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# 15 Equilibria of Other Reaction Classes

Figure 15.1 The mineral fluorite (CaF2) is deposited through a precipitation process. Note that pure fluorite is colorless, and that the color in this sample is due to the presence of other metals in the crystal.

### Chapter Outline

• 15.1 Precipitation and Dissolution
• 15.2 Lewis Acids and Bases
• 15.3 Multiple Equilibria

### Introduction

In Figure 15.1, we see a close-up image of the mineral fluorite, which is commonly used as a semiprecious stone in many types of jewelry because of its striking appearance. These solid deposits of fluorite are formed through a process called hydrothermal precipitation. In this process, the fluorite remains dissolved in solution, usually in hot water heated by volcanic activity deep below the earth, until conditions arise that allow the mineral to come out of solution and form a deposit. These deposit-forming conditions can include a change in temperature of the solution, availability of new locations to form a deposit such as a rock crevice, contact between the solution and a reactive substance such as certain types of rock, or a combination of any of these factors.

We previously learned about aqueous solutions and their importance, as well as about solubility rules. While this gives us a picture of solubility, that picture is not complete if we look at the rules alone. Solubility equilibrium, which we will explore in this chapter, is a more complex topic that allows us to determine the extent to which a slightly soluble ionic solid will dissolve, and the conditions under which precipitation (such as the fluorite deposit in Figure 15.1) will occur.

## 15.1 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

The preservation of medical laboratory blood samples, mining of sea water for magnesium, formulation of over-the-counter medicines such as Milk of Magnesia and antacids, and treating the presence of hard water in your home’s water supply are just a few of the many tasks that involve controlling the equilibrium between a slightly soluble ionic solid and an aqueous solution of its ions.

In some cases, we want to prevent dissolution from occurring. Tooth decay, for example, occurs when the calcium hydroxylapatite, which has the formula Ca5(PO4)3(OH), in our teeth dissolves. The dissolution process is aided when bacteria in our mouths feast on the sugars in our diets to produce lactic acid, which reacts with the hydroxide ions in the calcium hydroxylapatite. Preventing the dissolution prevents the decay. On the other hand, sometimes we want a substance to dissolve. We want the calcium carbonate in a chewable antacid to dissolve because the CO32− ions produced in this process help soothe an upset stomach.

In this section, we will find out how we can control the dissolution of a slightly soluble ionic solid by the application of Le Châtelier’s principle. We will also learn how to use the equilibrium constant of the reaction to determine the concentration of ions present in a solution.

### The Solubility Product Constant

Silver chloride is what’s known as a sparingly soluble ionic solid (Figure 15.2). Recall from the solubility rules in an earlier chapter that halides of Ag+ are not normally soluble. However, when we add an excess of solid AgCl to water, it dissolves to a small extent and produces a mixture consisting of a very dilute solution of Ag+ and Cl ions in equilibrium with undissolved silver chloride:

This equilibrium, like other equilibria, is dynamic; some of the solid AgCl continues to dissolve, but at the same time, Ag+ and Cl ions in the solution combine to produce an equal amount of the solid. At equilibrium, the opposing processes have equal rates.

Figure 15.2 Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag+ and Cl ions in equilibrium with undissolved silver chloride.

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

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

Some common solubility products are listed in Table 15.1 aaccording to their Ksp values, whereas a more extensive compilation of products appears in Appendix J. Each of these equilibrium constants is much smaller than 1 because the compounds listed are only slightly soluble. A small Ksp represents a system in which the equilibrium lies to the left, so that relatively few hydrated ions would be present in a saturated solution.

### Example 15.1

Writing Equations and Solubility Products

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

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

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

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

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

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

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

Solution

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

(a) BaSO4

(b) Ag2SO4

(c) Al(OH)3

(d) Pb(OH)Cl

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

### Ksp and Solubility

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

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

### Example 15.2

Calculation of Ksp from Equilibrium Concentrations

We began the chapter with an informal discussion of how the mineral fluorite ([link]) is formed. Fluorite, CaF2, is a slightly soluble solid that dissolves according to the equation:

The concentration of Ca2+ in a saturated solution of CaF2 is 2.1 × 10–4 M; therefore, that of F– is 4.2 × 10–4 M, that is, twice the concentration of Ca2+. What is the solubility product of fluorite?

Solution

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

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

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

In a saturated solution that is in contact with solid Mg(OH)2, the concentration of Mg2+ is 3.7 × 10–5 M. What is the solubility product for Mg(OH)2?

