chapter solubility and equilibria proofs

Think back to the last time you dissolved a solid in a liquid. One moment the solid has volume, shape and possibly colour; the next it is added to water and, with a little bit of stirring, it quickly seems to disappear.

Many different substances dissolve in water, so that water is often described as the universal solvent because of the wide range of substances it can dissolve. However, some ionic substances are sparingly soluble, and an equilibrium is established between the solid and the ions.

In this chapter, you will learn how to predict whether a compound is likely to be soluble, and relate the solubility of an ionic compound to its equilibrium constant.

ContentINQUIRY QUESTION

How does solubility relate to chemical equilibrium?By the end of this chapter, you will be able to:

• describe and analyse the processes involved in the dissolution of ionic compounds in water

• investigate the use of solubility equilibria by Aboriginal and Torres Strait Islander peoples when removing toxicity from foods, for example: AHC

- toxins in cycad fruit

• conduct an investigation to determine solubility rules, and predict and analyse the composition of substances when two ionic solutions are mixed, for example:

- potassium chloride and silver nitrate

- potassium iodide and lead nitrate

- sodium sulfate and barium nitrate (ACSCH065)

• derive equilibrium expressions for saturated solutions in terms of Ksp and calculate the solubility of an ionic substance from its Ksp value ICT N

• predict the formation of a precipitate given the standard reference values for Ksp

Chemistry Stage 6 Syllabus © NSW Education Standards Authority for and on behalf of the Crown in right of the State of NSW, 2017.

Solubility and equilibriaCHAPTER

Page

Proo

fs

MODULE 5 | EQUILIBRIUM AND ACID REACTIONS114

5.1 Process of dissolution of ionic compounds

CHEMISTRY INQUIRY CCT

Hard water: Do all dissolved ions affect soap?

COLLECT THIS …

• Epsom salts (magnesium sulfate)

• table salt

• three 600 mL water bottles

• dishwashing liquid

DO THIS …

1 Add 250 mL of water to each of the water bottles.

2 Add half a teaspoon of Epsom salts to one of the bottles. Seal the bottle and shake until the Epsom salts dissolve.

3 Add half a teaspoon of table salt to a different bottle. Seal the bottle and shake until the salt dissolves.

4 Add two drops of dishwashing liquid to each bottle.

5 Shake all the bottles.

RECORD THIS …

Record the height of the suds produced in each bottle.

Describe the effects of the different salts on the height of the suds.

REFLECT ON THIS …

1 Why was the height of the suds different?

2 Why did the salts dissolve?

3 Is there a difference in the amount of the salts that can be dissolved?

4 What effect would less soap suds have in households?

Water is an excellent solvent. This is one of its most important properties. Almost all biological processes and many industrial processes occur in water. These systems are known as aqueous environments.

When substances are dissolved in water, the particles are free to move throughout the solution. When two aqueous reactants are combined in a reaction vessel, the dissolved reactant particles mix freely. This increases the chances of the reactants coming into contact. Because of the increased movement of the reactant particles, interactions between reactants are generally much more effective than if the same reactants were mixed as solids.

In Figure 5.1.1 you can see how the deep purple colour of potassium permanganate spreads through the water as the solid dissolves. Eventually the liquid will appear completely purple as the particles continue to mix and move.

FIGURE 5.1.1 When solid potassium permanganate is added to water, it dissolves. The particles disperse into the solution and move around freely.

Page

Proo

fs

115CHAPTER 5 | SOLUBILITY AND EQUILIBRIA

Although many ionic substances are readily soluble in water, some are not. This section looks at what happens when ionic substances dissolve in water, and reviews the solubility of ions in water.

REVISION

Dissolution of ionic substances in water Year 11 Chapter 14GO TO ➤

Water molecules are polar: the oxygen atom has a partial negative charge and the hydrogen atoms have a partial positive charge. When an ionic substance is added to water, the hydrogen atoms of the water molecules are attracted to the negatively charged ions and the oxygen atoms of the water molecules are attracted to the positively charged ions. The attraction between the water molecules and the ions is known as ion–dipole attraction. When these ion–dipole attractions are strong enough, the ions are removed from the ionic lattice and enter the solution. When an ion is surrounded by water molecules, it is said to be hydrated.

Figure 5.1.2 shows the hydration of sodium and chloride ions. Note the orientation of the water molecules: the hydrogen atoms of the water molecules (white) are oriented towards the negatively charged chloride ions (green), whereas the oxygen atoms of the water molecules (red) are oriented towards the sodium ions (yellow).

ion-dipole attraction

FIGURE 5.1.2 Ion–dipole attraction between the ions and adjacent water molecules to form hydrated sodium and chloride ions

DISSOLUTION OF AN IONIC COMPOUND IN WATERSports drinks are used by athletes to replace the electrolytes lost in sweat (Figure 5.1.3). The electrolytes are dissolved ionic substances such as sodium chloride and potassium phosphate.

Sodium chloride is a typical ionic compound that exists as a solid at room temperature. Figure 5.1.4 shows the arrangement of sodium cations (Na+) and chloride anions (Cl−) in a three-dimensional ionic lattice. The ions are held together by strong electrostatic forces between the positive and negative charges of the ions.

chloride ionCl–

sodium ionNa+

FIGURE 5.1.4 A representation of the crystal lattice of the ionic compound sodium chloride.

FIGURE 5.1.3 Sports drinks are used by athletes to replace water and dissolved ionic solutes.

Page

Proo

fs


The process of separating positive and negative ions from a solid ionic compound to form hydrated ions when an ionic compound dissolves in water is called dissociation.

Although the ionic bonds within the lattice are strong, the ions can be pulled away from the lattice by the interactions of many water molecules.

In summary, when sodium chloride dissolves in water:• ionic bonds within the sodium chloride lattice are broken• hydrogen bonds between water molecules are broken• ion–dipole attractions form between ions and polar water molecules.

