7.2 Precipitation and Dissolution

Vishakha Monga; Paul Flowers; Klaus Theopold; William R. Robinson; Richard Langley

7.2 Precipitation and Dissolution

Learning Objectives

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:

$\text{AgCl}\left(s\right)\underset{\text{precipitation}}{\overset{\text{dissolution}}{\rightleftharpoons}}{\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Cl}^-}^{\text{}}\left(aq\right)$

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 7.2.1). 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 7.2.1 – 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 solubility equilibria such as this one is called the solubility product constant, K_sp, in this case:

$\text{AgCl}\left(s\right)$\rightleftharpoons${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Cl}^-}^{\text{}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\left[{\text{Ag}}^{\text{+}}\left(aq\right)\right]\left[{\text{Cl}^-}^{\text{}}\left(aq\right)\right]$

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.

Activity 7.2.1 – 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

(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

Solution

$\begin{array}{ccccc}\left(\text{a}\right)\phantom{\rule{0.2em}{0ex}}\text{AgI}\left(s\right)$\rightleftharpoons${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{I}}^{\text{-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & {\left[\text{Ag}}^{\text{+}}\right]\left[{\text{I}}^{\text{-}}\right]\hfill \\ \left(\text{b}\right)\phantom{\rule{0.2em}{0ex}}{\text{CaCO}}_{3}\left(s\right)$\rightleftharpoons${\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{CO}}_{3}{}^{\text{2-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & \left[{\text{Ca}}^{\text{2+}}{\right]\left[\text{CO}}_{3}{}^{\text{2-}}\right]\hfill \\ \left(\text{c}\right)\phantom{\rule{0.2em}{0ex}}{\text{Mg}\left(\text{OH}\right)}_{2}\left(s\right)$\rightleftharpoons${\text{Mg}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{\text{-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & \left[{\text{Mg}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}\hfill \\ \left(\text{d}\right)\phantom{\rule{0.2em}{0ex}}\text{Mg}\left({\text{NH}}_{\text{4}}\right){\text{PO}}_{\text{4}}\left(s\right)$\rightleftharpoons${\text{Mg}}^{\text{2+}}\left(aq\right)+{\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{PO}}_{4}{}^{\text{3-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & \left[{\text{Mg}}^{\text{2+}}{\right]\left[\text{NH}}_{4}{}^{\text{+}}\right]{\left[\text{PO}}_{4}{}^{\text{3-}}\right]\hfill \\ \left(\text{e}\right)\phantom{\rule{0.2em}{0ex}}{\text{Ca}}_5}\left({\text{PO}}_{4}\right)_3\text{OH}\left(s\right)$\rightleftharpoons${\text{5Ca}}^{\text{2+}}\left(aq\right)+{\text{3PO}}_{4}{}^{\text{3-}}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & {{\left[\text{Ca}}^{\text{2+}}\right]}^{5}\left[\text{P}{\text{O}}_{4}{}^{\text{3-}}{\right]}^{3}\left[{\text{OH}}^{\text{-}}\right]\hfill \end{array}$

Check Your Learning

Write the dissolution equation 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

$\begin{array}{ccccc}\left(\text{a}\right)\phantom{\rule{0.2em}{0ex}}{\text{BaSO}}_{4}\left(s\right)$\rightleftharpoons${\text{Ba}}^{\text{2+}}\left(aq\right)+{\text{SO}}_{4}{}^{\text{2-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & \left[{\text{Ba}}^{\text{2+}}{\right]\left[\text{SO}}_{4}{{}^{2}}^{\text{-}}\right];\hfill \\ \left(\text{b}\right)\phantom{\rule{0.2em}{0ex}}{\text{Ag}}_{2}{\text{SO}}_{4}\left(s\right)$\rightleftharpoons${\text{2Ag}}^{\text{+}}\left(aq\right)+{\text{SO}}_{4}{}^{\text{2-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & {\left[{\text{Ag}}^{\text{+}}\right]}^{2}\left[{\text{SO}}_{4}{}^{\text{2-}}\right];\hfill \\ \left(\text{c}\right)\phantom{\rule{0.2em}{0ex}}{\text{Al}\left(\text{OH}\right)}_{3}\left(s\right)$\rightleftharpoons${\text{Al}}^{\text{3+}}\left(aq\right)+{\text{3OH}}^{\text{-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & \left[{\text{Al}}^{\text{3+}}\right]{\left[\text{OH}}^{\text{-}}{\right]}^{\text{3}};\hfill \\ \left(\text{d}\right)\phantom{\rule{0.2em}{0ex}}\text{Pb(OH)Cl(s}\right)$\rightleftharpoons${\text{Pb}}^{\text{2+}}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)+{\text{Cl}}^{\text{-}}\left(aq\right)\hfill & & \hfill {K}_{\text{sp}}& =\hfill & \left[{\text{Pb}}^{\text{2+}}\right]{\left[\text{OH}}^{\text{-}}\right]\left[{\text{Cl}}^{\text{-}}\right]\hfill \end{array}$

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:

${\text{M}}_{p}{\text{X}}_{q}\left(s\right)$\rightleftharpoons$p{\text{M}}^{\text{m+}}\left(aq\right)+q{\text{X}}^{\text{n-}}\left(aq\right)$

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.

Activity 7.2.2 – Calculation of K_sp from Equilibrium Concentrations

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

${\text{CaF}}_{2}\left(s\right)$\rightleftharpoons${\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{2F}}^{\text{-}}\left(aq\right)$

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:

$\left[{\text{F}}^{-}\right]=\left(2\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{F}}^{-}\phantom{\rule{0.2em}{0ex}}\text{/}\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{Ca}}^{2+}\right)=\left(2\right)\left(2.15\phantom{\rule{0.2em}{0ex}} $\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4})\phantom{\rule{0.2em}{0ex}}M\right)=4.30\phantom{\rule{0.2em}{0ex}} $\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}M$

Substituting the ion concentrations into the K_sp expression gives:

${K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2+}}\right]{\left[{\text{F}}^{\text{-}}\right]}^{2}=\text{(2.15}\phantom{\rule{0.2em}{0ex}} \times\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)){\left(4.30\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)})^{2}=\text{3.98}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-11}$

Check Your Learning

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

${\text{Mg(OH)}}_{2}\left(s\right)$\rightleftharpoons${\text{Mg}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{\text{-}}\left(aq\right)$

Answer

8.99 $\times$ 10^–12

Activity 7.2.3 – 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.