### Example 15.3

Determination of Molar Solubility from Ksp

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

Solution

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

The reaction is:

First, write out the solubility product equilibrium constant expression:

Create an ICE table (as introduced in the chapter on fundamental equilibrium concepts), leaving the CuBr column empty as it is a solid and does not contribute to the Ksp:

At equilibrium:

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

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

1.2 × 10–8 M

### Example 15.4

Determination of Molar Solubility from Ksp, Part II

The Ksp of calcium hydroxide, Ca(OH)2, is 8.0 × 10–6. Calculate the molar solubility of calcium hydroxide.

Solution

The solubility product constant of calcium hydroxide is 8.0 × 10–6.

The reaction is

First, write out the solubility product equilibrium constant expression:

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

At equilibrium:

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

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

Note that solubility is not always given as a molar value. When the solubility of a compound is given in some unit other than moles per liter, we must convert the solubility into moles per liter (i.e., molarity) in order to use it in the solubility product constant expression. Example 15.5 shows how to perform those unit conversions before determining the solubility product equilibrium.

### Example 15.5

Determination of Ksp from Gram Solubility

Many of the pigments used by artists in oil-based paints (Figure) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO4, is 4.3 × 10–5 g/L. Determine the solubility product equilibrium constant for PbCrO4

Figure 15.3 Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow (PbCrO4), examples include Prussian blue (Fe7(CN)18), the reddish-orange color vermilion (HgS), and green color veridian (Cr2O3). (credit: Sonny Abesamis)

Solution

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

1. Use the molar mass of PbCrO

to convert the solubility of PbCrO4 in grams per liter into moles per liter

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

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

3. Solve. Ksp = [Pb2+][CrO42−] = (1.3 × 10–7)(1.3 × 10–7) = 1.7 × 10–14

The solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.2 grams per liter at 20 °C. What is its solubility product?

Answer: 1.4 × 10–4 (1.5 × 10–4 if we round the solubility to two digits before calculating Ksp)

### Example 15.6

Calculating the Solubility of Hg2Cl2

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

Calculate the molar solubility of Hg2Cl2.

Solution

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

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

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

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

3. Solve for x and the equilibrium concentrations. We substitute the equilibrium concentrations into the expression for Ksp and calculate the value of x

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

4.

Check the work. At equilibrium, Q = Ksp:

The calculations check.

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

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

### 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 Ksp of barium sulfate is 1.1 × 10–10, 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 15.4).

Figure 15.4 The suspension of barium sulfate coats the intestinal tract, which allows for greater visual detail than a traditional X-ray. (credit modification of work by “glitzy queen00”/Wikimedia Commons)

Further diagnostic testing can be done using barium sulfate and fluoroscopy. In fluoroscopy, a continuous X-ray is passed through the body so the doctor can monitor, on a TV or computer screen, the barium sulfate’s movement as it passes through the digestive tract. 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 (http://openstaxcollege.org/l/16barium)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:

We can establish this equilibrium either by adding solid calcium carbonate to water or by mixing a solution that contains calcium ions with a solution that contains carbonate ions. If we add calcium carbonate to water, the solid will dissolve until the concentrations are such that the value of the reaction quotient (Q=[Ca2+][CO32−]) is equal to the solubility product (Ksp = 4.8 × 10–9). If we mix a solution of calcium nitrate, which contains Ca2+ ions, with a solution of sodium carbonate, which contains CO32 ions, the slightly soluble ionic solid CaCO3 will precipitate, provided that the concentrations of Ca2+ and CO32− ions are such that Q is greater than Ksp for the mixture. The reaction shifts to the left and the concentrations of the ions are reduced by formation of the solid until the value of Q equals Ksp. A saturated solution in equilibrium with the undissolved solid will result. If the concentrations are such that Q is less than Ksp, then the solution is not saturated and no precipitate will form.

We can compare numerical values of Q with Ksp to predict whether precipitation will occur, as Example 15.7 shows. (Note: Since all forms of equilibrium constants are temperature dependent, we will assume a room temperature environment going forward in this chapter unless a different temperature value is explicitly specified.)

### Example 15.7

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:

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

Solution

This problem asks whether the reaction:

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

Because Q is greater than Ksp (Q = 5.4 × 10–8 is larger than Ksp = 2.1 × 10–13), we can expect the reaction to shift to the left and form solid magnesium hydroxide. Mg(OH)2(s) forms until the concentrations of magnesium ion and hydroxide ion are reduced sufficiently so that the value of Q is equal to Ksp.

Use the solubility product in Appendix J to determine whether CaHPO4 will precipitate from a solution with [Ca2+] = 0.0001 M and [HPO42] = 0.001 M.

