Library Guides: Chemistry Textbook: Application: Precipitation and Dissolution

Application: Precipitation and Dissolution

By the end of this section, you will be able to:

Write chemical equations and equilibrium expressions representing solubility equilibria
Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations

Solubility equilibria are established when the dissolution and precipitation of a solute species occur at equal rates. These equilibria underlie many natural and technological processes, ranging from tooth decay to water purification. An understanding of the factors affecting compound solubility is, therefore, essential to the effective management of these processes. This section applies previously introduced equilibrium concepts and tools to systems involving dissolution and precipitation.

The Solubility Product

Recall from the chapter on solutions that the solubility of a substance can vary from essentially zero (insoluble or sparingly soluble) to infinity (miscible). A solute with finite solubility can yield a saturated solution when it is added to a solvent in an amount exceeding its solubility, resulting in a heterogeneous mixture of the saturated solution and the excess, undissolved solute. For example, a saturated solution of silver chloride is one in which the equilibrium shown below has been established.

AgCl (s) ⇌_{precipitation}^{dissolution} {Ag}^{+} (a q) + {Cl}^{−} (a q)

In this solution, an excess of solid AgCl dissolves and dissociates to produce aqueous Ag⁺ and Cl^– ions at the same rate that these aqueous ions combine and precipitate to form solid AgCl (Figure 11.22). Because silver chloride is a sparingly soluble salt, the equilibrium concentration of its dissolved ions in the solution is relatively low.

Two beakers are shown with a bidirectional arrow between them. Both beakers are just over half filled with a clear, colorless liquid. The beaker on the left shows a cubic structure composed of alternating green and slightly larger grey spheres. Evenly distributed in the region outside, 11 space filling models are shown. These are each composed of a central red sphere with two smaller white spheres attached in a bent arrangement. In the beaker on the right, the green and grey spheres are no longer connected in a cubic structure. Nine green spheres, 10 grey spheres, and 11 red and white molecules are evenly mixed and distributed throughout the liquid in the beaker. — Figure 11.22 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 solubility equilibria such as this one is called the solubility product constant, K_sp, in this case

AgCl (s) ⇌ {Ag}^{+} (a q) + {Cl}^{−} (a q) K_{sp} = [{Ag}^{+} (a q)] [{Cl}^{−} (a q)]

Recall that only gases and solutes are represented in equilibrium constant expressions, so the K_sp does not include a term for the undissolved AgCl. A listing of solubility product constants for several sparingly soluble compounds is provided in Appendix J.

EXAMPLE 11.26

Writing Equations and Solubility Products Write the dissolution equation 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

(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

Solution $\begin{matrix} (a) AgI (s) ⇌ {Ag}^{+} (a q) + I^{−} (a q) & K_{sp} & = & {[Ag}^{+}] [I^{−}] \\ (b) {CaCO}_{3} (s) ⇌ {Ca}^{2+} (a q) + {CO}_{3}^{2−} (a q) & K_{sp} & = & [{Ca}^{2+} {] [CO}_{3}^{2−}] \\ (c) {Mg (OH)}_{2} (s) ⇌ {Mg}^{2+} (a q) + {2OH}^{−} (a q) & K_{sp} & = & [{Mg}^{2+}] {[{OH}^{−}]}^{2} \\ (d) Mg ({NH}_{4}) {PO}_{4} (s) ⇌ {Mg}^{2+} (a q) + {NH}_{4}^{+} (a q) + {PO}_{4}^{3−} (a q) & K_{sp} & = & [{Mg}^{2+} {] [NH}_{4}^{+}] {[PO}_{4}^{3−}] \\ (e) {Ca}_{5} ({PO}_{4}) 3 OH (s) ⇌ {5Ca}^{2+} (a q) + {3PO}_{4}^{3−} (a q) + {OH}^{−} (a q) & K_{sp} & = & {{[Ca}^{2+}]}^{5} [P O_{4}^{3−}]^{3} [{OH}^{−}] \end{matrix}$

Check Your LearningWrite the dissolution equation and the solubility product for each of the following slightly soluble compounds:

(a) BaSO₄

(b) Ag₂SO₄

(d) Pb(OH)Cl

Answer:

$\begin{matrix} (a) {BaSO}_{4} (s) ⇌ {Ba}^{2+} (a q) + {SO}_{4}^{2−} (a q) & K_{sp} & = & [{Ba}^{2+} {] [SO}_{4} {^{2}}^{−}]; \\ (b) {Ag}_{2} {SO}_{4} (s) ⇌ {2Ag}^{+} (a q) + {SO}_{4}^{2−} (a q) & K_{sp} & = & {[{Ag}^{+}]}^{2} [{SO}_{4}^{2−}]; \\ (c) {Al (OH)}_{3} (s) ⇌ {Al}^{3+} (a q) + {3OH}^{−} (a q) & K_{sp} & = & [{Al}^{3+}] {[OH}^{−}]^{3}; \\ (d) Pb(OH)Cl(s) ⇌ {Pb}^{2+} (a q) + {OH}^{−} (a q) + {Cl}^{−} (a q) & K_{sp} & = & [{Pb}^{2+}] {[OH}^{−}] [{Cl}^{−}] \end{matrix}$

K_sp and Solubility

The K_sp of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:

M_{p} X_{q} (s) ⇌ p M^{m+} (a q) + q X^{n−} (a q)

For cases such as these, one may derive K_sp values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound’s molar solubility, measured as moles of dissolved solute per liter of saturated solution.

EXAMPLE 11.27

Calculation of K_sp from Equilibrium Concentrations Fluorite, CaF₂, is a slightly soluble solid that dissolves according to the equation:

{CaF}_{2} (s) ⇌ {Ca}^{2+} (a q) + {2F}^{−} (a q)

The concentration of Ca²⁺ in a saturated solution of CaF₂ is 2.15 $\times$ 10^–4 M. What is the solubility product of fluorite?

Solution According to the stoichiometry of the dissolution equation, the fluoride ion molarity of a CaF₂ solution is equal to twice its calcium ion molarity:

[F^{-}] = (2 mol F^{-} / 1 mol {Ca}^{2 +}) = (2) (2.15 \times 10^{- 4} M) = 4.30 \times 10^{- 4} M

Substituting the ion concentrations into the K_sp expression gives

K_{sp} = [{Ca}^{2+}] {[F^{−}]}^{2} = (2.15 \times 10^{- 4}) {(4.30 \times 10^{- 4})}^{2} = 3.98 \times 10^{- 11}

Check Your LearningIn a saturated solution of Mg(OH)₂, the concentration of Mg²⁺ is 1.31 $\times$ 10^–4 M. What is the solubility product for Mg(OH)₂?

{Mg(OH)}_{2} (s) ⇌ {Mg}^{2+} (a q) + {2OH}^{−} (a q)

Answer:

8.99 $\times$ 10^–12

EXAMPLE 11.28

Determination of Molar Solubility from K_sp The K_sp of copper(I) bromide, CuBr, is 6.3 $\times$ 10^–9. Calculate the molar solubility of copper bromide.

SolutionThe dissolution equation and solubility product expression are

CuBr (s) ⇌ {Cu}^{+} (a q) + {Br}^{−} (a q)

K_{sp} = [{Cu}^{+}] [{Br}^{−}]

Following the ICE approach to this calculation yields the table

This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, “C u B r equilibrium arrow C u superscript positive sign plus B r superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, positive x, x. The third column has the following 0, positive x, x.

Substituting the equilibrium concentration terms into the solubility product expression and solving for x yields

K_{sp} = {[Cu}^{+}] [{Br}^{−}]

6.3 \times 10^{- 9} = (x) (x) = x^{2}

x = \sqrt{(6.3 \times 10^{- 9})} = 7.9 \times 10^{- 5} M

Since the dissolution stoichiometry shows one mole of copper(I) ion and one mole of bromide ion are produced for each moles of Br dissolved, the molar solubility of CuBr is 7.9 $\times$ 10^–5 M.

Check Your Learning The K_sp of AgI is 1.5 $\times$ 10^–16. Calculate the molar solubility of silver iodide.

Answer:

1.2 $\times$ 10^–8 M

EXAMPLE 11.29

Determination of Molar Solubility from K_spThe K_sp of calcium hydroxide, Ca(OH)₂, is 1.3 $\times$ 10^–6. Calculate the molar solubility of calcium hydroxide.