Solution

The dissolution equation and solubility product expression are:

$\text{CuBr}\left(s\right)$\rightleftharpoons${\text{Cu}}^{\text{+}}\left(aq\right)+{\text{Br}}^{\text{-}}\left(aq\right)$

${K}_{\text{sp}}=\left[{\text{Cu}}^{\text{+}}\right]\left[{\text{Br}}^{\text{-}}\right]$

Following the ICE approach to this calculation yields the table:

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

${K}_{\text{sp}}={\left[\text{Cu}}^{\text{+}}\right]\left[{\text{Br}}^{\text{-}}\right]$

$6.3\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-9}=\left(x\right)\left(x\right)={x}^{2}$

$x=\sqrt{\left(6.3\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-9}\right)}=\text{7.9}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}M$

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^–5M.

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^–8M

Activity 7.2.4 – Determination of Molar Solubility from K_sp

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

Solution

The dissolution equation and solubility product expression are:

${\text{Ca(OH)}}_{2}\left(s\right)$\rightleftharpoons${\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{\text{-}}\left(aq\right)$

${K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}$

The ICE table for this system is:

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

${K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}$

$1.3\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-6\text{}}=\left(x\right)\left(2x{\right)}^{2}=\left(x\right)\left(4{x}^{2}\right)=4{x}^{3}$

$x=\sqrt[3]{\phantom{\rule{0.2em}{0ex}}\frac{1.3\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-6}}{4}\phantom{\rule{0.2em}{0ex}}}=\text{7.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}M$

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^–3M.

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^–3M

Activity 7.2.5 – Determination of K_sp from Gram Solubility

Many of the pigments used by artists in oil-based paints (Figure 7.2.2) 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 7.2.2 – 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:

$\left[{\text{PbCrO}}_{4}]=\frac{4.6\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-6}{\text{g PbCrO}}_{4}}{1\phantom{\rule{0.2em}{0ex}}\text{L}}\phantom{\rule{0.4em}{0ex}}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.4em}{0ex}}=\frac{1{\phantom{\rule{0.2em}{0ex}}\text{mol PbCrO}}_{4}}{323.2{\phantom{\rule{0.2em}{0ex}}\text{g PbCrO}}_{4}}\phantom{\rule{0.2em}{0ex}}\\ =\phantom{\rule{0.2em}{0ex}}\frac{1.4\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-8}{\phantom{\rule{0.2em}{0ex}}\text{mol PbCrO}}_{4}}{1\phantom{\rule{0.2em}{0ex}}\text{L}}\phantom{\rule{0.2em}{0ex}}\\ =\phantom{\rule{0.2em}{0ex}}1.4\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-8}M$

The dissolution equation for this compound is:

${\text{PbCrO}}_{4}\left(s\right)$\rightleftharpoons${\text{Pb}}^{\text{2+}}\left(aq\right)+{\text{CrO}}_{4}{}^{\text{2-}}\left(aq\right)$

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

$\left[{\text{Pb}}^{\text{2+}}\right]=\left[{\text{CrO}}_{4}{}^{\text{2-}}\right]=\text{1.4}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-8}M$

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

Check Your Learning

The 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

Activity 7.2.6 – Calculating the Solubility of Hg₂Cl₂

Calomel, Hg₂Cl₂, is a compound composed of the diatomic ion of mercury(I), ${\text{Hg}}_{2}{}^{\text{2+}},$ 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:

${\text{Hg}}_{2}{\text{Cl}}_{2}\left(s\right)$\rightleftharpoons${\text{Hg}}_{2}{}^{\text{2+}}\left(aq\right)+{\text{2Cl}}^{\text{-}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\text{1.1}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-18}$

Calculate the molar solubility of Hg₂Cl₂.

Solution

The 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 ${\text{Hg}}_{2}{}^{\text{2+}}$ ions.

Following the ICE approach results in:

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

${K}_{\text{sp}}=\left[{\text{Hg}}_{\text{2}}{}^{\text{2+}}\right]{\left[{\text{Cl}}^{\text{-}}\right]}^{2}$

$1.1\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-18}=\left(x\right)\left(2x{\right)}^{2}$

$4{x}^{3}=\text{1.1}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-18}$

$x=\sqrt[3]{\left(\frac{1.1\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-18}}{4}\right)}=\text{6.5}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}M$

$\left[{\text{Hg}}_{2}{}^{\text{2+}}\right]=\text{6.5}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}M=\text{6.5}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}M$

$\left[{\text{Cl}}^{\text{-}}\right]=2x=\text{2(6.5}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-7}\right))=1.3\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}M$

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

Check Your Learning

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

Answer

1.2 $\times$ 10^–3M

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 7.2.3).

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 7.2.3 – 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

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

$\begin{array}{ccc}{\text{CaCO}}_{3}\left(s\right)$\rightleftharpoons${\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{CO}}_{3}^{\text{2-}}^{}\left(aq\right)& & \hfill {K}_{sp}=\left[{\text{Ca}}^{2+}\right]\left[{\text{CO}}_{3}{}^{2-}\right]=8.7\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-9}\end{array}$

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.

Activity 7.2.7 – 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:

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

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

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

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

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 $\left[{\text{HPO}}_{4}{}^{\text{2-}}\right]$ = 0.001 M.

Answer

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

Activity 7.2.8 – Precipitation of AgCl

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

Solution

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

$\text{AgCl}\left(s\right)$\rightleftharpoons${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{\text{-}}\left(aq\right)$

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:

$\frac{1}{2}\left(2.0\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)\phantom{\rule{0.2em}{0ex}}M=\text{1.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}M)$

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

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

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 ${\text{ClO}}_{4}{}^{\text{-}}?$ (Hint: Use the dilution equation to calculate the concentrations of potassium and perchlorate ions in the mixture.)

Solution

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

Activity 7.2.9 – 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 (Figure 7.2.4). 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 $\times$ 10^–3M. What concentration of ${\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}$ ion must be established before CaC₂O₄·H₂O begins to precipitate?

A photograph is shown of 6 vials of blood resting on and near a black and white document. Two of the vials have purple caps, three have tan caps, and one has a red cap. Each has a label and the vials with tan caps have a small amount of an off-white material present in a layer at the base of the vial. — **Figure 7.2.4 – 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:

${\text{CaC}}_{2}{\text{O}}_{4}\left(s\right)$\rightleftharpoons${\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\left(aq\right)$

For this reaction:

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

(see Appendix J)

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

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

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

$\left[{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\right]=\phantom{\rule{0.2em}{0ex}}\frac{1.96\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-8}}{2.2\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-3}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\text{8.9}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}M$

A concentration of $\left[{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\right]$ = 8.9 $\times$ 10^–6M is necessary to initiate the precipitation of CaC₂O₄ 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 AgNO₃ before Ag₂CrO₄ begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate.