An equation can be written to represent the dissociation process:

NaCl(s) Na (aq) Cl (aq)H O(l)2 → ++ −

Note that the formula of water sits above the arrow. This is because there is no direct reaction between the water and the sodium chloride. No chemical change occurs; only the state symbol for sodium chloride is altered from solid (s) to aqueous (aq), indicating it is now dissolved in water. (You may omit the H2O from this equation if you wish.)

It is important to note that the dissociation of ionic compounds is simply freeing ions from the lattice so that they move freely throughout the solution. This is different from the ionisation of molecular compounds such as gaseous hydrogen chloride, where new ions are formed by the reaction of the molecule with water.

Ionic substances dissolve by dissociation. Ion–dipole attractions are formed between the ions and water molecules.

CHEMFILE PSC

Blood plasmaThe human body is 66% water by weight. In biological organisms, water transports nutrients and soluble wastes, as well as essential reagents for chemical reactions to cells and tissues. The circulatory system of the body forms a transportation network (Figure 5.1.5). Essential nutrients, ions and molecules dissolved in water in the blood are carried to the organs, brain and tissues, and waste is carried away. This part of the blood is a component called plasma.

Plasma is a yellow substance that makes up over half of the blood volume in the body. Sodium chloride is present in blood plasma. Sodium and chloride ions play a vital role in many bodily functions, including regulating blood pressure and maintaining cell membrane potentials. Although plasma is more than 90% water, it can be used in the treatment of a range of conditions, including burns, haemophilia and immune disorders. Its medical versatility is due to the small amount of dissolved proteins that it contains, including coagulation factors, albumin and immunoglobulins. These proteins can be isolated from donated plasma and used in medical treatment. For instance, coagulation factors can be given to those who suffer from blood clotting conditions such as haemophilia.

Although blood plasma can be stored frozen for a year, it may take over one hundred donations to create one bag of plasma. It is for this reason that blood plasma donations are always needed.

FIGURE 5.1.5 Plasma transports many substances, including dissolved salts, proteins, glucose and lipids around the body via the circulatory system.

Page

Proo

fs


INSOLUBLE IONIC COMPOUNDSThe solubility of a substance is a measure of how much of the substance will dissolve in a given amount of solvent. For example, sodium chloride is very soluble in water. A saline drip containing sodium chloride is often used during medical procedures to replenish body fluids in patients (Figure 5.1.7). However, many ionic substances are far less soluble.

FIGURE 5.1.7 A saline drip is a solution of sodium chloride.

CHEMISTRY IN ACTION CCT IU

Prussian blue and radiation poisoningPrussia was a European state covering parts of modern Germany and Poland. In the early 18th century the first artificial dye was synthesised in Berlin, the Prussian capital. The deep blue pigment, chemically known as ferric ferrocyanide, with the formula Fe4[Fe(CN)6]3, was used to colour the jackets of the Prussian army (Figure 5.1.6). For this reason the pigment has long been known as Prussian blue. Variations of Prussian blue are variously known as Turnbull’s blue, bronze blue and Milori blue. Cyanide was first isolated from this pigment; its name comes from kuanos, the Greek word for dark blue.

Prussian blue also has a more surprising use: treating people with radiation sickness. In 1969 German pharmacologist Horst Heydlauf published an article in the European Journal of Pharmacology entitled ‘Ferric-cyanoferrate (II): An effective antidote in thallium poisoning’. Heydlauf’s breakthrough meant that thallium poisoning, once considered invariably fatal, could be treated. In Australia thallium was once sold as a rat poison called Thall-rat. Despite a string of thallium poisonings, Thall-rat was not banned until 1954, following several highly publicised cases of murder and attempted murder in NSW.

Prussian blue can also be used to treat caesium poisoning. In 1987 a cancer treatment device containing radioactive caesium was stolen from an abandoned hospital in Brazil and dismantled in the local community. As a result, 250 people were exposed to significant amounts of radiation. Prussian blue was used to treat these people, and only four died from the effects of radiation exposure.

FIGURE 5.1.6 Prussian field marshal August Neidhart von Gneisenau (1760–1831). The deep blue of his jacket comes from Prussian blue dye.

Page

Proo

fs


Insoluble ionic compounds do not dissolve in water because the energy required to separate the ions from the lattice is greater than the energy released when the ions are hydrated.

Although substances are usually described as either soluble or insoluble, this is a generalisation. Substances that are described as insoluble tend to dissolve very slightly.

Definition of solubilityIn chemistry, the term solubility has a specific meaning. It refers to the maximum amount of a solute (substance that is dissolved) that can be dissolved in a given quantity of a solvent (the common quantity often used is 100 g) at a certain temperature. Table 5.1.1 gives the solubility of some substances in 1 L of water at 25°C.

TABLE 5.1.1 Solubility of selected solutes at 25°C

Solute Solubility in water (mol L−1)

sodium chloride 6.3

calcium hydroxide 0.016

calcium carbonate 6.0 × 10−6

There are a number of ways that chemists describe the solubility of ionic substances. A simple version is as follows:• When a substance is soluble, more than 0.1 mol of the substance will dissolve

in 1 L of water.• When a substance is insoluble, less than 0.01 mol of the substance will dissolve

in 1 L of water. • When a substance is slightly soluble, 0.01–0.1 mol of the substance will dissolve

in 1 L of water. Using this scale, sodium chloride is soluble, calcium carbonate is insoluble and

calcium hydroxide is slightly soluble.

Solubility rules Sometimes when two ionic solutions are mixed the ions interact to form a new substance with a lower solubility. This substance may be an insoluble solid, called a precipitate. To determine whether two ionic solutions will form a precipitate, you can either conduct an experiment or use the solubility rules.

The solubility rules tell you which substances are soluble, and to what extent.

A substance is rarely ever completely soluble or insoluble. Its solubility can be considered to be one point on a solubility scale.

SKILLBUILDER WE

Mixing solutions to form a precipitateWhile this chapter explains how to use solubility rules to determine whether a precipitate is formed, you may want to confirm your working experimentally.

When mixing two solutions to form a precipitate, you need to remember the following points.

