Predicting Precipitation

Predicting Precipitation

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

$$CaCO_3(s)⇌Ca^{2+}(aq)+CO_3^{2-}(aq)\qquad K_{sp}=[Ca^{2+}][CO_3^{2-}]=8.7×10^{-19}$$

It is important to realize that this equilibrium is established in any aqueous solution containing Ca2+ and CO32– ions, not just in a solution formed by saturating water with calcium carbonate. Consider, for example, mixing aqueous solutions of the soluble compounds sodium carbonate and calcium nitrate. If the concentrations of calcium and carbonate ions in the mixture do not yield a reaction quotient, Qsp, that exceeds the solubility product, Ksp, then no precipitation will occur. If the ion concentrations yield a reaction quotient greater than the solubility product, then precipitation will occur, lowering those concentrations until equilibrium is established (Qsp = Ksp). The comparison of Qsp to Ksp to predict precipitation is an example of the general approach to predicting the direction of a reaction first introduced in the chapter on equilibrium. For the specific case of solubility equilibria:

Qsp < Ksp: the reaction proceeds in the forward direction (solution is not saturated; no precipitation observed)

Qsp > Ksp: the reaction proceeds in the reverse direction (solution is supersaturated; precipitation will occur)

This predictive strategy and related calculations are demonstrated in the next few example exercises.

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:

$$Mg(OH)_2(s)⇌Mg^{2+}(aq)+2OH^-(aq)\qquad K_{sp}=8.9×10^{-12}$$

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?


Calculation of the reaction quotient under these conditions is shown here:


Because Q is greater than Ksp (Q = $5.4×10^{-8}$ is larger than Ksp = $8.9×10^{-12}$), the reverse reaction will proceed, precipitating magnesium hydroxide until the dissolved ion concentrations have been sufficiently lowered, so that Qsp = Ksp.

Check Your Learning
Predict whether CaHPO4 will precipitate from a solution with [Ca2+] = 0.0001 M and $[HPO_4^{2-}]$= 0.001 M.


No precipitation of CaHPO4; Q = $1×10^{-7}$, which is less than Ksp (7 × 10–7)

Precipitation of AgCl
Does silver chloride precipitate when equal volumes of a $2.0×10^{-4}\;M$ solution of AgNO3 and a $2.0×10^{-4}\;M$ solution of NaCl are mixed?

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


The solubility product is $1.6×10^{-10}$ (see Appendix on solubility products).

AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO3 and NaCl is greater than Ksp. Because the volume doubles when equal volumes of AgNO3 and NaCl solutions are mixed, each concentration is reduced to half its initial value


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


AgCl will precipitate from the mixture until the dissolution equilibrium is established, with Q equal to Ksp.

Check Your Learning
Will KClO4 precipitate when 20 mL of a 0.050-M solution of K+ is added to 80 mL of a 0.50-M solution of $ClO_4^-$ (Hint: Use the dilution equation to calculate the concentrations of potassium and perchlorate ions in the mixture.)


No, Q = $4.0×10^{-3}$, which is less than Ksp = $1.05×10^{-2}$

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, $C_2O_4^{2-}$, for this purpose. At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC2O4·H2O (calcium oxalate monohydrate). The concentration of Ca2+ in a sample of blood serum is $2.2×10^{-3}\;M$. What concentration of $C_2O_4^{2-}$ 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)

SolutionThe equilibrium expression is:


For this reaction:


(see Appendix on solubility products)

Substitute the provided calcium ion concentration into the solubility product expression and solve for oxalate concentration:

$$Q=K_{sp}=[Ca^{2+}][C_2O_4^{2-}]=1.96×10^{-8}$$ $$(2.2×10^{-3})[C_2O_4^{2-}]=1.96×10^{-8}$$ $$[C_2O_4^{2-}]=\frac{1.96×10^{-8}}{2.2×10^{-3}}=8.9×10^{-6}\;M$$

A concentration of $[C_2O_4^{2-}]=8.9×10^{-6}\;M$ is necessary to initiate the precipitation of CaC2O4 under these conditions.

Check Your Learning
If a solution contains 0.0020 mol of $CrO_4^{2-}$ per liter, what concentration of Ag+ ion must be reached by adding solid AgNO3 before Ag2CrO4 begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.



Concentrations Following Precipitation
Clothing washed in water that has a manganese [Mn2+(aq)] concentration exceeding 0.1 mg/L ($1.8×10^{-6}\;M$) may be stained by the manganese upon oxidation, but the amount of Mn2+ in the water can be decreased by adding a base to precipitate Mn(OH)2. What pH is required to keep [Mn2+] equal to $1.8×10^{-6}\;M$?

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

$$Mn(OH)_2(s)⇌Mn^{2+}(aq)+2OH^-(aq)\qquad K_{sp}=2×10^{-13}$$

At equilibrium:






Calculate the pH from the pOH:

$$\text{pOH}=-log[OH^-]=-log(3.3×10^{-4})=3.48$$ $$\text{pH}=14.00-\text{pOH}=14.00-3.48=10.52$$

(final result rounded to one significant digit, limited by the certainty of the Ksp)

Check Your Learning
The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of Ca(OH)2. The concentration of Mg2+(aq) in sea water is $5.37×10^{-2}\;M$. Calculate the pH at which [Mg2+] is decreased to $1.0×10^{-5}\;M$



In solutions containing two or more ions that may form insoluble compounds with the same counter ion, an experimental strategy called selective precipitation may be used to remove individual ions from solution. By increasing the counter ion concentration in a controlled manner, ions in solution may be precipitated individually, assuming their compound solubilities are adequately different. In solutions with equal concentrations of target ions, the ion forming the least soluble compound will precipitate first (at the lowest concentration of counter ion), with the other ions subsequently precipitating as their compound’s solubilities are reached. As an illustration of this technique, the next example exercise describes separation of a two halide ions via precipitation of one as a silver salt.

The Role of Precipitation in Wastewater Treatment

Solubility equilibria are useful tools in the treatment of wastewater carried out in facilities that may treat the municipal water in your city or town ([link]). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions $(PO_4^{3-})$ are often present in the water discharged from manufacturing facilities. An abundance of phosphate causes excess algae to grow, which impacts the amount of oxygen available for marine life as well as making water unsuitable for human consumption.

Wastewater treatment facilities, such as this one, remove contaminants from wastewater before the water is released back into the natural environment. (credit: “eutrophication&hypoxia”/Wikimedia Commons)

One common way to remove phosphates from water is by the addition of calcium hydroxide, or lime, Ca(OH)2. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca5(PO4)3OH, which then precipitates out of the solution:


Because the amount of calcium ion added does not result in exceeding the solubility products for other calcium salts, the anions of those salts remain behind in the wastewater. The precipitate is then removed by filtration and the water is brought back to a neutral pH by the addition of CO2 in a recarbonation process. Other chemicals can also be used for the removal of phosphates by precipitation, including iron(III) chloride and aluminum sulfate.

View this site for more information on how phosphorus is removed from wastewater.

Precipitation of Silver Halides
A solution contains 0.00010 mol of KBr and 0.10 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgBr or solid AgCl?

The two equilibria involved are:

$$AgCl(s)⇌Ag^+(aq)+Cl^-(aq)\qquad K_{sp}=1.6×10^{-10}$$ $$AgBr(s)⇌Ag^+(aq)+Br^-(aq)\qquad K_{sp}=5.0×10^{-13}$$

If the solution contained about equal concentrations of Cl and Br, then the silver salt with the smaller Ksp (AgBr) would precipitate first. The concentrations are not equal, however, so the [Ag+] at which AgCl begins to precipitate and the [Ag+] at which AgBr begins to precipitate must be calculated. The salt that forms at the lower [Ag+] precipitates first.

AgBr precipitates when Q equals Ksp for AgBr

$$Q_{sp}=K_{sp}=[Ag^+][Br^-]=[Ag^+](0.00010)=5.0×10^{-13}$$ $$[Ag^+]=frac{5.0×10^{-13}}{0.00010}=5.0×10^{-9}\;M$$

AgBr begins to precipitate when [Ag+] is $5.0×10^{-9}$$.

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

$$Q_{sp}=K_{sp}=[Ag^+][Cl^-]=[Ag^+](0.10)=1.6×10^{-10}$$ $$[Ag^+]=\frac{1.6×10^{-10}}{0.10}=1.6×10^{-9}\;M$$

AgCl begins to precipitate when [Ag+] is $1.6×10^{-9}\;M$.

AgCl begins to precipitate at a lower [Ag+] than AgBr, so AgCl begins to precipitate first. Note the chloride ion concentration of the initial mixture was significantly greater than the bromide ion concentration, and so silver chloride precipitated first despite having a Ksp greater than that of silver bromide.

Check Your Learning
If silver nitrate solution is added to a solution which is 0.050 M in both Cl and Br ions, at what [Ag+] would precipitation begin, and what would be the formula of the precipitate?


[Ag+] = $1.0×10^{-11}\;M$; AgBr precipitates first