SolutionThe dissolution equation and solubility product expression are

{Ca(OH)}_{2} (s) ⇌ {Ca}^{2+} (a q) + {2OH}^{−} (a q)

K_{sp} = [{Ca}^{2+}] {[{OH}^{−}]}^{2}

The ICE table for this system is

Substituting terms for the equilibrium concentrations into the solubility product expression and solving for x gives

K_{sp} = [{Ca}^{2+}] {[{OH}^{−}]}^{2}

1.3 \times 10^{- 6 ​} = (x) (2 x)^{2} = (x) (4 x^{2}) = 4 x^{3}

x = \sqrt[3]{\frac{1.3 \times 10^{- 6}}{4}} = 7.0 \times 10^{- 3} M

As defined in the ICE table, x is the molarity of calcium ion in the saturated solution. The dissolution stoichiometry shows a 1:1 relation between moles of calcium ion in solution and moles of compound dissolved, and so, the molar solubility of Ca(OH)₂ is 6.9 $\times$ 10^–3 M.

Check Your Learning The K_sp of PbI₂ is 1.4 $\times$ 10^–8. Calculate the molar solubility of lead(II) iodide.

Answer:

1.5 $\times$ 10^–3 M

EXAMPLE 11.30

Determination of K_sp from Gram SolubilityMany of the pigments used by artists in oil-based paints (Figure 11.23) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO₄, is 4.6 $\times$ 10^–6 g/L. Determine the solubility product for PbCrO₄.

A photograph is shown of a portion of an oil painting which reveals colors of orange, brown, yellow, green, blue, and purple colors in its strokes. A few water droplets rest on the surface. — Figure 11.23 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 color vermilion (HgS), and green color veridian (Cr₂O₃). (credit: Sonny Abesamis)

Solution

Before calculating the solubility product, the provided solubility must be converted to molarity:

\begin{matrix} [{PbCrO}_{4}] = \frac{4.6 \times 10^{- 6} {g PbCrO}_{4}}{1 L} \times \frac{1 {mol PbCrO}_{4}}{323.2 {g PbCrO}_{4}} \\ = \frac{1.4 \times 10^{- 8} {mol PbCrO}_{4}}{1 L} \\ = 1.4 \times 10^{- 8} M \end{matrix}

The dissolution equation for this compound is

{PbCrO}_{4} (s) ⇌ {Pb}^{2+} (a q) + {CrO}_{4}^{2−} (a q)

The dissolution stoichiometry shows a 1:1 relation between the molar amounts of compound and its two ions, and so both [Pb²⁺] and $[{CrO}_{4}^{2−}]$ are equal to the molar solubility of PbCrO₄:

[{Pb}^{2+}] = [{CrO}_{4}^{2−}] = 1.4 \times 10^{- 8} M

K_sp = [Pb²⁺] $[{CrO}_{4}^{2−}]$ = (1.4 $\times$ 10^–8)(1.4 $\times$ 10^–8) = 2.0 $\times$ 10^–16

Check Your LearningThe solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.12 grams per liter at 20 °C. What is its solubility product?

Answer:

2.08 $\times$ 10^–4

EXAMPLE 11.31

Calculating the Solubility of Hg₂Cl₂ Calomel, Hg₂Cl₂, is a compound composed of the diatomic ion of mercury(I), ${Hg}_{2}^{2+},$ 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 has a very low solubility, as suggested by its very small K_sp:

{Hg}_{2} {Cl}_{2} (s) ⇌ {Hg}_{2}^{2+} (a q) + {2Cl}^{−} (a q) K_{sp} = 1.1 \times 10^{- 18}

Calculate the molar solubility of Hg₂Cl₂.

SolutionThe dissolution stoichiometry shows a 1:1 relation between the amount of compound dissolved and the amount of mercury(I) ions, and so the molar solubility of Hg₂Cl₂ is equal to the concentration of ${Hg}_{2}^{2+}$ ions

Following the ICE approach results in

Substituting the equilibrium concentration terms into the solubility product expression and solving for x gives

K_{sp} = [{Hg}_{2}^{2+}] {[{Cl}^{−}]}^{2}

1.1 \times 10^{- 18} = (x) (2 x)^{2}

4 x^{3} = 1.1 \times 10^{- 18}

x = \sqrt[3]{(\frac{1.1 \times 10^{- 18}}{4})} = 6.5 \times 10^{- 7} M

[{Hg}_{2}^{2+}] = 6.5 \times 10^{- 7} M = 6.5 \times 10^{- 7} M

[{Cl}^{−}] = 2 x = 2(6.5 \times 10^{- 7}) = 1.3 \times 10^{- 6} M

The dissolution stoichiometry shows the molar solubility of Hg₂Cl₂ is equal to $[{Hg}_{2}^{2+}],$ or 6.5 $\times$ 10^–7 M.