• Wear safety goggles. Your teacher or lab technician will tell you if you also need to wear gloves.

• Place the beaker you will be pouring the solutions into on a flat surface.

• Stand when pouring the solution.

• Slowly trickle the second solution down the inside of the beaker to form the best precipitate. If you add the solution quickly or do not trickle it down the inside of the beaker, the precipitate will trap bubbles and will not be as clear.

• Bring your eyes down to the beaker to observe the precipitate; never hold the beaker up to eye level.

• Ask your teacher or lab technician how to dispose of the remaining solutions and precipitate properly.

Page

Proo

fs


REVISION

In Year 11 you learned the SNAAP rule that helps to remember which anions and cations never form precipitates:

• Sodium (Na+)

• Nitrate (NO3−)

• Ammonium (NH4+)

• Acetate (CH3COO−)

• Potassium (K+)

When you write ionic equations, these are regarded as spectator ions. For other ionic compounds you can refer to a solubility table. Tables 5.1.2 and 5.1.3 identify some combinations of ions will form a precipitate, and to what extent.

TABLE 5.1.2 Relative solubilities of soluble ionic compounds

Soluble in water (>0.1 mol L−1 at 25°C)

Exceptions: insoluble (<0.1 mol L−1 at 25°C)

Exceptions: slightly soluble (0.01–0.1 mol L−1 at 25°C)

most chlorides (Cl−), bromides (Br−) and iodides (I−)

AgCl, AgBr, AgI, PbI2 PbCl2, PbBr2

all nitrates (NO3−) no exceptions no exceptions

all ammonium (NH4+) salts no exceptions no exceptions

all sodium (Na+) and potassium (K−) salts

no exceptions no exceptions

all acetates (CH3COO−) no exceptions no exceptions

most sulfates (SO42−) SrSO4, BaSO4, PbSO4 CaSO4, Ag2SO4

TABLE 5.1.3 Relative solubilities of insoluble ionic compounds

Insoluble in water (<0.1 mol dissolves in 1 L at 25°C)

Exceptions: soluble (>0.1 mol dissolves in 1 L at 25°C)*

Exceptions: slightly soluble (0.01−0.1 mol dissolves in 1 L at 25°C)

most hydroxides (OH−) NaOH, KOH, Ba(OH)2, NH4OH, AgOH

Ca(OH)2, Sr(OH)2

most carbonates (CO32−) Na2CO3, K2CO3, (NH4)2CO3 no exceptions

most phosphates (PO4

3−)Na3PO4, K3PO4, (NH4)3PO4 no exceptions

most sulfides (S2−) Na2S, K2S, (NH4)2S no exceptions

* NH4OH does not exist in significant amounts in an ammonia solution. Ammonium and hydroxide ions readily combine to form ammonia and water. AgOH readily decomposes to form a precipitate of silver oxide and water.

Year 11 Chapter 10GO TO ➤

Page

Proo

fs


Worked example 5.1.1

DETERMINING WHETHER A PRECIPITATE WILL FORM FROM TWO IONIC SOLUTIONS

Determine whether a precipitate will be formed when a solution of potassium chloride and a solution of silver nitrate are mixed. If so, identify the precipitate.

Thinking Working

Identify the ions that are present in each ionic solution. Potassium (K+), chloride (Cl−), silver (Ag+) and nitrate (NO3−)

Identify any spectator ions. Potassium (K+) and nitrate (NO3−) never form precipitates.

Identify the remaining ions. Chloride (Cl−) and silver (Ag+)

Use the solubility tables to determine if this combination of ions is insoluble.

Chloride (Cl−) and silver (Ag+) form an insoluble compound, so a precipitate of silver chloride will form.

Worked example: Try yourself 5.1.1

DETERMINING WHETHER A PRECIPITATE WILL FORM FROM TWO IONIC SOLUTIONS

Determine whether a precipitate will be formed when a solution of sodium hydroxide and a solution of barium acetate are mixed. If so, identify the precipitate.

5.1 Review

SUMMARY

• Many ionic substances are readily soluble in water.

• When ionic substances dissolve, ion–dipole attractions between the ions and water molecules cause the ionic lattice to dissociate. The ions are said to be hydrated.

• Ionic substances can be classified as soluble, slightly soluble or insoluble.

• The solubility rules are a guide to the solubility of substances.

KEY QUESTIONS

1 Sodium nitrate (NaNO3) and calcium hydroxide (Ca(OH)2) will dissociate when they dissolve in water. Write chemical equations to represent the dissolving process for each of these compounds.

2 Explain the process of the dissolution of potassium bromide.

3 Write the formulae for the ions produced when these compounds dissolve in water:a sodium carbonateb calcium nitratec potassium bromided iron(III) sulfatee copper(II) chloride

4 Which of the following substances would you expect to be soluble in water?A sodium sulfateB lead(II) carbonate

C magnesium hydroxideD silver nitrateE iron(III) chlorideF barium phosphateG ammonium chlorideH potassium carbonateI silver sulfide J calcium acetate

5 Predict whether a precipitate will form when the following solutions are mixed. If a precipitate forms, identify it:a silver nitrate and iron(III) chlorideb potassium hydroxide and aluminium sulfatec copper(II) bromide and ammonium phosphate d sodium sulfate and zinc acetatee lead(II) nitrate and iron(II) fluoridef magnesium iodide and calcium chloride.

Worked example 5.1.1 shows you how to use the solubility rules to determine if two ionic solutions will form a precipitate.

Page

Proo

fs


5.2 Solubility of ionic compounds and equilibriumIonic substances classified as insoluble may still dissolve to some extent. For example, limestone (CaCO3) is classified as insoluble in water. Limestone caves, such as those seen in Figure 5.2.1, are formed over a long period of time as CaCO3 is dissolved and redeposited.

FIGURE 5.2.1 Stalactites and stalagmites form when limestone (calcium carbonate) dissolves over very long periods of time.

The solubility of ionic compounds in water depends on the difference in the energy required to separate the ions from the lattice, and the energy released when the ions are hydrated. This section investigates the relationship between solubility and chemical equilibrium.