Answer

6.7 $\times$ 10^–5M

Activity 7.2.10 – Concentrations Following Precipitation

Clothing washed in water that has a manganese [Mn²⁺(aq)] concentration exceeding 0.1 mg/L (1.8 $\times$ 10^–6M) 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 $\times$ 10^–6M?

Solution

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

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

At equilibrium:

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

or

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

so

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

Calculate the pH from the pOH:

$\begin{array}{c}\text{pOH}=\text{−log}\left[{\text{OH}}^{\text{-}}\right]=\text{−log}\left(3.3\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}10^{-4}\right))=\text{3.48}\\ \text{pH}=\text{14.00}-\text{pOH}=\text{14.00}-\text{3.80}=\text{10.5}\end{array}$

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

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 $\times$ 10^–2M. Calculate the pH at which [Mg²⁺] is decreased to 1.0 $\times$ 10^–5M

Answer

10.97

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 (Figure 7.2.5). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions ${\text{(PO}}_{4}{}^{\text{3-}}\right))$ 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.

A color photograph is shown of a high volume wastewater treatment facility. Nineteen large circular pools of water undergoing treatment are visible across the center of the photograph. A building and parking lot are visible in the foreground. — Figure 7.2.5 – 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:

$5{\text{Ca}}^{\text{2+}}+{\text{3PO}}_{4}{}^{\text{3-}}+{\text{OH}}^{\text{-}}$\rightleftharpoons${\text{Ca}}_{5}{{\text{(PO}}_{4}\right))}_{\text{3}}\text{}\text{OH}\left(s\right)$

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.

Activity 7.2.11 – 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?

Solution

The two equilibria involved are:

$\begin{array}{cccccc}\hfill \text{AgCl}\left(s\right)$\rightleftharpoons${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{\text{-}}\left(aq\right)\hfill & & & \hfill {K}_{\text{sp}}& =\hfill & \text{1.6}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-10}\hfill \\ \hfill \text{AgBr}\left(s\right)$\rightleftharpoons${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Br}}^{\text{-}}\left(aq\right)\hfill & & & \hfill {K}_{\text{sp}}& =\hfill & \text{5.0}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-13}\hfill \end{array}$

If the solution contained about equal concentrations of Cl^– and Br^–, then the silver salt with the smaller K_sp (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 K_sp for AgBr:

$Q=\left[{\text{Ag}}^{\text{+}}\right]{\left[\text{Br}}^{\text{-}}\right]=\left[{\text{Ag}}^{+}\right]\text{(0.00010)}=\text{5.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-13}$

$\left[{\text{Ag}}^{\text{+}}\right]=\phantom{\rule{0.2em}{0ex}}\frac{\text{5.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-13}}{0.00010}\phantom{\rule{0.2em}{0ex}}=\text{5.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}M$

AgI begins to precipitate when [Ag⁺] is 5.0 $\times$ 10^–9M.

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

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

$\left[{\text{Ag}}^{\text{+}}\right]=\phantom{\rule{0.2em}{0ex}}\frac{1.6\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-10}}{0.10}=\text{1.6}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}M$

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

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 K_sp 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?

Answer

[Ag⁺] = 1.0 $\times$ 10^–11M; AgBr precipitates first

Common Ion Effect

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:

$\text{AgI}\left(s\right)$\rightleftharpoons${\text{Ag}}^{\text{+}}\left(aq\right)+{\text{I}}^{\text{-}}\left(aq\right)$

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:

${K}_{\text{sp}}=\left[{\text{Ag}}^{+}\right]\left[{\text{I}}^{-}\right]$

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.

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

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

Activity 7.2.12 – 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

(c) NaNO₃

(d) Mg(OH)₂

Solution

The solubility equilibrium is:

$\text{Mg}{\left(\text{OH}\right)}_{2}\left(s\right)$\rightleftharpoons${\text{Mg}}^{\text{2+}}\left(aq\right)+2{\text{OH}}^{\text{-}}\left(aq\right)$

(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.

(c) The concentration of OH^– is reduced as the OH^– reacts with the acid. The reaction shifts to the right direction.

(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:

$Q=\left[{\text{Mg}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}$

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 ${\text{CO}}_{3}{}^{\text{2-}}$ when each of the following are added to a saturated solution of NiCO₃?

(a) Ni(NO₃)₂

(b) KClO₄

(c) NiCO₃

(d) K₂CO₃

Answer

(a) mass of NiCO₃(s) increases, [Ni²⁺] increases, $\left[{\text{CO}}_{3}{}^{\text{2-}}\right]$ decreases

(b) no appreciable effect

(c) no effect except to increase the amount of solid NiCO₃

(d) mass of NiCO₃(s) increases, [Ni²⁺] decreases, $\left[{\text{CO}}_{3}{}^{\text{2-}}\right]$ increases

Activity 7.2.13 – Common Ion Effect

Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-M solution of cadmium bromide (CdBr₂). The K_sp of CdS is 1.0 $\times$ 10^–28.

Solution

This calculation can be performed using the ICE approach:

$\text{CdS}\left(s\right)$\rightleftharpoons${\text{Cd}}^{\text{2+}}\left(aq\right)+{\text{S}}^{\text{2-}}\left(aq\right)$

${K}_{\text{sp}}=\left[{\text{Cd}}^{\text{2+}}\right]\left[{\text{S}}^{\text{2-}}\right]=1.0\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-28}$

$\left(0.010+x\right)\left(x\right)=\text{1.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-28}$

Because K_sp is very small, assume x << 0.010 and solve the simplified equation for x:

$\left(0.010\right)\left(x\right)=\text{1.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-28}$

$x=\text{1.0}\phantom{\rule{0.2em}{0ex}}$\times$\phantom{\rule{0.2em}{0ex}}{10}^{-26}\phantom{\rule{0.2em}{0ex}}M$

The molar solubility of CdS in this solution is 1.0 $\times$ 10^–26M.

Check Your Learning

Calculate the molar solubility of aluminum hydroxide, Al(OH)₃, in a 0.015-M solution of aluminum nitrate, Al(NO₃)₃. The K_sp of Al(OH)₃ is 2 $\times$ 10^–32.

