6.4 – Equilibria of Slightly Soluble Ionic Compounds

Solubility

When a solid substance is mixed with a solvent in order to form a solution, there are several possible results. The determining factor for the result is the solubility of the substance, which is defined as the maximum possible concentration of the solute. The solubility rules help determine which substances are soluble, and to what extent.

Solubility Effects on Reactions

Depending on the solubility of a solute, there are three possible results: 1) if the solution has less solute than the maximum amount that it is able to dissolve (its solubility), it is a dilute unsaturated solution; 2) if the amount of solute is exactly the same amount as its solubility, it is saturated; 3) if there is more solute than is able to be dissolved, it is said to be supersaturated and the excess solute separates from the solution. If this separation process includes crystallization, it forms a precipitate. Precipitation lowers the concentration of the solute to the saturation in order to increase the stability of the solution.

We can use the following general rules for predicting the relative solubility of many ionic compounds:

Solubility Rules

1. Salts containing Group I elements (Li⁺, Na⁺, K⁺, Cs⁺, Rb⁺) are soluble. There are few exceptions to this rule. Salts containing the ammonium ion (NH₄⁺) are also soluble.

2. Salts containing nitrate ion (NO₃^–) are generally soluble.

3. Salts containing Cl^–, Br^–, or I^– are generally soluble. Important exceptions to this rule are halide salts of Ag⁺, Pb²⁺, and (Hg₂)²⁺. Thus, AgCl, PbBr₂, and Hg₂Cl₂ are insoluble.

4. Most silver salts are insoluble. AgNO₃ and Ag(C₂H₃O₂) are common soluble salts of silver; virtually all others are insoluble.

5. Most sulfate salts are soluble. Important exceptions to this rule include CaSO₄, BaSO₄, PbSO₄, Ag₂SO₄ and SrSO₄ .

6. Most hydroxide salts are only slightly soluble. Hydroxide salts of Group I elements are soluble. Hydroxide salts of Group II elements (Ca, Sr, and Ba) are slightly soluble. Hydroxide salts of transition metals and Al³⁺ are insoluble. Thus, Fe(OH)₃, Al(OH)₃, Co(OH)₂ are not soluble.

7. Most sulfides of transition metals are highly insoluble, including CdS, FeS, ZnS, and Ag₂S. Arsenic, antimony, bismuth, and lead sulfides are also insoluble.

8. Carbonates are frequently insoluble. Group II carbonates (CaCO₃, SrCO₃, and BaCO​₃) are insoluble, as are FeCO₃ and PbCO₃.

9. Chromates are frequently insoluble. Examples include PbCrO₄ and BaCrO₄.

10. Phosphates such as Ca₃(PO₄)₂ and Ag₃PO₄ are frequently insoluble.

11. Fluorides such as BaF_2, MgF₂ and PbF₂ are frequently insoluble.

Example 6.4.1 – Predicting Precipitation Reactions 

Titration

Predict the result of mixing reasonably concentrated solutions of the following ionic compounds. If precipitation is expected, write a balanced net ionic equation for the reaction.

(a) Potassium sulfate and barium nitrate

(b) Lithium chloride and silver acetate

(c) Lead nitrate and ammonium carbonate

Solution

a) The two possible products for this combination are KNO₃ and BaSO₄. The solubility guidelines indicate BaSO₄ is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in section 1.4, is:

Ba²⁺ (aq) + SO₄^2- (aq) → BaSO₄ (s)

b) The two possible products for this combination are LiC₂H₃O₂ and AgCl. The solubility guidelines indicate AgCl is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is:

Ag⁺ (aq) + Cl^– (aq) → AgCl (s)

c) The two possible products for this combination are PbCO₃ and NH₄NO₃. The solubility guidelines indicate PbCO₃ is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is:

Pb²⁺ (aq) + CO₃^2- (aq) → PbCO₃ (s)

Check Your Learning 6.4.1 – Predicting Precipitation Reactions 

Predict the result of mixing reasonably concentrated solutions of the following ionic compounds. If precipitation is expected, write a balanced net ionic equation for the reaction.