SATURATED SOLUTIONSIn the previous section you saw that sodium chloride dissociates in water according to the following equation:


Sodium chloride is highly soluble in water, so if a small amount of solid is added to a large volume of water the dissociation can be considered to go to completion. Solutions of sodium chloride are usually unsaturated solutions: solutions that contain less than the maximum amount of solute that can be dissolved at a particular temperature. In fact, at 25°C up to 36 g of sodium chloride can be added to 100 g of water before the sodium chloride stops dissolving. At this point the solution would be a saturated solution: a solution that contains the maximum amount of solute that can be dissolved at a particular temperature.

Some ionic compounds are classified as insoluble. For example, silver chloride is considered to be insoluble in water. However, this is not strictly true: 0.002 g of silver chloride will dissolve in 100 g of water at 25°C. So it is more accurate to call insoluble compounds sparingly soluble. Because only very small amounts of sparingly soluble salts dissolve, they readily form saturated solutions. For example, any amount greater than 0.002 g of silver chloride in 100 mL of water will form a saturated solution.

Unsaturated solutions are able to dissolve more solute. Saturated solutions are unable to dissolve any more solute at a particular temperature.

Sparingly soluble ionic compounds are those that only dissolve to a very small extent. Only very small amounts of these salts are needed to form a saturated solution.

Page

Proo

fs


If 1 g of silver chloride crystals is added to 100 mL of water, 0.002 g will dissolve and the remaining 0.998 g will remain as a solid. But if you could observe this on the atomic level, you would see that the shape of the ionic crystal lattice is constantly changing. Silver and chloride ions are continually breaking away from the lattice and returning to it. Because ions are leaving the lattice at the same rate that others are returning to it, the system is at equilibrium (Figure 5.2.2). This can be represented by an ionic equation:

�AgCl(s) Ag (aq) Cl (aq)++ −

Note that an equilibrium arrow is used. When a soluble substance dissolves it is assumed that the solid dissolves completely, so only the forward reaction occurs. But for sparingly soluble salts an equilibrium arrow shows that the rate of dissolution is the same as the rate of precipitation. The position of equilibrium lies very far to the left when sparingly soluble salts dissolve; although the rate of dissolution is the same as the rate of precipitation, most of the ions remain in the ionic lattice.

SOLUBILITY PRODUCTIn Chapter 3 you learned to use the equilibrium constant Keq. For the reaction:

aW + bX cY + dZThe equilibrium expression is:

Keq = [Y] [Z][W] [X]

c d

a b

Equilibrium expressions for saturated solutionsThe equation for the dissociation of silver chloride in water is:

� + −AgCl(s) Ag (aq) + Cl (aq)

The equilibrium expression for this reaction is therefore:

Keq = + −[Ag ][Cl ]

[AgCl]

The reaction is heterogeneous: the silver ions and the chloride ions are aqueous and the silver chloride is solid. You will recall from Chapter 3 that, for heterogenous equilibria, the concentration of a pure liquid or solid is assigned the value of 1. Therefore the silver chloride can be removed from the equilibrium expression.

This gives the solubility product, Ksp: Ksp = [Ag+][Cl−]

The product of the concentrations of the individual ions of a sparingly soluble salt present in a particular solution, whether it is saturated or not, is called the ion product. For example, the ion product of a silver chloride solution is represented by [Ag+][Cl−]. If the solution is saturated, the ion product is equal to Ksp.

If a solution is unsaturated, all of the solid will dissolve and no equilibrium will be established between the solid and the aqueous ions.


WRITING EQUILIBRIUM EXPRESSIONS IN TERMS OF Ksp

Write the equilibrium expression for a saturated solution of copper(II) hydroxide.

Thinking Working

Check the solubility tables to determine if this combination of ions is insoluble.

Copper (Cu2+) and hydroxide (OH−) are insoluble.

Write a balanced ionic equation. Cu(OH)2(s) � Cu2+(aq) + 2OH−(aq)

Write the expression for Ksp. Ksp = [Cu2+][OH−]2

Section 3.1 page 000GO TO ➤

In writing the equilibrium expression for a sparingly soluble salt, only the aqueous ions are included.

The solubility product is the product of the concentration of ions in a saturated solution of a sparingly soluble salt.

Ag+

Ag+ Ag+

Ag+

AgCl(s)

⇋

Cl– Cl–

Cl–Cl–

Ag+Cl–

FIGURE 5.2.2 In a saturated solution of AgCl, the Ag+ and Cl− ions are breaking away from the ionic lattice at the same rate as they are returning to it. The system is in an equilibrium state.

The ion product equals Ksp only when a solution is saturated.Pa

ge Pr

oofs



WRITING EQUILIBRIUM EXPRESSIONS IN TERMS OF Ksp

Write the equilibrium expression for a saturated solution of silver carbonate.

Calculating the solubility of an ionic compoundBecause the value of an equilibrium constant depends on temperature, Ksp values only apply to specific temperatures. Table 5.2.1 lists the values for a range of ionic compounds at 25°C.

TABLE 5.2.1 Ksp values for some common ionic substances at 25°C

Ionic substance Ksp (at 25°C)

AgCl 1.77 × 10−13

AgBr 5.35 × 10−13

BaSO4 1.08 × 10−10

CaCO3 3.36 × 10−9

PbCl2 1.70 × 10−5

Fe(OH)2 4.87 × 10−17

Ca3(PO4)2 2.07 × 10−29

Fe(OH)3 2.79 × 10−39

CHEMFILE S

Solubility and coral formationAlthough corals look like plants, they are actually invertebrate animals closely related to sea anemones. Corals take in carbonate and calcium ions from seawater, which they convert into calcium carbonate, depositing the precipitate as their exoskeleton. Coral reefs are formed when many thousands of corals are linked by calcium carbonate secreted by algae that live among them (Figure 5.2.3).