Answer:No precipitation of CaHPO4; Q = 1 × 10–7, which is less than Ksp

### Example 15.8

Precipitation of AgCl upon Mixing Solutions

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

(Note: The solution also contains Na+ and NO3− ions, but when referring to solubility rules, one can see that sodium nitrate is very soluble and cannot form a precipitate.)

Solution

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

The solubility product is 1.8 × 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 Ksp. The volume doubles when we mix equal volumes of AgNO3 and NaCl solutions, so each concentration is reduced to half its initial value. Consequently, immediately upon mixing, [Ag+] and [Cl] are both equal to:

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

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

Will KClO4 precipitate when 20 mL of a 0.050-M solution of K+ is added to 80 mL of a 0.50-M solution of ClO4? (Remember to calculate the new concentration of each ion after mixing the solutions before plugging into the reaction quotient expression.)

Answer:No, Q = 4.0 × 10–3, which is less than Ksp = 1.07 × 10–2

In the previous two examples, we have seen that Mg(OH)2 or AgCl precipitate when Q is greater than Ksp. In general, when a solution of a soluble salt of the Mm+ ion is mixed with a solution of a soluble salt of the Xn– ion, the solid, MpXq precipitates if the value of Q for the mixture of Mm+ and Xn– is greater than Ksp for MpXq. Thus, if we know the concentration of one of the ions of a slightly soluble ionic solid and the value for the solubility product of the solid, then we can calculate the concentration that the other ion must exceed for precipitation to begin. To simplify the calculation, we will assume that precipitation begins when the reaction quotient becomes equal to the solubility product constant.

### Example 15.9

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, C2O42−, for this purpose (Figure 15.5).. At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC2O4·H2O (which also contains water bound in the solid). The concentration of Ca2+ in a sample of blood serum is 2.2 × 10–3 M. What concentration of C2O42 ion must be established before CaC2O4·H2O begins to precipitate?

Figure 15.5 Anticoagulants can be added to blood that will combine with the Ca2+ ions in blood serum and prevent the blood from clotting. (credit: modification of work by Neeta Lind)

Solution

The equilibrium expression is:

For this reaction:

(see Appendix J)

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

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

A concentration of [C2O42−] = 1.0 × 10–6 M is necessary to initiate the precipitation of CaC2O4 under these conditions.

If a solution contains 0.0020 mol of CrO42− 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.

It is sometimes useful to know the concentration of an ion that remains in solution after precipitation. We can use the solubility product for this calculation too: If we know the value of Ksp and the concentration of one ion in solution, we can calculate the concentration of the second ion remaining in solution. The calculation is of the same type as that in Example 15.9—calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. However, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation.

### Example 15.10

Concentrations Following Precipitation

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

Solution

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

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

or

so

Now we calculate the pH from the pOH:

If the person doing laundry adds a base, such as the sodium silicate (Na4SiO4) in some detergents, to the wash water until the pH is raised to 10.20, the manganese ion will be reduced to a concentration of 1.8 × 10–6 M; at that concentration or less, the ion will not stain clothing.

The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of Ca(OH)2. The concentration of Mg2+(aq) in sea water is 5.37 × 10–2 M. Calculate the pH at which [Mg2+] is diminished to 1.0 × 10–5 M by the addition of Ca(OH)2.

Due to their light sensitivity, mixtures of silver halides are used in fiber optics for medical lasers, in photochromic eyeglass lenses (glass lenses that automatically darken when exposed to sunlight), and—before the advent of digital photography—in photographic film. Even though AgCl (Ksp = 1.6 × 10–10), AgBr (Ksp = 7.7 × 10–13), and AgI (Ksp = 8.3 × 10–17) are each quite insoluble, we cannot prepare a homogeneous solid mixture of them by adding Ag+ to a solution of Cl–, Br–, and I–; essentially all of the AgI will precipitate before any of the other solid halides form because of its smaller value for Ksp. However, we can prepare a homogeneous mixture of the solids by slowly adding a solution of Cl, Br, and I to a solution of Ag+.

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

### Chemistry in Everyday Life

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). Specifically, selective precipitation is used to remove contaminants from wastewater before it is released back into natural bodies of water. For example, phosphate ions (PO42−) 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.

Figure 15.6 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, known as lime, Ca(OH)2. The lime is converted into calcium carbonate, a strong base, in the water. As the water is made more basic, the calcium ions react with phosphate ions to produce hydroxylapatite, Ca5(PO4)3(OH), which then precipitates out of the solution:

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.