Answer

4 $\times$ 10^–11M

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–:

${\text{M}}_{p}{\text{X}}_{q}\left(s\right)$\rightleftharpoons$p{\text{M}}^{\text{m+}}\left(aq\right)+q{\text{X}}^{\text{n-}}\left(aq\right)$

the solubility product expression is:

${K}_{\text{sp}}={{\text{[M}}^{\text{m+}}]\right]}^{p}{{\text{[X}}^{\text{n-}}]\right]}^{q}$

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

${\text{M}}_{p}{\text{X}}_{q}\left(s\right)$\rightleftharpoons$p{\text{M}}^{\text{m+}}\left(aq\right)+q{\text{X}}^{\text{n-}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}={\left[\text{M}}^{\text{m+}}{\right]}^{p}{{\left[\text{X}}^{\text{n-}}\right]}^{q}$

Some useful links related to waster water treatment and its process and application in chemical engineering:

News Articles:

Water purification methods to get tap water:

End of Chapter Exercises

(1) Complete the changes in concentrations for each of the following reactions:

(1a) $\begin{array}{ccc}\text{AgI}\left(s\right)\phantom{\rule{0.2em}{0ex}}\rightarrow\phantom{\rule{0.2em}{0ex}}& {\text{Ag}}^{\text{+}}\left(aq\right)& +\phantom{\rule{0.2em}{0ex}}{\text{I}}^{\text{-}}\left(aq\right)\\ \\ \underset{_}{}\end{array}$

(1b) $\begin{array}{ccc}{\text{CaCO}}_{3}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& {\text{Ca}}^{\text{2+}}\left(aq\right)+& {\text{CO}}_{3}{^{2-}}^{\text{}}\left(aq\right)\\ \\ & & \underset{_}{}\end{array}$

(1c) $\begin{array}{ccc}\text{Mg}{\left(\text{OH}\right)}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& {\text{Mg}}^{\text{2+}}\left(aq\right)+& 2{\text{OH}^-}^{\text{}}\left(aq\right)\\ \\ & & \underset{_}{}\end{array}$

(1d) $\begin{array}{ccc}{\text{Mg}}_{3}\left({\text{PO}}_{4}{\right)}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& 3{\text{Mg}}^{\text{2+}}\left(aq\right)+& 2{\text{PO}}_{4}{^{3-}}^{\text{}}\left(aq\right)\\ \\ & & \underset{_}{}\end{array}$

(1e) $\begin{array}{cccc}{\text{Ca}}_{5}\left({\text{PO}}_{4}{\right)}_{3}\text{OH}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& 5{\text{Ca}}^{\text{2+}}\left(aq\right)+& 3{\text{PO}}_{4}{^{3-}}^{\text{}}\left(aq\right)+& {\text{OH}^-}^{\text{}}\left(aq\right)\\ \\ & & & \underset{_}{}\end{array}$

Solutions

(a) $\begin{array}{ccc}\text{AgI}\left(s\right)$\rightleftharpoons$& {\text{Ag}}^{\text{+}}\left(aq\right)+& {\text{I}}^{\text{-}}\left(aq\right)\\ \\ & x& \underset{_}{x}\end{array}$

(b) $\begin{array}{ccc}{\text{CaCO}}_{3}\left(s\right)$\rightleftharpoons$& {\text{Ca}}^{\text{2+}}\left(aq\right)+& {\text{CO}}_{3}{}^{\text{2-}}\left(aq\right)\\ \\ & \underset{_}{x}& x\end{array}$

(c) $\begin{array}{lll}\text{Mg}{\left(\text{OH}\right)}_{2}\left(s\right)$\rightleftharpoons$\hfill & {\text{Mg}}^{\text{2+}}\left(aq\right)\hfill & +\phantom{\rule{0.2em}{0ex}}2{\text{OH}}^{\text{-}}\left(aq\right)\hfill \\ \hfill & x\hfill & \underset{_}{\text{2}x}\hfill \end{array}$

(d) $\begin{array}{ccc}{\text{Mg}}_{3}{\left({\text{PO}}_{4}\right)}_{2}\left(s\right)$\rightleftharpoons$& {\text{3Mg}}^{\text{2+}}\left(aq\right)+& {\text{2PO}}_{4}{}^{\text{3-}}\left(aq\right)\\ \\ & \underset{_}{3x}& 2x\end{array}$

(e) $\begin{array}{cccc}{\text{Ca}}_{5}\left({\text{PO}}_{4}{\right)}_{3}\text{OH}\left(s\right)$\rightleftharpoons$& {\text{5Ca}}^{\text{2+}}\left(aq\right)+& {\text{3PO}}_{4}{}^{\text{3-}}\left(aq\right)+& {\text{OH}}^{\text{-}}\left(aq\right)\\ \\ & \underset{_}{5x}& \underset{_}{3x}& x\end{array}$

(2) Complete the changes in concentrations for each of the following reactions:

(2a) $\begin{array}{ccc}{\text{BaSO}}_{4}\left(s\right)\phantom{\rule{0.2em}{0ex}}\rightarrow\phantom{\rule{0.2em}{0ex}}& {\text{Ba}}^{\text{2+}}\left(aq\right)+& {\text{SO}}_{4}{}^{\text{2-}}\left(aq\right)\\ \\ & x& \underset{_}{x}\end{array}$

(2b) $\begin{array}{ccc}{\text{Ag}}_{2}{\text{SO}}_{4}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& 2{\text{Ag}}^{\text{+}}\left(aq\right)+& {\text{SO}}_{4}{}^{\text{2-}}\left(aq\right)\\ \\ & 2x & \underset{_}{x}\end{array}$

(2c) $\begin{array}{ccc}\text{Al}{\left(\text{OH}\right)}_{3}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& {\text{Al}}^{\text{3+}}\left(aq\right)+& 3{\text{OH}}^{\text{-}}\left(aq\right)\\ \\ & x& \underset{_}{3x}\end{array}$

(2d) $\begin{array}{cccc}\text{Pb}\left(\text{OH}\right)\text{Cl}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& {\text{Pb}}^{\text{2+}}\left(aq\right)+& {\text{OH}}^{\text{-}}\left(aq\right)+& {\text{Cl}}^{\text{-}}\left(aq\right)\\ \\ & x\phantom{\rule{0.2em}{0ex}}& x& \underset{_}{x}\end{array}$

(2e) $\begin{array}{ccc}{\text{Ca}}_{3}\left({\text{AsO}}_{4}{\right)}_{2}\left(s\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}& 3{\text{Ca}}^{\text{2+}}\left(aq\right)+& 2{\text{AsO}}_{4}{}^{\text{3-}}\left(aq\right)\\ \\ & 3x&\underset{_}{2x}\end{array}$

(3) How do the concentrations of Ag⁺ and ${\text{CrO}}_{4}{}^{\text{2-}}$ in a saturated solution above 1.0 g of solid Ag₂CrO₄ change when 100 g of solid Ag₂CrO₄ is added to the system? Explain.