Increasing atmospheric CO2 has caused the ocean to become more acidic. When pH decreases, it becomes harder for corals to get the calcium carbonate that they need to maintain their skeletons. If the acidity of the oceans reaches the level that some have forecasted, the calcium carbonate that holds reefs together could dissolve.

FIGURE 5.2.3 A clownfish among coral on the Great Barrier Reef. The reef is under threat from increasing acidity in the ocean.

Page

Proo

fs


CHEMISTRY IN ACTION CCT

Precipitation and the first photographsEnglishman William Henry Fox Talbot had many strings to his bow (Figure 5.2.4). He was a Member of Parliament, a mathematician and an astronomer. He also translated cuneiform tablets from the ancient city-state of Nineveh. However, it was his lack of talent in sketching that led him to invent the process that formed the foundation of modern photography.

While staying at Lake Como in Italy in 1833, Talbot was frustrated by his inability to reproduce sketches from nature faithfully and began to experiment with methods to assist him. By soaking a sheet in sodium chloride then brushing it with silver nitrate, a precipitate of silver chloride was formed on the surface. The sheet was then placed in a simple camera, called a camera obscura. Where light hit the sheet, the silver chloride would decompose, leaving dark silver metal behind. He further refined his method by ‘fixing’ the sheet, removing the silver. This reduced exposure times dramatically (Figure 5.2.5).

FIGURE 5.2.4 A photograph of Talbot taken in the1840s.

The daguerreotype was the other main form of early photography; however, it could only produce a single print. Talbot’s method produced a negative, which could be used to generate many copies of an image. Although many modifications have since been made to Talbot’s method, the fundamental aspects of it are still used in photography today.

FIGURE 5.2.5 The insolubility of silver salts made early photography possible. Talbot took this image of his third daughter, Matilda, in the 1840s.

MOLAR SOLUBILITYThe molar concentration of a salt in a saturated solution is called its molar solubility. If the molar solubility of a salt is known, it can be used to calculate Ksp.

The molar solubility of a salt is the number of moles that will dissolve in water to form 1 L of saturated solution.

The molar solubility of a sparingly soluble salt can also be regarded to be the number of moles that will dissolve per litre of water to form a saturated solution. Since the salt is sparingly soluble, there is negligible effect on the volume of water, so the volume of solution is essentially the same.

Page

Proo

fs


An ICE table (also called a reaction table) can be used to calculate the equilibrium concentration of ions and thus Ksp. You used ICE tables for homogeneous equilbria questions in Chapter 3, but ICE tables for the dissolution of sparingly soluble salts are slightly different: there are no concentrations entered for the solid, and the change in concentration for the aqueous ions is always positive.


CALCULATING Ksp FROM SOLUBILITY DATA

Silver chloride is a sparingly soluble salt. At a certain temperature, 1 L of water can dissolve 1.3 × 10−5 mol of AgCl. Calculate Ksp for silver chloride at this temperature.

Thinking Working

Construct an ICE table using each species in the balanced equation as the headings for the columns in the table. Insert three rows in the table labelled I (initial), C (change) and E (equilibrium):

Reactants Products

I

C

E

Initially, there is no Ag+ or Cl− present, so the initial amounts of both are zero.

When the maximum amount of silver chloride that can dissolve in 1 L of water (1.3 × 10−5 mol) is added, 1.3 × 10−5 mol of Ag+ ions and 1.3 × 10−5 mol of Cl− ions enter the solution.

Therefore, the amount of Ag+ and Cl− ions at equilibrium is 1.3 × 10−5 mol for both.

AgCl(s) Ag +(aq) + Cl−(aq)

I 0 0

C +1.3 × 10−5 mol +1.3 × 10−5 mol

E 1.3 × 10−5 mol 1.3 × 10−5 mol

Write the expression for Ksp and substitute the equilibrium concentrations.

Calculate the equilibrium constant, Ksp.

Since there is 1 L of solution:

[Ag+] = 1.3 × 10−5 mol L−1

[Cl−] = 1.3 × 10−5 mol L−1

Ksp = [Ag+][Cl−]

= (1.3 × 10−5) (1.3 × 10−5)

= 1.7 × 10−10


CALCULATING Ksp FROM SOLUBILITY DATA

Magnesium hydroxide is a white solid that is commonly used in antacids. At 25°C, 1 L of water can dissolve 1.12 × 10−4 mol of Mg(OH)2. Calculate Ksp for magnesium hydroxide at this temperature.

When comparing the relative solubility of salts, Ksp cannot always be used, because different salts may produce different numbers of ions in solution. Ksp can only be used to compare the solubility of salts that produce the same number of ions.

Silver chloride (AgCl) and calcium sulfate (CaSO4) both produce two ions when they dissociate, so you can compare their relative solubility by referring to their Ksp values. The salt with the smaller value of Ksp is the less soluble. The Ksp of silver chloride is 1.77 × 10−10, smaller than the Ksp of 4.93 × 10−5 of calcium sulfate hence silver chloride is less soluble that calcium sulfate.

Section 3.3 page 000GO TO ➤

Page

Proo

fs


However, to compare the relative solubility of salts that produce different numbers of ions in solution, the molar solubility needs to be calculated for each. For example, calcium sulfate (CaSO4) produces two ions when it dissociates, and lead(II) chloride (PbCl2) produces three ions when it dissociates. These salts have similar Ksp values but different molar solubilities, as shown in Table 5.2.2.

TABLE 5.2.2 Ksp (at 25°C) and molar solubility of calcium sulfate and lead(II) chloride

Ion Ksp (at 25°C) Molar solubility (mol L−1)

calcium sulfate 4.93 × 10−5 7.0 × 10−3

lead(II) chloride 1.70 × 10−5 1.62 × 10−2


CALCULATING MOLAR SOLUBILITY FROM Ksp

Lead(II) chloride has a Ksp value of 1.70 × 10−5 at 25°C. Calculate the molar solubility of lead(II) chloride at this temperature.