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

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

### Example 15.11

Precipitation of Silver Halides

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

Solution

The two equilibria involved are:

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

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

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

For AgCl: AgCl precipitates when Q equals Ksp for AgCl (1.8 × 10–10). When [Cl] = 0.10 M:

AgCl begins to precipitate when [Ag+] is 1.8 × 10–9 M.

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

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+] = 1.5 × 10–11 M; AgBr precipitates first

### Common Ion Effect

As we saw when we discussed buffer solutions, the hydronium ion concentration of an aqueous solution of acetic acid decreases when the strong electrolyte sodium acetate, NaCH3CO2, is added. We can explain this effect using Le Châtelier’s principle. The addition of acetate ions causes the equilibrium to shift to the left, decreasing the concentration of H3O+ to compensate for the increased acetate ion concentration. This increases the concentration of CH3CO2H:

Because sodium acetate and acetic acid have the acetate ion in common, the influence on the equilibrium is called the common ion effect.

The common ion effect can also have a direct effect on solubility equilibria. Suppose we are looking at the reaction where silver iodide is dissolved:

If we were to add potassium iodide (KI) to this solution, we would be adding a substance that shares a common ion with silver iodide. Le Châtelier’s principle tells us that when a change is made to a system at equilibrium, the reaction will shift to counteract that change. In this example, there would be an excess of iodide ions, so the reaction would shift toward the left, causing more silver iodide to precipitate out of solution.

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

### Example 15.12

Common Ion Effect

Calculate the molar solubility of cadmium sulfide (CdS) in a 0.010-M solution of cadmium bromide (CdBr2). The Ksp of CdS is 1.0 × 10–28.

Solution

The first thing you should notice is that the cadmium sulfide is dissolved in a solution that contains cadmium ions. We need to use an ICE table to set up this problem and include the CdBr2 concentration as a contributor of cadmium ions:

We can solve this equation using the quadratic formula, but we can also make an assumption to make this calculation much simpler. Since the Ksp value is so small compared with the cadmium concentration, we can assume that the change between the initial concentration and the equilibrium concentration is negligible, so that 0.010 + x ~ 0.010. Going back to our Ksp expression, we would now get:

Therefore, the molar solubility of CdS in this solution is 1.0 × 10–26 M

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

## Exercises

### Question 15.1

15.1

Complete the changes in concentrations for each of the following reactions:

(a)$AgI(s) \longrightarrow Ag^{+}(aq) + I^{−}(aq)$ x $\underline{\hspace{1cm}}$

(b) $CaCO_{3} (s) \longrightarrow Ca^{2+}(aq) + CO3 2\:^{−}(aq)$ $\underline{\hspace{1cm}}$ x

(c)$Mg(OH)_{2} (s) \longrightarrow Mg^{2+}(aq) + 2OH^{−}(aq)$ x $\underline{\hspace{1cm}}$

(d)$Mg_{3}(PO_{4} )_{2} (s) \longrightarrow 3Mg^{2+}(aq) + 2PO_{4}\:^ {3−}(aq)$ x $\underline{\hspace{1cm}}$

(e)$Ca5 (PO4 ) 3 OH(s) \longrightarrow 5Ca2+(aq) + 3PO4 3−(aq) + OH^{−}(aq)$ $\underline{\hspace{1cm}}$ $\underline{\hspace{1cm}}$ x

### Question 15.2

15.2

Complete the changes in concentrations for each of the following reactions:

(a)$BaSO_{4} (s) \longrightarrow Ba^{2+}(aq) + SO_{4}\: ^{2−}(aq)$ x $\underline{\hspace{1cm}}$

(b)$Ag_{2}SO_{4} (s) \longrightarrow 2Ag^{+}(aq) + SO_{4}\:^{ 2−}(aq)$ $\underline{\hspace{1cm}}$ x

(c)$Al(OH)_{3} (s) \longrightarrow Al^{3+}(aq) + 3OH^{−}(aq)$ x $\underline{\hspace{1cm}}$

(d)$Pb(OH)Cl(s) \longrightarrow Pb^{2+}(aq) + OH^{−}(aq) + Cl−(aq)$ $\underline{\hspace{1cm}}$ x $\underline{\hspace{1cm}}$

(e)$Ca_{3} (AsO_{4} ) 2 (s) \longrightarrow 3Ca^{2+}(aq) + 2AsO_4\:^{3−}(aq)$ 3x $\underline{\hspace{1cm}}$

### Question 15.3

15.3

How do the concentrations of $Ag^+$ and $CrO_4\:^{2−}$ in a saturated solution above 1.0 g of solid $Ag_{2}CrO_{4}$ change when 100 g of solid $Ag_{2}CrO_{4}$ is added to the system? Explain.