Solution

There is no change. A solid has an activity of 1 whether there is a little or a lot.

(4) How do the concentrations of Pb²⁺ and S^2– change when K₂S is added to a saturated solution of PbS?

(5) What additional information do we need to answer the following question: How is the equilibrium of solid silver bromide with a saturated solution of its ions affected when the temperature is raised?

Solution

The solubility of silver bromide at the new temperature must be known. Normally the solubility increases and some of the solid silver bromide will dissolve.

(6) Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: CoSO₃, CuI, PbCO₃, PbCl₂, Tl₂S, KClO₄?

(7) Which of the following slightly soluble compounds has a solubility greater than that calculated from its solubility product because of hydrolysis of the anion present: AgCl, BaSO₄, CaF₂, Hg₂I₂, MnCO₃, ZnS, PbS?

Solution

CaF₂, MnCO₃, and ZnS

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

(8a) PbCl₂

(8b) Ag₂S

(8c) Sr₃(PO₄)₂

(8d) SrSO₄

(9) Write the ionic equation for the dissolution and the K_sp expression for each of the following slightly soluble ionic compounds:

(9a) LaF₃

(9b) CaCO₃

(9c) Ag₂SO₄

(9d) Pb(OH)₂

Solution

(a) ${\text{LaF}}_{3}\left(s\right)$\rightleftharpoons${\text{La}}^{\text{3+}}\left(aq\right)+{\text{3F}}^{\text{-}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\left[{\text{La}}^{\text{3+}}\right]{\left[\text{F}}^{\text{-}}{\right]}^{3};$

(b) ${\text{CaCO}}_{3}\left(s\right)$\rightleftharpoons${\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{CO}}_{3}{}^{\text{2-}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2+}}\right]\left[{\text{CO}}_{3}{}^{\text{2-}}\right];$

(c) ${\text{Ag}}_{2}{\text{SO}}_{4}\left(s\right)$\rightleftharpoons${\text{2Ag}}^{\text{+}}\left(aq\right)+{\text{SO}}_{4}{}^{\text{2-}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}={\left[{\text{Ag}}^{\text{+}}\right]}^{2}\left[{\text{SO}}_{4}{}^{\text{2-}}\right];$

(d) ${\text{Pb(OH)}}_{2}\left(s\right)$\rightleftharpoons${\text{Pb}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{\text{-}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\left[{\text{Pb}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}$

(10) 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.

(10a) BaSiF₆, 0.026 g/100 mL (contains ${\text{SiF}}_{6}{}^{\text{2-}}$ ions)

(10b) Ce(IO₃)₄, 1.5 $\times$ 10^–2 g/100 mL

(10c) Gd₂(SO₄)₃, 3.98 g/100 mL

(10d) (NH₄)₂PtBr₆, 0.59 g/100 mL (contains ${\text{PtBr}}_{6}{}^{\text{2-}}$ ions)

(11) 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.

(11a) BaSeO₄, 0.0118 g/100 mL

(11b) Ba(BrO₃)₂·H₂O, 0.30 g/100 mL

(11c) NH₄MgAsO₄·6H₂O, 0.038 g/100 mL

(11d) La₂(MoO₄)₃, 0.00179 g/100 mL

Solution

(a) 1.77 $\times$ 10^–7; (b) 1.6 $\times$ 10^–6; (c) 2.2 $\times$ 10^–9; (d) 7.91 $\times$ 10^–22

(12) 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.

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

(13a) KHC₄H₄O₆

(13b) PbI₂

(13c) Ag₄[Fe(CN)₆], a salt containing the ${\text{Fe(CN)}}_{4}{}^{\text{-}}$ ion

(13d) Hg₂I₂

Solution

(a) 2 $\times$ 10^–2M; (b) 1.5 $\times$ 10^–3M; (c) 2.27 $\times$ 10^–9M; (d) 2.2 $\times$ 10^–10M

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

(14a) Ag₂SO₄

(14b) PbBr₂

(14c) AgI

(14d) CaC₂O₄·H₂O

(15) 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.

(15a) AgCl(s) in 0.025 M NaCl

(15b) CaF₂(s) in 0.00133 M KF

(15c) Ag₂SO₄(s) in 0.500 L of a solution containing 19.50 g of K₂SO₄

(15d) Zn(OH)₂(s) in a solution buffered at a pH of 11.45

Solution

(a) 6.4 $\times$ 10⁻⁹M = [Ag⁺], [Cl⁻] = 0.025 M. Check: $\frac{6.4\phantom{\rule{0.3em}{0ex}}\times\phantom{\rule{0.3em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}M}{0.025\phantom{\rule{0.2em}{0ex}}M}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}100%=\text{2.6}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-5}%;$ an insignificant change;

(b) 2.2 $\times$ 10⁻⁵M = [Ca²⁺], [F⁻] = 0.0013 M. Check: $\frac{\text{2.26}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}M}{0.00133\phantom{\rule{0.2em}{0ex}}M}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}100%=\text{1.70}%.$ This value is less than 5% and can be ignored.

(c) 0.2238 M = $\left[{\text{SO}}_{4}{}^{2\text{-}}\right];$ [Ag⁺] = 7.4 $\times$ 10^–3M. Check: $\frac{3.7\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-3}}{0.2238}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}100%=\text{1.64}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-2};$ the condition is satisfied.

(d) [OH^–] = 2.8 $\times$ 10^–3M; 5.7 $\times$ 10⁻¹²M = [Zn²⁺]. Check: $\frac{5.7\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-12}}{2.8\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-3}}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}100%=\text{2.0}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-7}%;$ x is less than 5% of [OH^–] and is, therefore, negligible.

(16) 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.

(16a) TlCl(s) in 1.250 M HCl

(16b) PbI₂(s) in 0.0355 M CaI₂

(16c) Ag₂CrO₄(s) in 0.225 L of a solution containing 0.856 g of K₂CrO₄

(16d) Cd(OH)₂(s) in a solution buffered at a pH of 10.995

(17) 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 it is not appropriate to neglect the changes in the initial concentrations of the common ions.