Check Your LearningDetermine the molar solubility of MgF₂ from its solubility product: K_sp = 6.4 $\times$ 10^–9.

Answer:

1.2 $\times$ 10^–3 M

HOW SCIENCES INTERCONNECT

Using Barium Sulfate for Medical Imaging

Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the K_sp of barium sulfate is 2.3 $\times$ 10^–8, very little of it dissolves as it coats the lining of the patient’s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray (Figure 11.24).

This figure contains one image. A black and white abdominal x-ray image is shown in which the intestinal tract of a person is clearly visible in white. — Figure 11.24 A suspension of barium sulfate coats the intestinal tract, permitting greater visual detail than a traditional X-ray. (credit modification of work by “glitzy queen00”/Wikimedia Commons)

Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn’s disease, and ulcers in addition to other conditions.

Visit this website for more information on how barium is used in medical diagnoses and which conditions it is used to diagnose.

Predicting Precipitation- an application of Reaction Quotient

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

\begin{matrix} {CaCO}_{3} (s) ⇌ {Ca}^{2+} (a q) + {CO}_{3} 2− (a q) & K_{s p} = [{Ca}^{2 +}] [{CO}_{3}^{2 -}] = 8.7 \times 10^{- 9} \end{matrix}

It is important to realize that this equilibrium is established in any aqueous solution containing Ca²⁺ and CO₃^2– 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, Q_sp, that exceeds the solubility product, K_sp, 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 (Q_sp = K_sp). The comparison of Q_sp to K_sp 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:

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

Q_sp > K_sp: 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.

EXAMPLE 11.32

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+} (a q) + {2OH}^{−} (a q) K_{sp} = 8.9 \times 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

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

Q = [{Mg}^{2+}] {[{OH}^{−}]}^{2} = (0.0537)({0.0010)}^{2} = 5.4 \times 10^{- 8}

Because Q is greater than K_sp (Q = 5.4 $\times$ 10^–8 is larger than K_sp = 8.9 $\times$ 10^–12), the reverse reaction will proceed, precipitating magnesium hydroxide until the dissolved ion concentrations have been sufficiently lowered, so that Q_sp = K_sp.

Check Your Learning Predict whether CaHPO₄ will precipitate from a solution with [Ca²⁺] = 0.0001 M and $[{HPO}_{4}^{2−}]$ = 0.001 M.

Answer:

No precipitation of CaHPO₄; Q = 1 $\times$ 10^–7, which is less than K_sp (7 × 10^–7)

EXAMPLE 11.33

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

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

AgCl (s) ⇌ {Ag}^{+} (a q) + {Cl}^{−} (a q)

The solubility product is 1.6 $\times$ 10^–10 (see Appendix J).

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

\frac{1}{2} (2.0 \times 10^{- 4}) M = 1.0 \times 10^{- 4} M

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

Q = [{Ag}^{+}] {[Cl}^{−}] = (1.0 \times 10^{- 4}) (1.0 \times 10^{- 4}) = 1.0 \times 10^{- 8} > K_{sp}

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

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.)

Answer:

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

Common Ion Effect- an application of LeChateliers Principle

Compared with pure water, the solubility of an ionic compound is less in aqueous solutions containing a common ion (one also produced by dissolution of the ionic compound). This is an example of a phenomenon known as the common ion effect, which is a consequence of the law of mass action that may be explained using Le ChÂtelier’s principle. Consider the dissolution of silver iodide:

AgI (s) ⇌ {Ag}^{+} (a q) + I^{−} (a q)

This solubility equilibrium may be shifted left by the addition of either silver(I) or iodide ions, resulting in the precipitation of AgI and lowered concentrations of dissolved Ag⁺ and I^–. In solutions that already contain either of these ions, less AgI may be dissolved than in solutions without these ions.

This effect may also be explained in terms of mass action as represented in the solubility product expression:

K_{sp} = [{Ag}^{+}] [I^{-}]

The mathematical product of silver(I) and iodide ion molarities is constant in an equilibrium mixture regardless of the source of the ions, and so an increase in one ion’s concentration must be balanced by a proportional decrease in the other.

LINK TO LEARNING

View this simulation to explore various aspects of the common ion effect.