### Question 15.4

15.3

How do the concentrations of $Pb^{2+}$ and $S^{2–}$ change when $K_{2}S[$ is added to a saturated solution of PbS?

### Question 15.5

15.5

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?

### Question 15.6

15.6

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: $CoSO_{3}, CuI, PbCO_{3}, PbCl_{2}, Tl_{2}S, KClO_{4}?$

### Question 15.7

15.7

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: $AgCl, BaSO_{4}, CaF_{2}, Hg_{2}I_{2}, MnCO_{3}, ZnS, PbS?$

### Question 15.8

15.8

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

$(a) PbCl_{2}$

$(b) Ag_{2}S$

$(c) Sr_{3}(PO_{4})_{2}$

$(d) SrSO_{4}$

### Question 15.9

15.9

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

$(a) LaF_3$

$(b) CaCO_3$

$(c) Ag_{2}SO_{4}$

$(d) Pb(OH)_2$

### Question 15.10

15.10

The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.

$(a) BaSiF_{6}, 0.026 g/100 \: mL \: (contains \: SiF_{6}\:^{2−} \:ions)$

$(b) Ce(IO_{3})_{4}, 1.5 × 10^{–2} \:g/100 \: mL$

$(c) Gd_{2}(SO_{4})_{3}, 3.98 \: g/100 \:mL$

$(d) (NH_{4})_{2}PtBr_{6}, 0.59 \:g/100 \:mL \: (contains \: PtBr_{6}\:^{2−} ions)$

### Question 15.11

15.11

The Handbook of Chemistry and Physics (http://openstaxcollege.org/l/16Handbook) gives solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.

$(a) BaSeO_{4}, 0.0118 \:g/100 \: mL$

$(b) Ba(BrO_{3})2∙H2O, 0.30 \:g/100 \:mL$

$(c) NH4MgAsO_{4}∙6H2O, 0.038 \:g/100 \: mL$

$(d) La2(MoO_{4})3, 0.00179 \: g/100 \: mL)$

### Question 15.12

15.12

Use solubility products and predict which of the following salts is the most soluble, in terms of moles per liter, in pure water: $CaF_{2}, Hg_{2}Cl_{2}, PbI_{2}, or Sn(OH)_{2}$

### Question 15.13

15.13

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

$(a) KHC_{4}H_{4}O_{6}$

$(b) PbI_{2}$

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

$(d) Hg_{2}I_{2}$

### Question 15.14

15.14

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

$(a) Ag_{2}SO_{4}$

$(b) PbBr_{2}$

(c) AgI

$(d) CaC_{2}O_{4}∙H_{2}O$

### Question 15.15

15.15

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.

(a) AgCl(s) in 0.025 M NaCl

$(b) CaF_{2}(s) in 0.00133 M KF$

$(c) Ag2SO_{4}(s) in 0.500 \:L \:of \:a \: solution \: containing \: 19.50 \: g \:of \: K_{2}SO_{4}$

$(d) Zn(OH)_{2}(s) \: in \: a \: solution \:buffered \:at \:a \:pH \:of \: 11.45$

### Question 15.16

15.16

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.

(a) TlCl(s) in 1.250 M HCl

$(b) PbI_{2}(s) in 0.0355 M CaI_{2}$

$(c) Ag_{2}CrO_{4}(s)$ in 0.225 L of a solution containing 0.856 g of K{2}CrO{4}

$(d) Cd(OH)_{2}(s)$ in a solution buffered at a pH of 10.995

### Question 15.17

15.17

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.

$(a) TlCl(s) in 0.025 M TlNO_{3}$

$(b) BaF_{2}(s) in 0.0313 M KF$

$(c) MgC_{2}O_{4}$ in 2.250 L of a solution containing 8.156 g of $Mg(NO_{3})_{2}$

$(d) Ca(OH)_{2}(s)$in an unbuffered solution initially with a pH of 12.700

### Question 15.18

15.18

Explain why the changes in concentrations of the common ions in Exercise 15.17 can be neglected.

### Question 15.19

15.19

Explain why the changes in concentrations of the common ions in Exercise 15.18 cannot be neglected.

### Question 15.20

15.20

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

### Question 15.21

15.21

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.