(17a) TlCl(s) in 0.025 M TlNO₃

(17b) BaF₂(s) in 0.0313 M KF

(17c) MgC₂O₄ in 2.250 L of a solution containing 8.156 g of Mg(NO₃)₂

(17d) Ca(OH)₂(s) in an unbuffered solution initially with a pH of 12.700

Solution

(a) [Cl^–] = 7.6 $\times$ 10⁻³M

Check: $\frac{7.6\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-3}}{0.025}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}100%=30%$

This value is too large to drop x. Therefore solve by using the quadratic equation:

[Ti⁺] = 3.1 $\times$ 10^–2M
[Cl^–] = 6.1 $\times$ 10^–3

(b) [Ba²⁺] = 7.7 $\times$ 10^–4M

Check: $\frac{7.7\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-4}}{0.0313}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}100%=2.4%$

Therefore, the condition is satisfied.

[Ba²⁺] = 7.7 $\times$ 10^–4M
[F^–] = 0.0321 M;

(c) Mg(NO₃)₂ = 0.02444 M

$\left[{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\right]=2.9\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{\text{−5}}$

Check: $\frac{2.9\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{\text{−5}}}{0.02444}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}100%=0.12%$

The condition is satisfied; the above value is less than 5%.

$\left[{\text{C}}_{2}{\text{O}}_{4}{}^{\text{2-}}\right]=2.9\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{\text{−5}}\phantom{\rule{0.2em}{0ex}}M$

[Mg²⁺] = 0.0244 M

(d) [OH^–] = 0.0501 M

[Ca²⁺] = 3.15 $\times$ 10^–3

Check: $\frac{3.15\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{\text{−3}}}{0.050}\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}100%=6.28%$

This value is greater than 5%, so a more exact method, such as successive approximations, must be used.

[Ca²⁺] = 2.8 $\times$ 10^–3M
[OH^–] = 0.053 $\times$ 10^–2M

(18) Explain why the changes in concentrations of the common ions in (Figure) can be neglected.

(19) Explain why the changes in concentrations of the common ions in (Figure) cannot be neglected.

Solution

The changes in concentration are greater than 5% and thus exceed the maximum value for disregarding the change.

(20) Calculate the solubility of aluminum hydroxide, Al(OH)₃, in a solution buffered at pH 11.00.

(21) Refer to Appendix J for solubility products for calcium salts. Determine which of the calcium salts listed is most soluble in moles per liter and which is most soluble in grams per liter.

Solution

CaSO₄∙2H₂O is the most soluble Ca salt in mol/L, and it is also the most soluble Ca salt in g/L.

(22) 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 7.2.3). 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.

(23) Public Health Service standards for drinking water set a maximum of 250 mg/L (2.60 $\times$ 10^–3M) of ${\text{SO}}_{4}{}^{\text{2-}}$ because of its cathartic action (it is a laxative). Does natural water that is saturated with CaSO₄ (“gyp” water) as a result or passing through soil containing gypsum, CaSO₄·2H₂O, meet these standards? What is the concentration of ${\text{SO}}_{4}{}^{\text{2-}}$ in such water?

Solution

4.8 $\times$ 10^–3M = $\left[{\text{SO}}_{4}{}^{\text{2-}}\right]$ = [Ca²⁺]; Since this concentration is higher than 2.60 $\times$ 10^–3M, “gyp” water does not meet the standards.

(24) Perform the following calculations:

(24a) Calculate [Ag⁺] in a saturated aqueous solution of AgBr.

(24b) What will [Ag⁺] be when enough KBr has been added to make [Br^–] = 0.050 M?

(24c) What will [Br^–] be when enough AgNO₃ has been added to make [Ag⁺] = 0.020 M?

(25) The solubility product of CaSO₄·2H₂O is 2.4 $\times$ 10^–5. What mass of this salt will dissolve in 1.0 L of 0.010 M ${\text{SO}4}_{}{}^{\text{2-}}?$

Solution

Mass (CaSO₄·2H₂O) = 0.72 g/L

(26) Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products).

(26a) TlCl

(26b) BaF₂

(26c) Ag₂CrO₄

(26d) CaC₂O₄·H₂O

(26e) the mineral anglesite, PbSO₄

(27) Assuming that no equilibria other than dissolution are involved, calculate the concentrations of ions in a saturated solution of each of the following (see Appendix J for solubility products):

(27a) AgI

(27b) Ag₂SO₄

(27c) Mn(OH)₂

(27d) Sr(OH)₂·8H₂O

(27e) the mineral brucite, Mg(OH)₂

Solution

(a) [Ag⁺] = [I^–] = 1.3 $\times$ 10^–5M

(b) [Ag⁺] = 2.88 $\times$ 10^–2M, $\left[{\text{SO}}_{4}{}^{\text{2-}}\right]$ = 1.44 $\times$ 10^–2M

(c) [Mn²⁺] = 3.7 $\times$ 10^–5M, [OH^–] = 7.4 $\times$ 10^–5M

(d) [Sr²⁺] = 4.3 $\times$ 10^–2M, [OH^–] = 8.6 $\times$ 10^–2M

(e) [Mg²⁺] = 1.3 $\times$ 10^–4M, [OH^–] = 2.6 $\times$ 10^–4M

(28) The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate K_sp for each of the slightly soluble solids indicated:

(28a) AgBr: [Ag⁺] = 5.7 $\times$ 10^–7M, [Br^–] = 5.7 $\times$ 10^–7M

(28b) CaCO₃: [Ca²⁺] = 5.3 $\times$ 10^–3M, $\left[{\text{CO}}_{3}{}^{\text{2-}}\right]$ = 9.0 $\times$ 10^–7M

(28c) PbF₂: [Pb²⁺] = 2.1 $\times$ 10^–3M, [F^–] = 4.2 $\times$ 10^–3M

(28d) Ag₂CrO₄: [Ag⁺] = 5.3 $\times$ 10^–5M, 3.2 $\times$ 10^–3M

(28e) InF₃: [In³⁺] = 2.3 $\times$ 10^–3M, [F^–] = 7.0 $\times$ 10^–3M

(29) The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate K_sp for each of the slightly soluble solids indicated:

(29a) TlCl: [Tl⁺] = 1.21 $\times$ 10^–2M, [Cl^–] = 1.2 $\times$ 10^–2M

(29b) Ce(IO₃)₄: [Ce⁴⁺] = 1.8 $\times$ 10^–4M, $\left[{\text{IO}}_{3}{}^{\text{-}}\right]$ = 2.6 $\times$ 10^–13M

(29c) Gd₂(SO₄)₃: [Gd³⁺] = 0.132 M, $\left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]$ = 0.198 M

(29d) Ag₂SO₄: [Ag⁺] = 2.40 $\times$ 10^–2M, $\left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]$ = 2.05 $\times$ 10^–2M

(29e) BaSO₄: [Ba²⁺] = 0.500 M, $\left[{\text{SO}}_{\text{4}}{}^{\text{2-}}\right]$ = 4.6 $\times$ 10⁻⁸M

Solution

(a) 1.7 $\times$ 10^–4

(b) 8.2 $\times$ 10^–55

(c) 1.35 $\times$ 10^–4

(d) 1.18 $\times$ 10^–5

(e) 1.08 $\times$ 10^–10

(30) Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for K_sp values.)