(a) Calcium bromide and sodium carbonate

(b)Ammonium chromate and silver nitrate

(c) Cesium chloride and magnesium iodide

Answer

(a) Ca²⁺ (aq) + CO₃^2– (aq) → CaCO₃ (s)

(b) 2 Ag⁺ (aq) + CrO₄^2– (aq) →Ag₂CrO₄ (s)

(c) No reaction (the two possible salts, cesium iodide and magnesium chloride, are soluble)

The Solubility Product Constant

By applying the above solubility rules, we can make qualitative predictions about precipitation reactions. However, these precipitates still exist in a dynamic equilibrium with aqueous ions, since these are technically sparingly (or slightly) soluble compounds that dissolve partially in aqueous solutions. For example, according to the solubility rules, silver chloride, AgCl, is classified as a sparingly soluble ionic solid (Figure 6.4.1). However, when we add an excess of solid AgCl to water, it dissolves to a small extent and produces a mixture consisting of a very dilute solution of Ag⁺ and Cl^– ions in equilibrium with undissolved silver chloride:

AgCl (s) ⇌ Ag⁺ (aq) + Cl^– (aq)

This equilibrium, like other equilibria, is dynamic; some of the solid AgCl is continuously dissolving, but at the same time, Ag⁺ and Cl^– ions in the solution combine to produce an equal amount of the solid. At equilibrium, the opposing processes have equal rates, and macroscopically, the concentrations of dissolved ions and mass of remaining solid appear constant.

Figure 6.4.1. Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag⁺ and Cl^– ions in equilibrium with undissolved silver chloride.

The equilibrium constant for the equilibrium between a slightly soluble ionic solid and a solution of its ions is the solubility product (K_sp) of the solid. Recall from Chapter 4 that we use an ion’s concentration as an approximation of its activity in a dilute solution. For silver chloride, at equilibrium:

AgCl (s) ⇌ Ag⁺ (aq) + Cl^– (aq)

When looking at dissolution reactions such as this, the solid is listed as a reactant, whereas the ions are listed as products. The solubility product constant, as with every equilibrium constant expression, is written as the product of the concentrations of each of the ions, raised to the power of their stoichiometric coefficients. Here, the solubility product constant is equal to [Ag⁺] x [Cl^–] when a solution of silver chloride is in equilibrium with undissolved AgCl. There is no denominator representing the reactants in this equilibrium expression since the reactant is a pure solid; therefore [AgCl] does not appear in the expression for K_sp.

Some common solubility products are listed in Table 6.4.1, according to their K_sp values, whereas a more extensive compilation of products appears in Appendix L. Each of these equilibrium constants is much smaller than 1 because the compounds listed are only slightly soluble. A small K_sp represents a system in which the equilibrium lies to the left, so that relatively few hydrated ions would be present in a saturated solution.

Table 6.4.1 Common Solubility Products and their Equilibrium Constants

Example 6.4.2 – Writing Equations and Solubility Products

Write the ionic equation for the dissolution and the solubility product expression for each of the following slightly soluble ionic compounds:

(a) AgI, silver iodide, a solid with antiseptic properties

(b) CaCO₃, calcium carbonate, the active ingredient in many over-the-counter chewable antacids

(c) Mg(OH)₂, magnesium hydroxide, the active ingredient in Milk of Magnesia

(d) Mg(NH₄)PO₄, magnesium ammonium phosphate, an essentially insoluble substance used in tests for magnesium

(e) Ca₅(PO₄)₃OH, the mineral apatite, a source of phosphate for fertilizers

(Hint: When determining how to break (d) and (e) up into ions, refer to the list of polyatomic ions in the section on chemical nomenclature.)

Solution

(a) AgI (s) ⇌ Ag⁺ (aq) + I^– (aq)

K_sp = [Ag⁺][I^–]

(b) CaCO₃ (aq) ⇌ Ca²⁺ (aq) + CO₃^2- (aq)

K_sp = [Ca²⁺][CO₃^2-]

(c) Mg(OH)₂ (s) ⇌ Mg²⁺ (aq) + 2OH^– (aq)

K_sp = [Mg²⁺(aq)][OH^–]²

(d) Mg(NH₄)PO₄ (s) ⇌ Mg⁺ (aq) + NH₄⁺ (aq) + PO₄^3- (aq)

K_sp = [Mg²⁺][NH₄⁺][PO₄^3-]

(e) Ca₅(PO₄)₃OH (s) ⇌ 5Ca²⁺ (aq) + 3PO₄^3- (aq) + OH^– (aq)

K_sp = [Ca²⁺]⁵[PO₄^3-]³[OH^–]

Check Your Learning 6.4.1 – Writing Equations and Solubility Products

Write the ionic equation for the dissolution and the solubility product for each of the following slightly soluble compounds:

(a) BaSO₄

(b) Ag₂SO₄

(c) Al(OH)₃

(d) Pb(OH)Cl

Answer

(a) BaSO₄ (s) ⇌ Ba²⁺ (aq) + SO₄^2- (aq)

K_sp = [Ba²⁺][SO₄^2-]

(b) Ag₂SO₄ (s) ⇌ 2Ag⁺ (aq) + SO₄^2- (aq)

K_sp = [Ag⁺]²[SO₄^2-]

(c) Al(OH)₃ (s) ⇌ Al³⁺ (aq) + 3OH^– (aq)

K_sp = [Al³⁺][OH^–]³

(d) Pb(OH)Cl (s) ⇌ Pb²⁺ (aq) + OH^– (aq)+ Cl^– (aq)

K_sp = [Pb²⁺][OH^–][Cl^–]

Now we will extend the discussion of K_sp and show how the solubility product constant is determined from the solubility of its ions, as well as how K_sp can be used to determine the molar solubility of a substance.

The Ion Product

The ion product (Q) of a salt is the product of the concentrations of the ions in solution raised to the same powers as in the solubility product expression. It is analogous to the reaction quotient (Q) discussed for equilibria in Chapter 4. Whereas K_sp describes equilibrium concentrations, the ion product describes concentrations that are not necessarily equilibrium concentrations.

As summarized in Figure 6.4.2, there are three possible conditions for an aqueous solution of an ionic solid:

• Q < K_sp The solution is unsaturated, and more of the ionic solid, if available, will dissolve.
• Q = K_sp The solution is saturated and at equilibrium.
• Q > K_sp The solution is supersaturated, and ionic solid will precipitate.

Figure 6.4.2. The relationship between Q and K_sp. If Q is less than K_sp, the solution is unsaturated and more solid will dissolve until the system reaches equilibrium (Q = K_sp). If Q is greater than K_sp, the solution is supersaturated and solid will precipitate until Q = K_sp. If Q = K_sp, the rate of dissolution is equal to the rate of precipitation; the solution is saturated, and no net change in the amount of dissolved solid will occur.

The process of calculating the value of the ion product and comparing it with the magnitude of the solubility product is a straightforward way to determine whether a solution is unsaturated, saturated, or supersaturated. More important, the ion product tells chemists whether a precipitate will form when solutions of two soluble salts are mixed.

K_sp and Solubility

Recall that the definition of solubility is the maximum possible concentration of a solute in a solution at a given temperature and pressure. We can determine the solubility product of a slightly soluble solid from that measure of its solubility at a given temperature and pressure, provided that the only significant reaction that occurs when the solid dissolves is its dissociation into solvated ions, that is, the only equilibrium involved is:

M_pX_q (s) ⇌ pM^m+ (aq) + qX^n– (aq)

Equation 6.4.1 – Dissolution of a slightly soluble solid.

In this case, we calculate the solubility product by taking the solid’s solubility expressed in units of moles per liter (mol/L), known as its molar solubility.

Example 6.4.3 – Calculation of K_sp from Equilibrium Concentrations

Fluorite, CaF₂, is a slightly soluble solid that dissolves according to the equation:

CaF₂ (s) ⇌ Ca²⁺ (aq) + 2F^– (aq)

The concentration of Ca²⁺ in a saturated solution of CaF₂ is 2.15 × 10^–4 M; therefore, that of F^– is 4.30 × 10^–4 M, that is, twice the concentration of Ca²⁺. What is the solubility product of fluorite?

Solution

First, write out the K_sp expression, then substitute in concentrations and solve for K_sp:

CaF₂ (s) ⇌ Ca²⁺ (aq) + 2F^– (aq)

A saturated solution is a solution at equilibrium with the solid. Thus:

K_sp = [Ca²⁺][F^–]² = (2.1×10^-4)(4.2×10^-4)² = 3.7×10^-11

As with other equilibrium constants, we do not include units with K_sp.

Check Your Learning 6.4.2 – Calculation of K_sp from Equilibrium Concentrations

In a saturated solution that is in contact with solid Mg(OH)₂, the concentration of Mg²⁺ is 1.31 × 10^–4 M. What is the solubility product for Mg(OH)₂?

Mg(OH)₂ (s) ⇌ Mg²⁺ (aq) + 2OH^– (aq)

Answer

K_sp = 8.99 × 10^–12

Example 6.4.4 – Determination of Molar Solubility from K_sp

The K_sp of copper(I) bromide, CuBr, is 6.3 × 10^–9. Calculate the molar solubility of copper bromide.

Solution

The solubility product constant of copper(I) bromide is 6.3 × 10^–9.

The reaction is:

CuBr (s) ⇌ Cu⁺ (aq) + Br^– (aq)

First, write out the solubility product equilibrium constant expression:

K_sp = [Cu⁺][Br^–]

Create an ICE table (as introduced in Chapter 4), leaving the CuBr column empty as it is a solid (its activity is equal to one) and thus does not appear in the K_sp expression:

CuBr (s) ⇌ Cu⁺ (aq) + Br^– (aq)

At equilibrium:

K_sp = [Cu⁺][Br^–]

6.3×10^-9 = (x)(x)=x²

x = sqrt(6.3×10^-9) = 7.9×10^-5

Therefore, the molar solubility of CuBr is 7.9 × 10^–5 M.

Check Your Learning 6.4.3 – Determination of Molar Solubility from K_sp

The K_sp of AgI is 1.5 × 10^–16. Calculate the molar solubility of silver iodide.

Answer

1.2 × 10^–8 M

Example 6.4.5 – Determination of Molar Solubility from K_sp, Part II

The K_sp of calcium hydroxide, Ca(OH)₂, is 1.3 × 10^–6. Calculate the molar solubility of calcium hydroxide.

Solution

The solubility product constant of calcium hydroxide is 1.3 × 10^–6.

The reaction is:

Ca(OH)₂ (s) ⇌ Ca²⁺ (aq) + 2 OH^– (aq)

First, write out the solubility product equilibrium constant expression:

K_sp = [Ca²⁺][OH^–]²

Create an ICE table, leaving the Ca(OH)₂ column empty as it is a solid and does not contribute to the K_sp:

Ca(OH)₂ (s) ⇌ Ca²⁺ (aq) + 2OH^– (aq)

At equilibrium:

K_sp = [Ca²⁺][OH^–]²

1.3×10^-6 = (x)(2x)² = (x)(4x²) = 4x³

x = cbrt(1.3×10^-6/4) = 6.9×10^-3

Therefore, the molar solubility of Ca(OH)₂ is 1.3 × 10^–2 M.

Check Your Learning 6.4.4 – Determination of Molar Solubility from K_sp, Part II

The K_sp of PbI₂ is 1.4 × 10^–8. Calculate the molar solubility of lead(II) iodide.

Answer

1.5 × 10^–3 M

Note that solubility is not always given as a molar value. When the solubility of a compound is given in some unit other than moles per liter, we must convert the solubility into moles per liter (i.e., molarity) in order to use it in the solubility product constant expression. The following example shows how to perform those unit conversions before determining the solubility product equilibrium.

Example 6.4.6 – Determination of K_sp from Gram Solubility

Many of the pigments used by artists in oil-based paints (Figure 6.4.3) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO₄, is 4.6 × 10^–6 g/L. Determine the solubility product equilibrium constant for PbCrO₄.

Figure 6.4.3. Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow (PbCrO₄), examples include Prussian blue (Fe₇(CN)₁₈), the reddish-orange colour vermilion (HgS), and green colour veridian (Cr₂O₃). (credit: Sonny Abesamis)

Solution

We are given the solubility of PbCrO₄ in grams per litre. If we convert this solubility into moles per litre, we can find the equilibrium concentrations of Pb²⁺ and CrO₄²⁻, then K_sp:

Use the molar mass of PbCrO₄ (323.2g/mol) to convert the solubility of PbCrO₄ in grams per litre into moles per litre:

[PbCrO₄] = (4.6×10^-6 g PbCrO₄/ 1 L) × (1 mol PbCrO₄ /  323.2 g PbCrO₄)

= 1.4×10^-8 mol PbCrO₄ / 1 L

[PbCrO₄] = 1.4×10^-8 M

The chemical equation for the dissolution indicates that 1 mol of PbCrO₄ gives 1 mol of Pb²⁺_(aq) and 1 mol of CrO₄²⁻_(aq):

PbCrO₄ (s) ⇌ Pb²⁺ (aq) + CrO₄^2- (aq)

Thus, at equilibrium, both [Pb²⁺] and [CrO₄²⁻] are equal to the molar solubility of PbCrO₄:

[Pb²⁺(aq)] = [CrO₄^2-(aq)] = 1.4×10^-8 M

Solve.

K_sp = [Pb²⁺][CrO₄²⁻] = (1.4 × 10^–8)(1.4 × 10^–8) = 2.0 × 10^–16

Check Your Learning 6.4.5 – Determination of K_sp from Gram Solubility

The solubility of thallium(I) chloride (TlCl), an intermediate formed when thallium is being isolated from ores, is 3.46 grams per litre at 20°C. What is its solubility product?

Answer

1.7 × 10^–4

Example 6.4.7 – Calculating the Solubility of Hg₂Cl₂

Calomel, Hg₂Cl₂, is a compound composed of the diatomic ion of mercury(I), Hg₂²⁺, and chloride ions, Cl^–. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel is quite insoluble:

Hg₂Cl₂ (s) ⇌ Hg₂²⁺ (aq) + 2Cl^– (aq)

Calculate the molar solubility of Hg₂Cl₂.

Solution

The molar solubility of Hg₂Cl₂ is equal to the concentration of Hg₂²⁺ ions because for each 1 mol of Hg₂Cl₂ that dissolves, 1 mol of Hg₂²⁺ forms:

Determine the direction of change. Before any Hg₂Cl₂ dissolves, Q is zero, and the reaction will shift to the right to reach equilibrium.

Determine x and equilibrium concentrations. Concentrations and changes are given in the following ICE table:

Hg₂Cl₂ (s) ⇌ Hg₂²⁺(aq) + 2Cl^– (aq)

Note that the change in the concentration of Cl^– (2x) is twice as large as the change in the concentration of Hg₂²⁺ (x) because 2 mol of Cl^– forms for each 1 mol of Hg₂²⁺ that forms. Hg₂Cl₂ is a pure solid, so it does not appear in the calculation.

Solve for x and the equilibrium concentrations. We substitute the equilibrium concentrations into the expression for K_sp and calculate the value of x:

K_sp = [Hg₂²⁺][Cl^–]²

1.1×10^-18 =(x)(2x)²

4x³=1.1×10^-18

x = cbrt(1.1×10^-18/4) = 6.5×10^-7 M

[Hg₂²⁺] = 6.5×10^-7 M = 6.5×10^-7 M

[Cl^–] = 2x = 2(6.5×10^-7 M) = 1.3×10^-6 M

The molar solubility of Hg₂Cl₂ is equal to [Hg₂²⁺], or 6.5 × 10^–7 M.

Check the work. At equilibrium, Q = K_sp:

Q = [Hg₂²⁺][Cl^–]² = (6.5×10^-7)(1.3×10^-6)² = 1.1×10^-18

The calculations check out.

Check Your Learning 6.4.6 – Calculating the Solubility of Hg₂Cl₂

Determine the molar solubility of MgF₂ from its solubility product: K_sp = 6.4 × 10^–9.

Answer

1.2 × 10^–3 M

Tabulated K_sp values can also be compared to reaction quotients calculated from experimental data to tell whether a solid will precipitate in a reaction under specific conditions: Q equals K_sp at equilibrium; if Q is less than K_sp, the solid will dissolve until Q equals K_sp; if Q is greater than K_sp, precipitation will occur at a given temperature until Q equals K_sp.

The Common Ion Effect and Solubility

What happens if we try to dissolve a sparingly soluble salt in an aqueous solution that already contains one of the salt’s ions? Following Section 6.1, this would be an example of the common ion effect. According to Le Chatelier’s Principle, the presence of product ion shifts the equilibrium towards the solid reactant, and we expect the solubility of the salt to be reduced.

Example 6.4.8 – The Common Ion Effect and Solubility

If we attempt to dissolve lead(II) chloride in a 0.100 M sodium chloride solution instead of in water, what is the equilibrium concentration of the lead(II) ions, as compared to pure water?

First, let us determine the solubility of PbCl₂ in pure water. We can define x to be the concentration of the lead(II) ions at equilibrium, and use an ICE table:

PbCl₂ (s) ⇌ Pb²⁺(aq) + 2Cl^–(aq)

We substitute the equilibrium concentrations into the expression for K_sp and calculate the value of x:

K_sp = [Pb²⁺][Cl^–]²

1.7×10^-5 = (x)(2x)²

4x³ = 1.7×10^-5

x = cbrt(1.7×10^-5/4) = 0.016 M

Now, to find the solubility of lead (II) chloride in 0.100 M NaCl, we repeat this process, only we take into account the presence of the common ion. In this example, the common ion is the chloride ion. Since NaCl is a soluble salt, we assume 100% ionization and thus [Cl^–]_i = 0.100 M.

PbCl₂ (s) ⇌ Pb²⁺(aq) + 2Cl^–(aq)

Again, we substitute the equilibrium concentrations into the expression for K_sp and calculate the value of x, and to simplify the calculation, we assume that the value of x is small:

K_sp = [Pb²⁺][Cl^–]²

1.7×10^-5 = (x)(0.100+2x)²

1.7×10^-5 = (x)(0.100)²

x = 1.7×10^-3 M

Finally, let’s compare the obtained values:

Original solution:

[Pb²⁺] = 0.016 M

In 0.100 M NaCl solution:

[Pb²⁺] = 0.0017 M

The concentration of the lead(II) ions has decreased by a factor of about 10, confirming that the common ion effect inhibits the solubility of a sparingly soluble salt. If more concentrated solutions of sodium chloride are used, the solubility decreases further.

Check Your Learning 6.4.7 – The Common Ion Effect and Solubility

BaSO₄ is a sparingly soluble salt that ionizes into Ba²⁺ and SO₄^2-:

BaSO₄ (s) ⇌ Ba²⁺(aq) + SO₄^2-(aq)

K_sp = 1.1×10^-10

Find the concentration of each ion at equilibrium for a saturated solution of barium sulfate. Next, lithium sulfate (Li₂SO₄) is added to the reaction vessel until its concentration is 0.500 M. What is the concentration of barium and sulfate ions after the addition of lithium sulfate?

Answer

Before addition of Li₂SO₄:

[Ba²⁺] = 1.0 x 10^-5 M , [SO₄^2-] = 1.0 x 10^-5 M

After addition of Li₂SO₄:

[Ba²⁺] = 2.2 x 10^-10 M , [SO₄^2-] = 0.500 M

Check Your Learning 6.4.8 – The Common Ion Effect and Solubility

Calculate the solubility of silver carbonate in a 0.25 M solution of sodium carbonate. The K_sp of silver carbonate is 8.46 × 10⁻¹² at 25°C.

Answer

2.9 × 10⁻⁶ M (versus 1.3 × 10⁻⁴ M in pure water)

Effect of pH on solubility

For certain salts, hydronium and hydroxide ion concentrations can influence solubility, as a special case of the common ion effect. Taking the concept of hydrolysis, introduced in Chapter 5, and applying Le Chatelier’s Principle to sparingly soluble salts, we can qualitatively predict that:

• ACIDIC salts will be less soluble in acid (pH < 7), and more soluble in base (pH > 7).
• BASIC salts will be less soluble in base (pH > 7), and more soluble in acid (pH < 7).
• NEUTRAL salts have solubilities unaffected by pH.

Consider the example of cobalt (II) hydroxide. It dissolves in the following equilibrium:

Co(OH)₂ (s) ⇋ Co²⁺ (aq) + 2 OH^– (aq)

K_sp = 2.5 x 10^-16 = [Co²⁺][OH^–]²

We can compare the solubility of this compound in pure water (where [OH^–] = 1.0 x 10^–7) versus its solubility in a pH = 6 buffer solution (where [OH^–] = 1.0 x 10^–8), using the following two ICE tables:

From the above calculations, we see that Co(OH)₂, originally only slightly soluble in neutral water, becomes much more soluble under acidic conditions. Why is this so? Complete the exercise below to illustrate the reason.

Check Your Learning 6.4.9 – The Common Ion Effect and Solubility

The equilibrium for the dissolution of cobalt (II) hydroxide is:

Co(OH)₂ (s) ⇋ Co²⁺ (aq) + 2 OH^– (aq) K_sp = 5.9 x 10^–15

In acidic solution, each hydroxide ion produced is neutralized by a hydronium ion. As we saw in Chapter 5, the net ionic equation for the acid-base neutralization is the inverse of the autoionization of water:

OH^– (aq) + H₃O⁺ (aq) ⇋ 2H₂O(l) K = 1/K_w = 1.0 x 10¹⁴

Combine the two equilibria provided, taking note of stoichiometry, to find an overall chemical equation for the dissolution of cobalt (II) hydroxide in acidic solution, as well as the value of the equilibrium constant. (Recall the rules for manipulating equilibrium constants from Chapter 4).

Answer:

Co(OH)₂ (s) + 2H₃O⁺ (aq) ⇋ Co²⁺ (aq) + 4H₂O (l), K = 5.9 x 10¹³

Predicting Precipitation

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

CaCO₃ (s) ⇌ Ca²⁺ (aq) + CO₃^2- (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 (Q = [Ca²⁺][CO₃²⁻]) is equal to the solubility product (K_sp = 8.7 × 10^–9). If we mix a solution of calcium nitrate, which contains Ca²⁺ ions, with a solution of sodium carbonate, which contains CO₃²⁻ ions, the slightly soluble ionic solid CaCO₃ will precipitate, provided that the concentrations of Ca²⁺ and CO₃²⁻ ions are such that Q is greater than K_sp 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 K_sp. A saturated solution in equilibrium with the undissolved solid will result. If the concentrations are such that Q is less than K_sp, then the solution is not saturated and no precipitate will form.

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

Example 6.4.9 – 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) ⇌ Mg²⁺ (aq) + 2OH^– (aq)

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:

Mg(OH)₂ (s) ⇌ Mg²⁺ (aq) + 2OH^– (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 K_sp. Calculation of the reaction quotient under these conditions is shown here:

Q = [Mg²⁺][OH^–]² = (0.0537)(0.0010)² = 5.4 x 10^-8

Because Q is greater than K_sp (Q = 5.4 × 10^-8 is larger than K_sp = 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 K_sp.

Check Your Learning 6.4.10 – Precipitation of Mg(OH)₂

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 ClO₄⁻? (Remember to calculate the new concentration of each ion after mixing the solutions before plugging into the reaction quotient expression.)

Answer

No, Q = 4.0 × 10^–3, which is less than K_sp = 1.05 × 10^–2

In the previous example, we have seen that a precipitate forms when Q is greater than K_sp. In general, according to Equation 6.4.1, 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_pX_q precipitates if the value of Q for the mixture of M^m+ and X^n– is greater than K_sp for M_pX_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.

Example 6.4.10 – 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, C₂O₄²⁻, for this purpose (Figure 6.4.5). At sufficiently high concentrations, the calcium and oxalate ions form solid calcium oxalate monohydrate, CaC₂O₄•H₂O. The concentration of Ca²⁺ in a sample of blood serum is 2.2 × 10^–3 M. What concentration of C₂O₄²⁻ ion must be established before CaC₂O₄•H₂O begins to precipitate?

Figure 6.4.5. 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:

CaC₂O₄ (s) ⇌ Ca²⁺ (aq) + C₂O₄^2- (aq)

For this reaction:

K_sp = [Ca²⁺][C₂O₄^2-] = 1.96×10^-8   (see Appendix L)

Solid CaC₂O₄ does not begin to form until Q equals K_sp. Because we know K_sp and [Ca²⁺], we can solve for the concentration of C₂O₄²⁻ that is necessary to produce the first trace of solid precipitate:

Q = K_sp = [Ca²⁺(aq)][C₂O₄^2-(aq)] = 1.96×10^-8

(2.2×10^-3)[C₂O₄^2-(aq)] = 1.96×10^-8

[C₂O₄^2-(aq)] = 1.96×10^-8/2.2×10^-3 = 8.9×10^-6 M

A concentration of [C₂O₄²⁻] = 8.9 × 10^–6 M is necessary to initiate the precipitation of CaC₂O₄ under these conditions.

Check Your Learning 6.4.11 – Precipitation of Calcium Oxalate

If a solution contains 0.0020 mol of CrO₄²⁻ per litre, 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.

Answer

4.5 × 10^–9 M

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 the previous example—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 6.4.11 – Concentrations Following Precipitation

Clothing washed in water that has an aqueous manganese (II) ion (Mn²⁺) 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 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^–6 M?

Solution

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

Mn(OH)₂ (s) ⇌ Mn²⁺ (aq) + 2OH^– (aq)

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

K_sp = [Mn²⁺][OH^–]²

Or:

(1.8×10^-6)[OH^–]² = 2×10^-3

So:

[OH^–] = 3.3×10^-4 M

Now we calculate the pH from the pOH:

pOH = -log[OH^–] = -log(3.3×10^-4) = 3.48

pH = 14.00-pOH = 14.00-3.80 = 10.52

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^–6 M; at that concentration or less, the ion will not stain clothing.

Check Your Learning 6.4.12 – Concentrations Following Precipitation

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 diminished to 1.0 × 10^–5 M by the addition of Ca(OH)₂.

Answer

pH = 10.97

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 (the compound with the smaller molar solubility) generally precipitates first when we add a precipitating agent to a solution containing both anions (or both cations). When the molar solubilities 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.

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 K_sp.

Example 6.4.12 – 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:

AgCl (s) ⇌ Ag⁺ (aq) + Cl^– (aq), K_sp = 1.6×10^-10

AgI (s) ⇌ Ag⁺ (aq) + I^– (aq), K_sp = 1.5×10^-16

In this example, both salts form the same number of ions (1 mol of dissolved solid produces 1 mol of cations and 1 mol of anions) and so in both cases, the K_sp value is calculated as the square of the molar solubility (i.e. if the molar solubility is x, K_sp = x²). Consequently, we can compare these K_sp values directly to get a qualitative sense of the order of precipitation. Note that K_sp values may NOT be directly compared if the number of ions formed in the dissolution equilibrium is not identical (for example, it would not be appropriate to compare the K_sp value of AgCl to that of Ag₂SO₄).

Fortunately, in this case, since the K_sp values identically correlate to each solid’s molar solubility (K_sp = x²), that means we can deduce that if the solution contained about equal concentrations of Cl^– and I^–, then the silver salt with the smallest K_sp (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 K_sp for AgI (1.5 × 10^–16). When [I^–] =

0.0010 M:

Q = [Ag⁺][I^–] = [Ag⁺](0.0010) = 1.5×10^-16

[Ag⁺] = (1.8×10^-10)(0.10) = 1.6×10^-9 M

AgI begins to precipitate when [Ag⁺] is 1.5 × 10^–13 M.

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

Q =[Ag⁺][Cl^–] = [Ag⁺](0.10) = 1.6×10^-10

[Ag+] = 1.8×10^-10(0.10) = 1.6×10^-9 M

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

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

precipitate first.