LINK TO LEARNING

View this simulation to see how the common ion effect works with different concentrations of salts.

EXAMPLE 11.34

Common Ion Effect on Solubility What is the effect on the amount of solid Mg(OH)₂ and the concentrations of Mg²⁺ and OH^– when each of the following are added to a saturated solution of Mg(OH)₂?

(a) MgCl₂

(b) KOH

(d) Mg(OH)₂

Solution The solubility equilibrium is

Mg {(OH)}_{2} (s) ⇌ {Mg}^{2+} (a q) + 2 {OH}^{−} (a q)

(a) The reaction shifts to the left to relieve the stress produced by the additional Mg²⁺ ion, in accordance with Le Châtelier’s principle. In quantitative terms, the added Mg²⁺ causes the reaction quotient to be larger than the solubility product (Q > K_sp), and Mg(OH)₂ forms until the reaction quotient again equals K_sp. At the new equilibrium, [OH^–] is less and [Mg²⁺] is greater than in the solution of Mg(OH)₂ in pure water. More solid Mg(OH)₂ is present.

(b) The reaction shifts to the left to relieve the stress of the additional OH^– ion. Mg(OH)₂ forms until the reaction quotient again equals K_sp. At the new equilibrium, [OH^–] is greater and [Mg²⁺] is less than in the solution of Mg(OH)₂ in pure water. More solid Mg(OH)₂ is present.

(a) Adding a common ion, Mg²⁺, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of hydroxide ion and increasing the amount of undissolved magnesium hydroxide.

(b) Adding a common ion, OH^–, will increase the concentration of this ion and shift the solubility equilibrium to the left, decreasing the concentration of magnesium ion and increasing the amount of undissolved magnesium hydroxide.

(c) The added compound does not contain a common ion, and no effect on the magnesium hydroxide solubility equilibrium is expected.

(d) Adding more solid magnesium hydroxide will increase the amount of undissolved compound in the mixture. The solution is already saturated, though, so the concentrations of dissolved magnesium and hydroxide ions will remain the same.

Q = [{Mg}^{2+}] {[{OH}^{−}]}^{2}

Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of Q, and no shift is required to restore Q to the value of the equilibrium constant.

Check Your Learning What is the effect on the amount of solid NiCO₃ and the concentrations of Ni²⁺ and ${CO}_{3}^{2−}$ when each of the following are added to a saturated solution of NiCO₃

(a) Ni(NO₃)₂

(b) KClO₄

(d) K₂CO₃

Answer:

(a) mass of NiCO₃(s) increases, [Ni²⁺] increases, $[{CO}_{3}^{2−}]$ decreases; (b) no appreciable effect; (c) no effect except to increase the amount of solid NiCO₃; (d) mass of NiCO₃(s) increases, [Ni²⁺] decreases, $[{CO}_{3}^{2−}]$ increases;

Key Concepts and Summary

The equilibrium constant for an equilibrium involving the precipitation or dissolution of a slightly soluble ionic solid is called the solubility product, K_sp, of the solid. For a heterogeneous equilibrium involving the slightly soluble solid M_pX_q and its ions M^m+ and X^n–:

M_{p} X_{q} (s) ⇌ p M^{m+} (a q) + q X^{n−} (a q)

the solubility product expression is:

K_{sp} = {{[M}^{m+}]}^{p} {{[X}^{n−}]}^{q}

The solubility product of a slightly soluble electrolyte can be calculated from its solubility; conversely, its solubility can be calculated from its K_sp, provided the only significant reaction that occurs when the solid dissolves is the formation of its ions.

A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product.

Key Equations

$M_{p} X_{q} (s) ⇌ p M^{m+} (a q) + q X^{n−} (a q) K_{sp} = {[M}^{m+}]^{p} {{[X}^{n−}]}^{q}$

Glossary

common ion effect: effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base

molar solubility: solubility of a compound expressed in units of moles per liter (mol/L)

selective precipitation: process in which ions are separated using differences in their solubility with a given precipitating reagent

solubility product constant (K_sp): equilibrium constant for the dissolution of an ionic compound

Chemistry Textbook

Application: Precipitation and Dissolution

The Solubility Product

Ksp and Solubility

Predicting Precipitation- an application of Reaction Quotient

Common Ion Effect- an application of LeChateliers Principle

Key Concepts and Summary

Key Equations

Glossary

K_sp and Solubility