(30a) KClO₄: [K⁺] = 0.01 M, $\left[{\text{ClO}}_{4}{}^{\text{-}}\right]$ = 0.01 M

(30b) K₂PtCl₆: [K⁺] = 0.01 M, $\left[{\text{PtCl}}_{\text{6}}{}^{\text{2-}}\right]$ = 0.01 M

(30c) PbI₂: [Pb²⁺] = 0.003 M, [I^–] = 1.3 $\times$ 10^–3M

(30d) Ag₂S: [Ag⁺] = 1 $\times$ 10^–10M, [S^2–] = 1 $\times$ 10^–13M

(31) Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for K_sp values.)

(31a) CaCO₃: [Ca²⁺] = 0.003 M, $\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]$ = 0.003 M

(31b) Co(OH)₂: [Co²⁺] = 0.01 M, [OH^–] = 1 $\times$ 10^–7M

(31c) CaHPO₄: [Ca²⁺] = 0.01 M, $\left[{\text{HPO}}_{\text{4}}{}^{\text{2-}}\right]$ = 2 $\times$ 10^–6M

(31d) Pb₃(PO₄)₂: [Pb²⁺] = 0.01 M, $\left[{\text{PO}}_{\text{4}}{}^{\text{3-}}\right]$ = 1 $\times$ 10^–13M

Solution

(a) CaCO₃ does precipitate. (b) The compound does not precipitate. (c) The compound does not precipitate. (d) The compound precipitates.

(32) Calculate the concentration of Tl⁺ when TlCl just begins to precipitate from a solution that is 0.0250 M in Cl^–.

(33) Calculate the concentration of sulfate ion when BaSO₄ just begins to precipitate from a solution that is 0.0758 M in Ba²⁺.

Solution

3.03 $\times$ 10⁻⁷M.

(34) Calculate the concentration of Sr²⁺ when SrF₂ starts to precipitate from a solution that is 0.0025 M in F^–.

(35) Calculate the concentration of ${\text{PO}}_{4}{}^{\text{3-}}$ when Ag₃PO₄ starts to precipitate from a solution that is 0.0125 M in Ag⁺.

Solution

9.2 $\times$ 10⁻¹³M

(36) Calculate the concentration of F^– required to begin precipitation of CaF₂ in a solution that is 0.010 M in Ca²⁺.

(37) Calculate the concentration of Ag⁺ required to begin precipitation of Ag₂CO₃ in a solution that is 2.50 $\times$ 10^–6M in ${\text{CO}}_{3}{}^{\text{2-}}.$

Solution

[Ag⁺] = 1.8 $\times$ 10^–3M

(38) What [Ag⁺] is required to reduce $\left[{\text{CO}}_{3}{}^{\text{2-}}\right]$ to 8.2 $\times$ 10^–4M by precipitation of Ag₂CO₃?

(39) What [F^–] is required to reduce [Ca²⁺] to 1.0 $\times$ 10^–4M by precipitation of CaF₂?

Solution

6.3 $\times$ 10^–4

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

Perform these calculations for nickel(II) carbonate.

(41a) With what volume of water must a precipitate containing NiCO₃ be washed to dissolve 0.100 g of this compound? Assume that the wash water becomes saturated with NiCO₃ (K_sp = 1.36 $\times$ 10^–7).

(41b) If the NiCO₃ were a contaminant in a sample of CoCO₃ (K_sp = 1.0 $\times$ 10^–12), what mass of CoCO₃ would have been lost? Keep in mind that both NiCO₃ and CoCO₃ dissolve in the same solution.

Solution

(a) 2.25 L; (b) 7.2 $\times$ 10^–7 g

(42) Iron concentrations greater than 5.4 $\times$ 10^–6M in water used for laundry purposes can cause staining. What [OH^–] is required to reduce [Fe²⁺] to this level by precipitation of Fe(OH)₂?

(43) A solution is 0.010 M in both Cu²⁺ and Cd²⁺. What percentage of Cd²⁺ remains in the solution when 99.9% of the Cu²⁺ has been precipitated as CuS by adding sulfide?

Solution

100% of it is dissolved

(44) A solution is 0.15 M in both Pb²⁺ and Ag⁺. If Cl^– is added to this solution, what is [Ag⁺] when PbCl₂ begins to precipitate?

(45) What reagent might be used to separate the ions in each of the following mixtures, which are 0.1 M with respect to each ion? In some cases, it may be necessary to control the pH. (Hint: Consider the K_sp values given in Appendix J.)

(45a) ${\text{Hg}}_{2}{}^{\text{2+}}$ and Cu²⁺

(45b) ${\text{SO}}_{4}{}^{\text{2-}}$ and Cl^–

(45c) Hg²⁺ and Co²⁺

(45d) Zn²⁺ and Sr²⁺

(45e) Ba²⁺ and Mg²⁺

(45f) ${\text{CO}}_{3}{}^{\text{2-}}$ and OH^–

Solution

(a) ${\text{Hg}}_{2}{}^{\text{2+}}$ and Cu²⁺: Add ${\text{SO}}_{4}{}^{\text{2-}}.$

(b) ${\text{SO}}_{4}{}^{\text{2-}}$ and Cl^–: Add Ba²⁺.

(c) Hg²⁺ and Co²⁺: Add S^2–.

(d) Zn²⁺ and Sr²⁺: Add OH^– until [OH^–] = 0.050 M.

(e) Ba²⁺ and Mg²⁺: Add ${\text{SO}}_{4}{}^{\text{2-}}.$

(f) ${\text{CO}}_{3}{}^{\text{2-}}$ and OH^–: Add Ba²⁺.

(46) A solution contains 1.0 $\times$ 10^–5 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?

(47) A solution contains 1.0 $\times$ 10^–2 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

AgI will precipitate first.

(48) The calcium ions in human blood serum are necessary for coagulation (Figure 7.2.4). Potassium oxalate, K₂C₂O₄, is used as an anticoagulant when a blood sample is drawn for laboratory tests because it removes the calcium as a precipitate of CaC₂O₄·H₂O. It is necessary to remove all but 1.0% of the Ca²⁺ in serum in order to prevent coagulation. If normal blood serum with a buffered pH of 7.40 contains 9.5 mg of Ca²⁺ per 100 mL of serum, what mass of K₂C₂O₄ is required to prevent the coagulation of a 10 mL blood sample that is 55% serum by volume? (All volumes are accurate to two significant figures. Note that the volume of serum in a 10-mL blood sample is 5.5 mL. Assume that the K_sp value for CaC₂O₄ in serum is the same as in water.)