### Question 15.22

15.22

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 15.4). This use of $BaSO_{4}$ is possible because of its low solubility. Calculate the molar solubility of $BaSO_{4}$ and the mass of barium present in 1.00 L of water saturated with $BaSO_{4}$

### Question 15.23

15.23

Public Health Service standards for drinking water set a maximum of 250 mg/L (2.60 × 10^{–3} M) of $SO_{4}\:^{2−}$ because of its cathartic action (it is a laxative). Does natural water that is saturated with $CaSO_{4}$ (“gyp” water) as a result or passing through soil containing gypsum, $CaSO_{4}$$2H_{2}O$, meet these standards? What is $SO_{4}\:^{2−}$in such water?

### Question 15.24

15.24

Perform the following calculations:

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

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

(c) What will $[Br^– ]$ be when enough $AgNO_{3}$ has been added to make $[Ag^+ ]$ = 0.020 M?

15.25

### Question 15.59

15.59

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

(a) $MgCO_{3} \: K_{sp} = 3.5 × 10^{−8}$

(b) $CaCO_{3} \: K_{sp} = 4.2 × 10^{−7}$

(c) $SrCO_{3} \: K_{sp} = 3.9 × 10^{−9}$

(d) $BaCO_{3} \: K_{sp} = 4.4 × 10^{−5}$

(e) $MnCO_{3} \: K_{sp} = 5.1 × 10^{−9}$

### Question 15.60

15.60

How many grams of $Zn(CN)_{2}(s)$ (117.44 g/mol) would be soluble in 100 mL of $H_{2}O$? 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 × $10^{–16}$

## 15.2 Lewis Acids and Bases

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

• • Explain the Lewis model of acid-base chemistry
• • Write equations for the formation of adducts and complex ions
• • Perform equilibrium calculations involving formation constants

In 1923, G. N. Lewis proposed a generalized definition of acid-base behavior in which acids and bases are identified by their ability to accept or to donate a pair of electrons and form a coordinate covalent bond.

A coordinate covalent bond (or dative bond) occurs when one of the atoms in the bond provides both bonding electrons. For example, a coordinate covalent bond occurs when a water molecule combines with a hydrogen ion to form a hydronium ion. A coordinate covalent bond also results when an ammonia molecule combines with a hydrogen ion to form an ammonium ion. Both of these equations are shown here.

A Lewis acid is any species (molecule or ion) that can accept a pair of electrons, and a Lewis base is any species (molecule or ion) that can donate a pair of electrons.

A Lewis acid-base reaction occurs when a base donates a pair of electrons to an acid. A Lewis acid-base adduct, a compound that contains a coordinate covalent bond between the Lewis acid and the Lewis base, is formed. The following equations illustrate the general application of the Lewis concept.

The boron atom in boron trifluoride, BF3, has only six electrons in its valence shell. Being short of the preferred octet, BF3 is a very good Lewis acid and reacts with many Lewis bases; a fluoride ion is the Lewis base in this reaction, donating one of its lone pairs:

In the following reaction, each of two ammonia molecules, Lewis bases, donates a pair of electrons to a silver ion, the Lewis acid:

Nonmetal oxides act as Lewis acids and react with oxide ions, Lewis bases, to form oxyanions:

Many Lewis acid-base reactions are displacement reactions in which one Lewis base displaces another Lewis base from an acid-base adduct, or in which one Lewis acid displaces another Lewis acid:

The last displacement reaction shows how the reaction of a Brønsted-Lowry acid with a base fits into the Lewis concept. A Brønsted-Lowry acid such as HCl is an acid-base adduct according to the Lewis concept, and proton transfer occurs because a more stable acid-base adduct is formed. Thus, although the definitions of acids and bases in the two theories are quite different, the theories overlap considerably.

Many slightly soluble ionic solids dissolve when the concentration of the metal ion in solution is decreased through the formation of complex (polyatomic) ions in a Lewis acid-base reaction. For example, silver chloride dissolves in a solution of ammonia because the silver ion reacts with ammonia to form the complex ion Ag(NH3)2+. The Lewis structure of the Ag(NH3)2+ ion is:

The equations for the dissolution of AgCl in a solution of NH3 are:

Aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion Al(OH)4−. The Lewis structure of the Al(OH)4− ion is:

The equations for the dissolution are:

A complex ion consists of a central atom, typically a transition metal cation, surrounded by ions, or molecules called ligands. These ligands can be neutral molecules like H2O or NH3, or ions such as CN or OH. Often, the ligands act as Lewis bases, donating a pair of electrons to the central atom. The ligands aggregate themselves around the central atom, creating a new ion with a charge equal to the sum of the charges and, most often, a transitional metal ion. This more complex arrangement is why the resulting ion is called a complex ion. The complex ion formed in these reactions cannot be predicted; it must be determined experimentally. The types of bonds formed in complex ions are called coordinate covalent bonds, as electrons from the ligands are being shared with the central atom. Because of this, complex ions are sometimes referred to as coordination complexes. This will be studied further in upcoming chapters.

