previous two examples, we have seen that Mg(OH) ₂ or AgCl precipitate when Q is greater than K _sp . In general, when a solution of a soluble salt of the M ^m+ ion is mixed with a solution of a soluble salt of the X ^n– ion, the solid, M _p X _q precipitates if the value of Q for the mixture of M ^m+ and X ^n– is greater than K _sp for M _p X _q . Thus, if we know the concentration of one of the ions of a slightly soluble ionic solid and the value for the solubility product of the solid, then we can calculate the concentration that the other ion must exceed for precipitation to begin. To simplify the calculation, we will assume that precipitation begins when the reaction quotient becomes equal to the solubility product constant.

Precipitation of calcium oxalate

Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, ${\text{C}}_2{\text{O}}_4{}^{\text{2−}},$ for this purpose ( [link] ). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC ₂ O ₄ ·H ₂ O (which also contains water bound in the solid). The concentration of Ca ²⁺ in a sample of blood serum is 2.2 $\times$ 10 ^–3 M . What concentration of ${\text{C}}_2{\text{O}}_4{}^{\text{2−}}$ ion must be established before CaC ₂ O ₄ ·H ₂ O begins to precipitate?

Solution

The equilibrium expression is:

${\text{CaC}}_2{\text{O}}_4(s)\rightleftharpoons {\text{Ca}}^{\text{2+}}(aq)+{\text{C}}_2{\text{O}}_4{}^{\text{2−}}(aq)$

For this reaction:

$K_{\text{sp}}=[{\text{Ca}}^{\text{2+}}][{\text{C}}_2{\text{O}}_4{}^{\text{2−}}]=1.96\times {10}^{-8}$

(see Appendix J )

CaC ₂ O ₄ does not appear in this expression because it is a solid. Water does not appear because it is the solvent.

Solid CaC ₂ O ₄ does not begin to form until Q equals K _sp . Because we know K _sp and [Ca ²⁺ ], we can solve for the concentration of ${\text{C}}_2{\text{O}}_4{}^{\text{2−}}$ that is necessary to produce the first trace of solid:

$Q=K_{\text{sp}}=[{\text{Ca}}^{\text{2+}}]{[\text{C}}_2{\text{O}}_4{}^{\text{2−}}]=\text{1.96}\times {10}^{-8}$

$(2.2\times {10}^{-3})[{\text{C}}_{\text{2}}{\text{O}}_4{}^{\text{2−}}]=\text{1.96}\times {10}^{-8}$

$[{\text{C}}_2{\text{O}}_4{}^{\text{2−}}]=\frac{1.96\times {10}^{-8}}{2.2\times {10}^{-3}}=\text{8.9}\times {10}^{-6}$

A concentration of $[{\text{C}}_2{\text{O}}_4{}^{\text{2−}}]$ = 8.9 $\times$ 10 ^–6 M is necessary to initiate the precipitation of CaC ₂ O ₄ under these conditions.

Check your learning

If a solution contains 0.0020 mol of ${\text{CrO}}_4{}^{\text{2−}}$ per liter, what concentration of Ag ⁺ ion must be reached by adding solid AgNO ₃ before Ag ₂ CrO ₄ begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.

Answer:

4.5 $\times$ 10 ^–9 M

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It is sometimes useful to know the concentration of an ion that remains in solution after precipitation. We can use the solubility product for this calculation too: If we know the value of K _sp and the concentration of one ion in solution, we can calculate the concentration of the second ion remaining in solution. The calculation is of the same type as that in [link] —calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. However, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation.