Chemistry » Equilibria of Other Reaction Classes » Precipitation and Dissolution

Predicting Precipitation

Predicting Precipitation

The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is:

$${\text{CaCO}}_{3}(s)⇌{\text{Ca}}^{\text{2+}}(aq)+{\text{CO}}_{3}{}^{\text{2−}}(aq)$$

We can establish this equilibrium either by adding solid calcium carbonate to water or by mixing a solution that contains calcium ions with a solution that contains carbonate ions. If we add calcium carbonate to water, the solid will dissolve until the concentrations are such that the value of the reaction quotient $$(\text{Q}=[{\text{Ca}}^{2+}][{\text{CO}}_{3}{}^{\text{2−}}])$$ is equal to the solubility product (Ksp = 8.7 $$×$$ 10–9). If we mix a solution of calcium nitrate, which contains Ca2+ ions, with a solution of sodium carbonate, which contains $${\text{CO}}_{3}{}^{\text{2−}}$$ ions, the slightly soluble ionic solid CaCO3 will precipitate, provided that the concentrations of Ca2+ and $${\text{CO}}_{3}{}^{\text{2−}}$$ ions are such that Q is greater than Ksp for the mixture. The reaction shifts to the left and the concentrations of the ions are reduced by formation of the solid until the value of Q equals Ksp. A saturated solution in equilibrium with the undissolved solid will result. If the concentrations are such that Q is less than Ksp, then the solution is not saturated and no precipitate will form.

We can compare numerical values of Q with Ksp to predict whether precipitation will occur, as the example below shows. (Note: Since all forms of equilibrium constants are temperature dependent, we will assume a room temperature environment going forward in this tutorial unless a different temperature value is explicitly specified.)

Example

Precipitation of Mg(OH)2

The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of lime, Ca(OH)2, a readily available inexpensive source of OH ion:

$${\text{Mg(OH)}}_{2}(s)⇌{\text{Mg}}^{\text{2+}}(aq)+{\text{2OH}}^{\text{−}}(aq)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\text{8.9}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-12}$$

The concentration of Mg2+(aq) in sea water is 0.0537 M. Will Mg(OH)2 precipitate when enough Ca(OH)2 is added to give a [OH] of 0.0010 M?

Solution

This problem asks whether the reaction:

$${\text{Mg(OH)}}_{2}(s)⇌{\text{Mg}}^{\text{2+}}(aq)+{\text{2OH}}^{\text{−}}(aq)$$

shifts to the left and forms solid Mg(OH)2 when [Mg2+] = 0.0537 M and [OH] = 0.0010 M. The reaction shifts to the left if Q is greater than Ksp. Calculation of the reaction quotient under these conditions is shown here:

$$Q=[{\text{Mg}}^{\text{2+}}]{[{\text{OH}}^{\text{−}}]}^{2}=\text{(0.0537)(}{\text{0.0010)}}^{2}=\text{5.4}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}$$

Because Q is greater than Ksp (Q = 5.4 $$×$$ 10–8 is larger than Ksp = 8.9 $$×$$ 10–12), we can expect the reaction to shift to the left and form solid magnesium hydroxide. Mg(OH)2(s) forms until the concentrations of magnesium ion and hydroxide ion are reduced sufficiently so that the value of Q is equal to Ksp.

Example

Precipitation of AgCl upon Mixing Solutions

Does silver chloride precipitate when equal volumes of a 2.0 $$×$$ 10–4M solution of AgNO3 and a 2.0 $$×$$ 10–4M solution of NaCl are mixed?

(Note: The solution also contains Na+ and $${\text{NO}}_{3}{}^{\text{−}}$$ ions, but when referring to solubility rules, one can see that sodium nitrate is very soluble and cannot form a precipitate.)

Solution

The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is:

$$\text{AgCl}(s)⇌{\text{Ag}}^{\text{+}}(aq)+{\text{Cl}}^{\text{−}}(aq)$$

The solubility product is 1.6 $$×$$ 10–10 (see this appendix).

AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO3 and NaCl is greater than Ksp. The volume doubles when we mix equal volumes of AgNO3 and NaCl solutions, so each concentration is reduced to half its initial value. Consequently, immediately upon mixing, [Ag+] and [Cl] are both equal to:

$$\cfrac{1}{2}(2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4})\phantom{\rule{0.2em}{0ex}}M=\text{1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}M$$

The reaction quotient, Q, is momentarily greater than Ksp for AgCl, so a supersaturated solution is formed:

$$Q=[{\text{Ag}}^{\text{+}}]{[\text{Cl}}^{\text{−}}]=\text{(1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4})(1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4})=\text{1.0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}>{K}_{\text{sp}}$$

Since supersaturated solutions are unstable, AgCl will precipitate from the mixture until the solution returns to equilibrium, with Q equal to Ksp.

In the previous two examples, we have seen that Mg(OH)2 or AgCl precipitate when Q is greater than Ksp. In general, when a solution of a soluble salt of the Mm+ ion is mixed with a solution of a soluble salt of the Xn– ion, the solid, MpXq precipitates if the value of Q for the mixture of Mm+ and Xn– is greater than Ksp for MpXq. 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.

Example

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 (see the figure below). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC2O4·H2O (which also contains water bound in the solid). The concentration of Ca2+ in a sample of blood serum is 2.2 $$×$$ 10–3M. What concentration of $${\text{C}}_{2}{\text{O}}_{4}{}^{\text{2−}}$$ ion must be established before CaC2O4·H2O begins to precipitate?

Anticoagulants can be added to blood that will combine with the Ca2+ ions in blood serum and prevent the blood from clotting. (credit: modification of work by Neeta Lind)

Solution

The equilibrium expression is:

$${\text{CaC}}_{2}{\text{O}}_{4}(s)⇌{\text{Ca}}^{\text{2+}}(aq)+{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2−}}(aq)$$

For this reaction:

$${K}_{\text{sp}}=[{\text{Ca}}^{\text{2+}}][{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2−}}]=1.96\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}$$

(see this appendix)

CaC2O4 does not appear in this expression because it is a solid. Water does not appear because it is the solvent.

Solid CaC2O4 does not begin to form until Q equals Ksp. Because we know Ksp and [Ca2+], 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:

$$Q={K}_{\text{sp}}=[{\text{Ca}}^{\text{2+}}]{[\text{C}}_{2}{\text{O}}_{4}{}^{\text{2−}}]\phantom{\rule{0.2em}{0ex}}=\text{1.96}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}$$

$$(2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3})[{\text{C}}_{\text{2}}{\text{O}}_{4}{}^{\text{2−}}]=\text{1.96}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}$$

$$[{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2−}}]=\phantom{\rule{0.2em}{0ex}}\cfrac{1.96\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}}{2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\text{8.9}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}$$

A concentration of $$[{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2−}}]$$ = 8.9 $$×$$ 10–6M is necessary to initiate the precipitation of CaC2O4 under these conditions.

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 Ksp 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 the example above—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.

Example

Concentrations Following Precipitation

Clothing washed in water that has a manganese [Mn2+(aq)] concentration exceeding 0.1 mg/L (1.8 $$×$$ 10–6M) may be stained by the manganese upon oxidation, but the amount of Mn2+ in the water can be reduced by adding a base. If a person doing laundry wishes to add a buffer to keep the pH high enough to precipitate the manganese as the hydroxide, Mn(OH)2, what pH is required to keep [Mn2+] equal to 1.8 $$×$$ 10–6M?

Solution

The dissolution of Mn(OH)2 is described by the equation:

$${\text{Mn(OH)}}_{2}(s)⇌{\text{Mn}}^{\text{2+}}(aq)+{\text{2OH}}^{\text{−}}(aq)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\text{2}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-13}$$

We need to calculate the concentration of OH when the concentration of Mn2+ is 1.8 $$×$$ 10–6M. From that, we calculate the pH. At equilibrium:

$${K}_{\text{sp}}=[{\text{Mn}}^{\text{2+}}]{[{\text{OH}}^{\text{−}}]}^{2}$$

or

$$(1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}){[{\text{OH}}^{\text{−}}]}^{2}=\text{2}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-13}$$

so

$$[{\text{OH}}^{\text{−}}]=\text{3.3}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}M$$

Now we calculate the pH from the pOH:

$$\begin{array}{c}\text{pOH}=\text{−log}[{\text{OH}}^{\text{−}}]=\text{−log}(3.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}10-4)=\text{3.48}\\ \text{pH}=\text{14.00}-\text{pOH}=\text{14.00}-\text{3.80}=\text{10.52}\end{array}$$

If the person doing laundry adds a base, such as the sodium silicate (Na4SiO4) in some detergents, to the wash water until the pH is raised to 10.52, the manganese ion will be reduced to a concentration of 1.8 $$×$$ 10–6M; at that concentration or less, the ion will not stain clothing.

Due to their light sensitivity, mixtures of silver halides are used in fiber optics for medical lasers, in photochromic eyeglass lenses (glass lenses that automatically darken when exposed to sunlight), and—before the advent of digital photography—in photographic film. Even though AgCl (Ksp = 1.6 $$×$$ 10–10), AgBr (Ksp = 5.0 $$×$$ 10–13), and AgI (Ksp = 1.5 $$×$$ 10–16) are each quite insoluble, we cannot prepare a homogeneous solid mixture of them by adding Ag+ to a solution of Cl, Br, and I; essentially all of the AgI will precipitate before any of the other solid halides form because of its smaller value for Ksp. However, we can prepare a homogeneous mixture of the solids by slowly adding a solution of Cl, Br, and I to a solution of Ag+.

When two anions form slightly soluble compounds with the same cation, or when two cations form slightly soluble compounds with the same anion, the less soluble compound (usually, the compound with the smaller Ksp) generally precipitates first when we add a precipitating agent to a solution containing both anions (or both cations). When the Ksp values of the two compounds differ by two orders of magnitude or more (e.g., 10–2 vs. 10–4), almost all of the less soluble compound precipitates before any of the more soluble one does. This is an example of selective precipitation, where a reagent is added to a solution of dissolved ions causing one of the ions to precipitate out before the rest.

Selective precipitation can also be used in qualitative analysis. In this method, reagents are added to an unknown chemical mixture in order to induce precipitation. Certain reagents cause specific ions to precipitate out; therefore, the addition of the reagent can be used to determine whether the ion is present in the solution.

Resource:

View this simulation to study the process of salts dissolving and forming saturated solutions and precipitates for specific compounds, or compounds for which you select the charges on the ions and the Ksp

Example

Precipitation of Silver Halides

A solution contains 0.0010 mol of KI and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgI or solid AgCl?

Solution

The two equilibria involved are:

$$\text{AgCl}(s)⇌{\text{Ag}}^{\text{+}}(aq)+{\text{Cl}}^{\text{−}}(aq)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\text{1.6}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$$

$$\text{AgI}(s)⇌{\text{Ag}}^{\text{+}}(aq)+{\text{I}}^{\text{−}}(aq)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\text{1.5}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-16}$$

If the solution contained about equal concentrations of Cl and I, then the silver salt with the smallest Ksp (AgI) would precipitate first. The concentrations are not equal, however, so we should find the [Ag+] at which AgCl begins to precipitate and the [Ag+] at which AgI begins to precipitate. The salt that forms at the lower [Ag+] precipitates first.

For AgI: AgI precipitates when Q equals Ksp for AgI (1.5 $$×$$ 10–16). When [I] = 0.0010 M:

$$Q=[{\text{Ag}}^{\text{+}}]{[\text{I}}^{\text{−}}]=[{\text{Ag}}^{+}]\text{(0.0010)}=\text{1.5}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-16}$$

$$[{\text{Ag}}^{\text{+}}]=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1.5}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-16}}{0.10}\phantom{\rule{0.2em}{0ex}}=\text{1.6}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}$$

AgI begins to precipitate when [Ag+] is 1.6 $$×$$ 10–9M.

For AgCl: AgCl precipitates when Q equals Ksp for AgCl (1.6 $$×$$ 10–10). When [Cl] = 0.10 M:

$${Q}_{\text{sp}}=[{\text{Ag}}^{\text{+}}]{[\text{Cl}}^{\text{−}}]=[{\text{Ag}}^{\text{+}}]\text{(0.10)}=\text{1.6}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$$

$$[{\text{Ag}}^{\text{+}}]=\phantom{\rule{0.2em}{0ex}}\cfrac{1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}}{0.10}=\text{1.6}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}M$$

AgCl begins to precipitate when [Ag+] is 1.6 $$×$$ 10–9M.

AgI begins to precipitate at a lower [Ag+] than AgCl, so AgI begins to precipitate first.

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