Thinking Working

Construct a reaction table using each species in the balanced equation as the headings for the columns in the table. Insert three rows in the table labelled I (initial), C (change) and E (equilibrium):

Reactants Products

I

C

E

Initially, there is no Pb2+ or Cl− present. When the lead(II) chloride is added to water, lead(II) ions and chloride ions enter the water in a 1:2 ratio. This can be represented by s.

PbCl2(s) Pb2+(aq) + 2Cl−(aq)

I 0 0

C +s +2s

E s 2s

Write the expression for Ksp and substitute s for the equilibrium concentrations.

Calculate the molar solubility, including the correct units.

Ksp = [Pb2+][Cl−]2

s s1.70 10 ( )(2 )5 2× =−

s1.70 10 45 3× =−

× =− s4.25 10 6 3

Take the cube root of both sides:

s 1.62 10 2= × −

The molar solubility of lead(II) chloride is 1.62 × 10−2 mol L−1


CALCULATING MOLAR SOLUBILITY FROM Ksp

Zinc hydroxide has a Ksp value of 3.8 × 10−17 at 25°C. Calculate the molar solubility of zinc hydroxide at this temperature.

Predicting the formation of a precipitateEarlier in this section you saw that the ion product is the product of the concentration of ions of a sparingly soluble salt present in a solution. If the ion product is equal to or greater than Ksp the solution is saturated, and a precipitate will form. If the ion product is less than Ksp the solution is unsaturated, and no precipitate will form. This is summarised in Table 5.2.3.

Page

Proo

fs


TABLE 5.2.3 Predicting the formation of a precipitate by comparing ion product and Ksp

Value of ion product compared with Ksp

Saturated or unsaturated? Precipitate formed?

ion product < Ksp unsaturated no

ion product > Ksp saturated yes

Therefore, by calculating the ion product and comparing this value to Ksp, you can predict whether or not a precipitate will form.


PREDICTING THE FORMATION OF A PRECIPITATE GIVEN Ksp

Barium sulfate has a Ksp value of 1.08 × 10−10 at 25°C. If 1.0 × 10−3 g of barium sulfate is added to 1 L of water at 25°C, predict whether a precipitate will form.

Thinking Working

Calculate the concentrations for the ions at equilibrium.

Use the formulae:

=n mM

and =c nV

There is 1.0 × 10−3 g of barium sulfate.(BaSO )

4.3 10 mol

nmM4

1.0 10233.43

6

3

=

== ×

×

−

−

c

=

4.3 10 molL

nV4.3 10

16 1

6

=

= ×

×

− −

−

Calculate the ion product. Assume that each mole of barium sulfate (BaSO4) dissociates to form 1 mol of barium ions and 1 mol of sulfate ions.

Therefore, when 4.3 × 10−3 mol of barium sulfate is added to 1 L of water, 4.3 × 10−3 mol of barium ions and 4.3 × 10−3 mol of sulfate ions are formed.

=

= × ×

×=

+ −

− −

−

ion product Ba SO

(4.3 10 )(4.3 10 )

1.8 10

24

2

6 6

11

Compare the ion product to the value of Ksp.

The ion product is 1.8 × 10−11. This value is smaller than the Ksp value of 1.08 × 10−11 at 25°C.

No precipitate will form.


PREDICTING THE FORMATION OF A PRECIPITATE GIVEN Ksp

Calcium fluoride has a Ksp value of 3.2 × 10−11 at 25°C. If 1.0 × 10−3 g of calcium fluoride is added to 1 L of water at 25°C, predict whether a precipitate will form.

Sometimes chemists have to prepare a solution that contains more than one salt. In this case they can calculate whether any combination of ions is insoluble at a given concentration. If one combination of ions is insoluble, they can predict whether a precipitate will form by comparing the Ksp value of the salt to the ion product.

Page

Proo

fs


+ ADDITIONAL

Supersaturated solutionsIt is possible for a solution to have an ion product that is greater than Ksp. A supersaturated solution is an unstable solution that contains more dissolved solute than a saturated solution. If this type of solution is disturbed, some of the solute will separate from the solvent as a solid.

Figure 5.2.6 shows a supersaturated solution of sodium acetate. Supersaturated sodium acetate is prepared by cooling a saturated solution very carefully so that solid crystals do not form. Adding a small seed crystal to the supersaturated solution causes the solute to crystallise (form solid crystals) so that a saturated solution remains.

FIGURE 5.2.6 Crystals of sodium acetate form after a seed crystal is added to a supersaturated solution of the compound


PREDICTING THE FORMATION OF A PRECIPITATE WHEN SOLUTIONS ARE MIXED

20 mL of 0.10 mol L−1 silver acetate and 20 mL of 0.050 mol L−1 sodium chloride are mixed together. Predict whether a precipitate will form and, if so, identify the precipitate.

Thinking Working

Identify the ions that are present in each solution, and determine whether any combination of these ions is insoluble.

Silver (Ag+), acetate (CH3COO−), sodium (Na+) and chloride (Cl−) ions are present. Silver and chloride ions combine to form an insoluble compound, so they could form a precipitate.

Identify the value of Ksp for the compound that could form a precipitate.

The value of Ksp for silver chloride is 1.77 × 10−10.

Calculate the concentration of the ions that would form the precipitate.

The concentrations of the solutions and the final volume of the mixture are used to determine the concentration of these ions in the mixture.

Use the formula:

=c V c V1 1 2 2

[Ag ] c VV1 1

2

+ =

×= 0.10 0.020

0.040

= 0.050 mol L−1

[Cl ]=c VV1 1

2

−

= 0.050 0.020

0.040×

= 0.025 mol L−1

Calculate the ion product for this salt. + −ion product = [Ag ][Cl ]

= (0.050)(0.025) = 1.3 × 10−3

Compare this to the value of Ksp for the salt.

This is greater than the value of Ksp for silver chloride. A precipitate will form.


PREDICTING THE FORMATION OF A PRECIPITATE WHEN SOLUTIONS ARE MIXED

5.0 mL of 0.010 mol L−1 barium chloride and 20 mL of 0.010 mol L−1 sodium sulfate are mixed together. Predict whether a precipitate will form, and, if so, identify the precipitate.