(49) About 50% of urinary calculi (kidney stones) consist of calcium phosphate, Ca₃(PO₄)₂. The normal mid range calcium content excreted in the urine is 0.10 g of Ca²⁺ per day. The normal mid range amount of urine passed may be taken as 1.4 L per day. What is the maximum concentration of phosphate ion that urine can contain before a calculus begins to form?

Solution

1.5 $\times$ 10⁻¹²M

(50) The pH of normal urine is 6.30, and the total phosphate concentration $\left({\left[\text{PO}}_{4}{}^{\text{3-}}\right]$ + $\left[{\text{HPO}}_{4}{}^{\text{2-}}\right]$ + $\left[{\text{H}}_{2}{\text{PO}}_{4}{}^{\text{-}}\right]$ + [H₃PO₄]) is 0.020 M. What is the minimum concentration of Ca²⁺ necessary to induce kidney stone formation? (See (Figure) for additional information.)

(51) Magnesium metal (a component of alloys used in aircraft and a reducing agent used in the production of uranium, titanium, and other active metals) is isolated from sea water by the following sequence of reactions:

${\text{Mg}}^{\text{2+}}\left(aq\right)+{\text{Ca(OH)}}_{2}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{Mg(OH)}}_{2}\left(s\right)+{\text{Ca}}^{\text{2+}}\left(aq\right)$

${\text{Mg(OH)}}_{2}\left(s\right)+\text{2HCl(}aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{MgCl}}_{2}\left(s\right)+{\text{2H}}_{2}\text{O}\left(l\right)$

${\text{MgCl}}_{2}\left(l\right)\stackrel{\phantom{\rule{0.7em}{0ex}}\text{electrolysis}\phantom{\rule{0.7em}{0ex}}}{\to }\text{Mg}\left(s\right)+{\text{Cl}}_{2}\left(g\right)$

Sea water has a density of 1.026 g/cm³ and contains 1272 parts per million of magnesium as Mg²⁺(aq) by mass. What mass, in kilograms, of Ca(OH)₂ is required to precipitate 99.9% of the magnesium in 1.00 $\times$ 10³ L of sea water?

Solution

3.99 kg

(52) Hydrogen sulfide is bubbled into a solution that is 0.10 M in both Pb²⁺ and Fe²⁺ and 0.30 M in HCl. After the solution has come to equilibrium it is saturated with H₂S ([H₂S] = 0.10 M). What concentrations of Pb²⁺ and Fe²⁺ remain in the solution? For a saturated solution of H₂S we can use the equilibrium:

${\text{H}}_{2}\text{S}\left(aq\right)+{\text{2H}}_{2}\text{O}\left(l\right)$\rightleftharpoons${\text{2H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{S}}^{\text{2-}}\left(aq\right)\phantom{\rule{4em}{0ex}}K=1.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-26}$

(Hint: The $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ changes as metal sulfides precipitate.)

Perform the following calculations involving concentrations of iodate ions:

(52a) The iodate ion concentration of a saturated solution of La(IO₃)₃ was found to be 3.1 $\times$ 10^–3 mol/L. Find the K_sp.

(52b) Find the concentration of iodate ions in a saturated solution of Cu(IO₃)₂ (K_sp = 7.4 $\times$ 10^–8).

Solution

(a) 3.1 $\times$ 10^–11; (b) [Cu²⁺] = 2.6 $\times$ 10^–3; $\left[{\text{IO}}_{3}{}^{\text{-}}\right]$ = 5.3 $\times$ 10^–3

(53) Calculate the molar solubility of AgBr in 0.035 M NaBr (K_sp = 5 $\times$ 10^–13).

(54) How many grams of Pb(OH)₂ will dissolve in 500 mL of a 0.050-M PbCl₂ solution (K_sp = 1.2 $\times$ 10^–15)?

Solution

1.8 $\times$ 10^–5 g Pb(OH)₂

(55) Use the simulation from the earlier Link to Learning to complete the following exercise. Using 0.01 g CaF₂, give the K_sp values found in a 0.2-M solution of each of the salts. Discuss why the values change as you change soluble salts.

(56) 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 $\times$ 10^–12. Include the ionic reaction and the expression for K_sp in your answer. (K_w = 1 $\times$ 10^–14 = [H₃O⁺][OH^–])

Solution

${\text{Mg(OH)}}_{2}\left(s\right)$\rightleftharpoons${\text{Mg}}^{\text{2+}}+{\text{2OH}}^{\text{-}}\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\left[{\text{Mg}}^{\text{2+}}\right]{\left[{\text{OH}}^{\text{-}}\right]}^{2}$
1.23 $\times$ 10⁻³ g Mg(OH)₂

(57) Two hypothetical salts, LM₂ and LQ, have the same molar solubility in H₂O. If K_sp for LM₂ is 3.20 $\times$ 10^–5, what is the K_sp value for LQ?

(58) The carbonate ion concentration is gradually increased in a solution containing divalent cations of magnesium, calcium, strontium, barium, and manganese. Which of the following carbonates will form first? Which of the following will form last? Explain.

(58a) ${\text{MgCO}}_{3}\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=3.5\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-8}$

(58b) ${\text{CaCO}}_{3}\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=4.2\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-7}$

(58c) ${\text{SrCO}}_{3}\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=3.9\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-9}$

(58d) ${\text{BaCO}}_{3}\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=4.4\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-5}$

(58e) ${\text{MnCO}}_{3}\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=5.1\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-9}$

Solution

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.

(59) How many grams of Zn(CN)₂(s) (117.44 g/mol) would be soluble in 100 mL of H₂O? Include the balanced reaction and the expression for K_sp in your answer. The K_sp value for Zn(CN)₂(s) is 3.0 $\times$ 10^–16.

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

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Inorganic Chemistry for Chemical Engineers Copyright © 2020 by Vishakha Monga; Paul Flowers; Klaus Theopold; William R. Robinson; and Richard Langley is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

The Solubility Product

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Ksp and Solubility

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Using Barium Sulfate for Medical Imaging

Predicting Precipitation

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The Role of Precipitation in Wastewater Treatment

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Common Ion Effect

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Key Concepts and Summary

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K_sp and Solubility