The equilibrium constant for the reaction of the components of a complex ion to form the complex ion in solution is called a formation constant (Kf) (sometimes called a stability constant). For example, the complex ion Cu(CN)2 is shown here:

It forms by the reaction:

At equilibrium:

The inverse of the formation constant is the dissociation constant (Kd), the equilibrium constant for the decomposition of a complex ion into its components in solution. We will work with dissociation constants further in the exercises for this section. Appendix K and Table 15.2 are tables of formation constants. In general, the larger the formation constant, the more stable the complex; however, as in the case of Ksp values, the stoichiometry of the compound must be considered.

As an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of Ag+ ([Ag+] = 1.3 × 10–5 M):

However, if NH3 is present in the water, the complex ion, Ag(NH3)2+, can form according to the equation:

with

The large size of this formation constant indicates that most of the free silver ions produced by the dissolution of AgCl combine with NH3 to form Ag(NH3)2+. As a consequence, the concentration of silver ions, [Ag+], is reduced, and the reaction quotient for the dissolution of silver chloride, [Ag+][Cl], falls below the solubility product of AgCl:

More silver chloride then dissolves. If the concentration of ammonia is great enough, all of the silver chloride dissolves.

### Example 15.13

Dissociation of a Complex Ion

Calculate the concentration of the silver ion in a solution that initially is 0.10 M with respect to Ag(NH3)2+.

Solution

We use the familiar path to solve this problem:

1. Determine the direction of change. The complex ion Ag(NH3)2+ is in equilibrium with its components, as represented by the equation:

We write the equilibrium as a formation reaction because Appendix K lists formation constants for complex ions. Before equilibrium, the reaction quotient is larger than the equilibrium constant [Kf = 1.6 × 107, and

it is infinitely large], so the reaction shifts to the left to reach equilibrium.

2. Determine x and equilibrium concentrations. We let the change in concentration of Ag+ be x. Dissociation of 1 mol of Ag(NH3)2+ gives 1 mol of Ag+ and 2 mol of NH3, so the change in [NH3] is 2x and that of Ag(NH3)2+ is –x. In summary:

3. Solve for x and the equilibrium concentrations. At equilibrium:

Both Q and Kf are much larger than 1, so let us assume that the changes in concentrations needed to reach equilibrium are small. Thus 0.10 – x is approximated as 0.10:

Because only 1.2% of the Ag(NH3)2+ dissociates into Ag+ and NH3, the assumption that x is small is justified.

Now we determine the equilibrium concentrations:

The concentration of free silver ion in the solution is 0.0012 M.

Check the work. The value of Q calculated using the equilibrium concentrations is equal to Kf within the error associated with the significant figures in the calculation.

Calculate the silver ion concentration, [Ag+], of a solution prepared by dissolving 1.00 g of AgNO3 and 10.0 g of KCN in sufficient water to make 1.00 L of solution. (Hint: Because Q < Kf, assume the reaction goes to completion then calculate the [Ag+] produced by dissociation of the complex.)

## Exercises

### Question 15.61

15.61

Under what circumstances, if any, does a sample of solid AgCl completely dissolve in pure water?

### Question 15.62

15.62

Explain why the addition of $NH_{3}$ or $HNO_{3}$ to a saturated solution of $Ag_{2}CO_{3}$ in contact with solid $Ag_{2}CO_{3}$ increases the solubility of the solid.

### Question 15.63

15.63

Calculate the cadmium ion concentration, $[Cd^{2+}]$, in a solution prepared by mixing 0.100 L of 0.0100 M Cd$(NO_{3})_{2}$ with 1.150 L of 0.100 $NH_{3}$ (aq).

### Question 15.64

15.64

Explain why addition of $NH_3$ or $HNO_3$ to a saturated solution of $Cu(OH)_{2}$ in contact with solid $Cu(OH)_2$ increases the solubility of the solid

### Question 15.65

15.65

Sometimes equilibria for complex ions are described in terms of dissociation constants, Kd. For the complex ion $AlF_{6}\:^{3−}$ the dissociation reaction is:

$AlF{6}\:^{3−} ⇌ Al^{3+}$ + $6F^{−}$ and $K_{d} = \frac{[Al^{3+}][F^{−}] ^{6}}{[AlF_{6}\:^{3−}]} = 2 × 10^{−24}$

### Question 15.66

15.66

Using the value of the formation constant for the complex ion $Co(NH_{3})_{6}\:^{2+}$ , calculate the dissociation constant