The common ion effectIn this chemistry course you have examined only the equilibria of salts dissolved in pure water. However, it is possible to dissolve salts into solutions that already contain dissolved ions. If the salt added and the existing solution have an ion in common, it is called a common ion. The common ion effect greatly decreases the solubility of ionic compounds, meaning that they are far less soluble in solutions with a common ion than they are in pure water. Consider the dissolution of AgCl:

�AgCl(s) Ag (aq) + Cl (aq)+ −

Page

Proo

fs


If solid AgCl is added to a solution containing NaCl, then Le Châtelier’s principle indicates that the position of equilibrium will shift to the left, because of the chloride ions that are already present from the dissolved NaCl. Therefore the solubility of silver chloride is reduced by the presence of the common ion.

If a sparingly soluble salt is added to a solution with a common ion, then it can be assumed that the concentration of the common ion comes entirely from the ions that are already present in the solution, and none from the sparingly soluble salt. This can greatly simplify calculations that involve a common ion.


CALCULATING THE EFFECT OF A COMMON ION ON THE MOLAR SOLUBILITY OF AN IONIC COMPOUND

Lead(II) chloride has a Ksp value of 1.70 × 10−5 at 25°C. Calculate its molar solubility in a 0.20 mol L−1 solution of NaCl.

Thinking Working

Construct a reaction table using each species in the balanced equation as the headings for the columns in the table. Insert three rows in the table labelled I (initial), C (change) and E (equilibrium):

Reactants � Products

I

C

E

Initially, there is no Pb2+ present and 0.20 mol L−1 of Cl−. When the lead(II) chloride is added to water, lead(II) ions and chloride ions enter the water in a 1:2 ratio. This can be represented by s. For 1 L of solution:

++ −PbCl (s) Pb (aq) 2Cl (aq)22�

I 0 0.20

C +s +2s

E s 0.20 + 2s

Assume that the solute that is being added makes a small contribution of the common ion in the solution. Assume that all of the common ion comes from the solute already in solution.

Since the amount of chloride ions contributed from dissolving lead(II) chloride is so small, it can be assumed that all of the chloride ions come from the NaCl solution, hence this value is used for the equilibrium concentration of Cl−.

+ −PbCl (s) Pb (aq) + 2Cl (aq)22�

I 0 0.20

C +s 0

E s 0.20

Write the expression for Ksp and substitute s for the equilibrium concentrations.

Calculate the molar solubility, including the correct units.

= + −K [Pb ][Cl ]sp

2 2

s1.70 10 ( ) (0.20)5 2× =−

s 1.70 10

(0.20)

5

2= × −

= 4.3 × 10−4 mol L−1

The molar solubility of lead(II) chloride in 0.20 mol L−1 NaCl is 4.3 × 10−4 mol L−1.

Note that this is lower than the molar solubility of lead(II) chloride in water calculated in Worked example 5.2.3 (1.62 × 10−2 mol L−1).


CALCULATING THE EFFECT OF A COMMON ION ON THE MOLAR SOLUBILITY OF AN IONIC COMPOUND

Silver bromide has a Ksp value of 5.35 × 10−13 at 25°C. Calculate its solubility in a 0.50 mol L−1 solution of NaBr.

The common ion effect greatly reduces the solubility of salts.

Page

Proo

fs


DETOXIFYING FOODIn many parts of the world there are nutrient-rich plants that contain toxins in their seeds or fruits, and Australia is no different. However, the ingenuity of Aboriginal and Torres Strait Islander people has enabled them to detoxify these foods and render them edible.

CycadsIn Year 11 chemistry you were introduced to the techniques used by Indigenous Australians to detoxify the seeds of the cycad, which contain the toxin cycasin. The seeds are crushed to expose the inner kernel and increase the surface area, and then either soaked in water to leach out the water-soluble toxins, or fermented until they are no longer toxic.

Aboriginal people in New South Wales used two leaching methods to detoxify cycads. Briefly roasting the seeds and then leaching them in water for a short period of time yielded cycad seeds that, although free of poison, did not keep well and had to be eaten quickly. Leaching the fruit for longer periods meant that the seeds did not need prior roasting and lasted longer. Another technique is ageing, where seeds that have lain underneath the trees for some months are collected, or the seeds are dried and stored or buried for several months.

Bitter yamThe Tiwi islands, off the coast of the Northern Territory, are thought to have been inhabited for at least 15 000 years. The Tiwi people, in their relative isolation, have developed a unique culture. One of their principal ceremonies is the kulama, or yam ceremony. The bitter yam (Dioscorea bulbifera), shown in Figure 5.2.7, contains oxalates that would be toxic if consumed. In the Tiwi islands the kulama ceremony ritualises the detoxification of the yam.

FIGURE 5.2.7 The bitter yam, Dioscorea bulbifera, occurs naturally in Africa and Asia, as well as Northern Australia.

Tiwi creation stories explain how, in the creation time, a boobook owl man and barn owl woman performed the first kulama. The white-bellied sea-eagle joined the ceremony; according to the Tiwi creation story, it still wears its ceremonial head paint. The success of this kulama led the sea-eagle to conduct another kulama where bitter yam was prepared for the first time, demonstrating to the Tiwi how to prepare this otherwise toxic plant, and these processes have been followed ever since.

Year 11 Chapter 10GO TO ➤

CHEMFILE AHC

The Cicada Woman In Arnhem land the Gunwinggu people also detoxify the bitter yam, which they call mangindjeg. According to their beliefs, the Cicada Woman, Ngalgindjeg-Ngalgindjeg, demonstrated the correct way to prepare mangindjeg to the Gunwinggu. She used the bone of a kangaroo to cut up the roots, which she roasted in an oven and then leached in water. Since then, the cicada’s call tells the Gunwinggu that mangindjeg are ready to be dug up.

Cycad seeds are often mistaken for fruits because they have a fleshy outer layer, which is removed before they are prepared for eating.