Check Your Learning 6.4.13 – Precipitation of Silver Halides

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?

Answer

[Ag⁺] = 1.0 × 10^–11 M; AgBr precipitates first

Questions

★ Questions

1. Write the ionic equation for the dissolution and the K_sp expression for each of the following slightly soluble ionic compounds:

(a) LaF₃

(b) CaCO₃

(c) Ag₂SO₄

(d) Pb(OH)₂

2. The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.

(a) BaSeO₄, 0.0118 g/100 mL

(b) Ba(BrO₃)₂·H₂O, 0.30 g/100 mL

(c) NH₄MgAsO₄·6H₂O, 0.038 g/100 mL

(d) La₂(MoO₄)₃, 0.00179 g/100 mL

3. Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: CaF₂, Hg₂Cl₂, PbI₂, or Sn(OH)₂.

4. Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:

(a) KHC₄H₄O₆

(b) PbI₂

(c) Ag₄[Fe(CN)₆], a salt containing the Fe(CN)₄⁻ ion

(d) Hg₂I₂

★★ Questions

5. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected.

(a) AgCl(s) in 0.025 M NaCl

(b) CaF₂(s) in 0.00133 M KF

(c) Ag₂SO₄(s) in 0.500 L of a solution containing 19.50 g of K₂SO₄

(d) Zn(OH)₂(s) in a solution buffered at a pH of 11.45

6. Most barium compounds are very poisonous; however, barium sulfate is often administered internally as an aid in the X-ray examination of the lower intestinal tract (Figure 6.4.4). This use of BaSO₄ is possible because of its low solubility. Calculate the molar solubility of BaSO₄ and the mass of barium present in 1.00 L of water saturated with BaSO₄.

7. The solubility product of CaSO₄·2H₂O is 2.4 × 10^–5. What mass of this salt will dissolve in 1.0 L of 0.010 M SO₄²⁻?

8. Which of the following carbonates will form first? Which of the following will form last? Explain.

(a) MgCO₃ K_sp = 3.5 × 10⁻⁸

(b) CaCO₃ K_sp = 4.2 × 10⁻⁷

(c) SrCO₃ K_sp = 3.9 × 10⁻⁹

(d) BaCO₃ K_sp = 4.4 × 10⁻⁵

(e) MnCO₃ K_sp = 5.1 × 10⁻⁹

9. How many grams of Milk of Magnesia, Mg(OH)₂ (s) (58.3 g/mol), would be soluble in 200 mL of water. K_sp = 7.1 × 10^–12. Include the ionic reaction and the expression for K_sp in your answer. (K_w = 1 × 10^–14 = [H₃O⁺][OH^–])

10. A volume of 0.800 L of a 2 × 10^–4 M Ba(NO₃)₂ solution is added to 0.200 L of 5 × 10^–4 M Li₂SO₄. Does BaSO₄ precipitate? Explain your answer.

11. Calculate the concentration of PO₄³⁻ when Ag₃PO₄ starts to precipitate from a solution that is 0.0125 M in Ag⁺.

Answers

1. (a) LaF₃ (s) ⇌ La³⁺ (aq) + 3F^– (aq) K_sp = [La³⁺][F^–]³

(b) CaCO₃ (s) ⇌ Ca³⁺ (aq) + CO₃^2- (aq) K_sp  =[Ca²⁺][CO₃^2-]

(c) Ag₂SO₄ (s) ⇌ 2Ag⁺ (aq) + SO₄^2- (aq) K_sp = [Ag⁺]²[SO₄^2-]

(d) Pb(OH)₂ (s) ⇌ Pb²⁺ (aq) + 2OH^– (aq) K_sp  = [Pb²⁺][OH^–]²

2. (a) 1.77 × 10^–7; (b) 1.6 × 10^–6; (c) 2.2 × 10^–9; (d) 7.91 × 10^–22

3. PbI₂ is the most soluble.

4. (a) 2 × 10^–2 M; (b) 1.5 × 10^–3 M; (c) 2.27 × 10^–9 M; (d) 2.2 × 10^–10 M

5. (a) 6.4×10^-9 M = [Ag⁺], [Cl-] = 0.025 M

6. The molar solubility of BaSO₄ is 1.51 × 10^–4 M and the mass of barium present in 1 L of water in 0.020 g.

7. Mass (CaSO₄·2H₂O) = 0.72 g/L

8. MnCO₃ will form first, since it has the smallest K_sp value it is the least soluble. MnCO₃ will be the last to precipitate, it has the largest K_sp value.

10. No precipitation occurs because the value of the solubility product (2.3 x 10^-8) is greater than the ionic product (1.6 x 10^-8).

11. 9.2 × 10⁻¹³ M