### Question 15.67

15.67

Using the dissociation constant, $Kd = 7.8 × 10^{–18},$ calculate the equilibrium concentrations of $Cd^{2+}$ and $CN^{–}$ in a 0.250-M solution of $Cd(CN)4_\: ^{2−}.$

### Question 15.68

15.68

Using the dissociation constant, $Kd = 3.4 × 10^{–15}$, calculate the equilibrium concentrations of $Zn^{2+}$ and $OH^–$ in a 0.0465-M solution of $Zn(OH)_{4}\: ^{2−}.$

### Question 15.69

15.69

Using the dissociation constant, $Kd = 2.2 × 10^{–34}$, calculate the equilibrium concentrations of $Co^{3+}$ and $NH_3$ in a 0.500-M solution of $Co(NH_3)_{6}\: ^{3+} .$

### Question 15.70

15.70

Using the dissociation constant, $K_{d} = 1 × 10^{–44},$ calculate the equilibrium concentrations of $Fe^{3+}$ and $CN^{–}$ in a 0.333 M solution of $Fe(CN)_6^\:{3−}$

### Question 15.71

15.71

Calculate the mass of potassium cyanide ion that must be added to 100 mL of solution to dissolve $2.0 × 10^{–2}$ mol of silver cyanide, AgCN.

### Question 15.72

15.72

Calculate the minimum concentration of ammonia needed in 1.0 L of solution to dissolve $3.0 × 10^{–3}$ mol of silver bromide.

### Question 15.73

15.73

A roll of 35-mm black and white photographic film contains about 0.27 g of unexposed AgBr before developing. What mass of $Na_{2}S_{2}O_{3}$$5H_{2}O$ (sodium thiosulfate pentahydrate or hypo) in 1.0 L of developer is required to dissolve the AgBr as $Ag(S_{2} O_{3})_{2} \:^{3−} (K_{f} = 4.7 × 10^{13})?$

### Question 15.74

15.74

We have seen an introductory definition of an acid: An acid is a compound that reacts with water and increases the amount of hydronium ion present. In the chapter on acids and bases, we saw two more definitions of acids: a compound that donates a proton (a hydrogen ion, $H^+$ ) to another compound is called a Brønsted-Lowry acid, and a Lewis acid is any species that can accept a pair of electrons. Explain why the introductory definition is a macroscopic definition, while the Brønsted-Lowry definition and the Lewis definition are microscopic definitions.

### Question 15.75

15.75

Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:

$(a) CO_{2} + OH^{−} \longrightarrow HCO_{3}^{–}$ $(b) B(OH)_{3} + OH^{−} \longrightarrow B(OH)_{4}^{–}$ $(c) I^{−} + I_{2} \longrightarrow I_{3}\:^{–}$ $(d) AlCl_{3} + Cl^{−} \longrightarrow AlCl_{4}^{−}$ (use Al-Cl single bonds) $(e) O ^{2−} + SO_{3} \longrightarrow SO_{4}\:^{2−}$

### Question 15.76

15.76

Write the Lewis structures of the reactants and product of each of the following equations, and identify the Lewis acid and the Lewis base in each:

$(a) CS_{2} + SH^{−} \longrightarrow HCS_{3}\:^{–}$ $(b) BF_{3} + F^{−} \longrightarrow BF_{4}^{–}$ $(c) I^{−} + SnI_{2} \longrightarrow SnI_{3} \:^{–}$ $(d) Al(OH)_{3} + OH^{−} \longrightarrow Al(OH)_{4}\:^{–}$ $(e) F^{−} + SO_{3} \longrightarrow SFO_{3}\:^{–}$

### Question 15.77

15.77

Using Lewis structures, write balanced equations for the following reactions:

$(a) CS_{2} + SH^{−} \longrightarrow HCS_{3}\:^{–}$ $(b) BF_{3} + F^{−} \longrightarrow BF_{4}\:^{–}$ $(c) I^{−} + SnI_{2} \longrightarrow SnI_{3}\:^{–}$ $(d) Al(OH)_{3} + OH^{−} \longrightarrow Al(OH)_{4}\:^{–}$ $(e) F^{−} + SO_{3} \longrightarrow SFO_{3}\:^{–}$

### Question 15.78

15.78

Calculate $[HgCl_{4}\:^{2−}]$ in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100-M $HgCl_{2}$ solution

### Question 15.79

15.79

In a titration of cyanide ion, 28.72 mL of 0.0100 M $AgNO_3$ is added before precipitation begins. [The reaction of $Ag^+$