Page

Proo

fs


Towards the end of the wet season, a ring around the moon signals to the Tiwi that the moon man is performing kulama and that they, too, should begin. The yams are picked according to specific rules, with care being taken not to damage the roots. The yams are placed in running water to remove the oxalates by leaching. While this is taking place, an earth oven is prepared. When the oven is ready, the yams are roasted. The three days of the kulama ceremony feature singing, dancing and body painting. Such is the importance of kulama to the Tiwi that the ceremonial dance circle is a major feature of their art.

CHEMISTRY IN ACTION CCT CC L

The Marsh testArsenic has been used as a poison since the time of the Roman emperor Nero. Colourless, odourless and tasteless, arsenic could go undetected in food and drink. The symptoms of arsenic poisoning—cramps, vomiting and diarrhoea—could easily be mistaken for food poisoning. And because arsenic accumulates in the body, small doses could be given over long periods of time to slowly kill the victim.

In the 19th century arsenic was widely used in industry and homes. It was a component of Scheele’s green, a pigment used in wallpaper, fabrics and food dyes. Powdered arsenic was available as a rat poison, meaning it was cheap and readily accessible, making it a poisoner’s first choice.

In 1832 John Bodle was accused of poisoning his grandfather George with arsenic added to coffee grains. James Marsh was a chemist called upon by the prosecution to demonstrate that samples taken from George Bodle contained arsenic. He carried out the standard test of the day: adding hydrogen sulfide in the presence of hydrochloric acid. If a yellow precipitate formed, then the presence of arsenic would be confirmed.

The yellow precipitate, used as a pigment in Egypt and Asia in ancient times, is arsenic sulfide, also known as King’s yellow or orpiment (Figure 5.2.9). It is a naturally occurring form of arsenic, and was called zarnikh by the Persians, from their word for gold. This became arsenikon in Greek, and led to the English name. The name orpiment is from the Latin for gold pigment.

Marsh carried out the test and the bright yellow precipitate appeared. However, in the time it took for the sample to get to court, it deteriorated and the colour faded. The defence were able to convince the jury that there was no arsenic present and Bodle was acquitted. Even worse for Marsh, Bodle later admitted his guilt, knowing that English law at the time meant that he could not be tried again. Frustrated, Marsh set about devising a better test.

The Marsh test for arsenic, as it is now known, was first used successfully in France in 1840 in the case of Madame Lafarge, who was convicted of poisoning of her husband. The test was used for more than 100 years (Figure 5.2.10). Today neutron bombardment can be used to detect minute amounts of arsenic.

FIGURE 5.2.9 Arsenic sulfide was used in ancient Egypt as a yellow pigment.

FIGURE 5.2.10 Arsenic detection apparatus used in the 19th century.

Page

Proo

fs


5.2 ReviewSUMMARY

• Salts that are considered insoluble do dissolve to a very small extent. These salts are best described as sparingly soluble.

• The solubility product, Ksp, is the product of the concentration of ions in a saturated solution of a sparingly soluble salt.

• The molar solubility of a salt is the number of moles that will dissolve per litre of water to form a saturated solution. Molar solubility can be used to calculate Ksp and vice versa.

• Ksp can be used to determine whether a precipitate will form.

• The common ion effect greatly reduces the solubility of salts.

• Aboriginal and Torres Strait Islander people use a range of methods to detoxify poisonous foods.

KEY QUESTIONS

1 Write equilibrium expressions for the dissolution of the following ionic compounds:a CaSO4

b ZnCO3

c PbI2d Fe(OH)3

2 The molar solubility of copper(II) hydroxide (Cu(OH)2) was found to be 3.8 × 10−7 mol L−1 at a certain temperature. Calculate the value of Ksp for Cu(OH)2 at this temperature.

3 Calculate the molar solubility of strontium sulfate (SrSO4) if its Ksp value at 25°C is 3.2 × 10−7.

4 If 4.5 g of lead(II) bromide is added to a litre of water, will a precipitate form? The value of Ksp for PbBr2 at 25°C is 6.60 × 10−6

5 Determine the molar solubility of silver chloride in a 0.15 mol L−1 solution of NaCl.

6 State the toxin present in cycad seeds, and describe how Indigenous Australians detoxify the seeds.

Page

Proo

fs


Chapter review

KEY TERMS

aqueouscommon ion effectdissociationhydrated insolubleion production–dipole attraction

molar solubilityprecipitatesaturated solutionslightly solublesparingly solublesolubilitysolubility product

solubility tablesolublesolutesolutionsolventspectator ionunsaturated solution

unsaturated solutionREVIEW QUESTIONS

1 a What is the name given to the process that ionic solids undergo when dissolving in water?

b What ions will be produced when each of the following compounds is added to water? i Cu(NO3)2ii ZnSO4

iii (NH4)3PO4

2 What ions would be produced when each of the following compounds is added to water? a potassium carbonate b lead(II) nitrate c sodium hydroxide d sodium sulfate e magnesium chloride f zinc nitrate g potassium sulfide h iron(III) nitrate.

3 Write equations to show the dissociation of the following compounds when they are added to water:a magnesium acetate b silver nitrate c potassium bromide d barium hydroxide e sodium phosphate.

4 What type of attraction forms between ions and water molecules when an ionic compound dissociates in water?A ionic bondB ion–dipole attractionC dispersion forceD dipole–dipole attraction

5 Briefly explain why the arrangement of water molecules around dissolved magnesium ions is different from that around dissolved chloride ions.

6 a Write the formulae of three carbonate compounds that are soluble in water.

b Write the formulae of three carbonate compounds that are insoluble in water.

7 a Write the formulae of three sulfate compounds that are soluble in water.

b Write the formulae of three sulfate compounds that are insoluble in water.

8 Using the information in Table 5.1.1 on page XXX, calculate the mass of sodium chloride that can be added to 150 mL of water before it stops dissolving.

9 Use the solubility tables to identify the precipitate formed, if any, when the following solutions are mixed.a silver acetate and potassium chloride b sodium hydroxide and iron(II) chloride c sodium iodide and lead(II) nitrate d barium bromide and potassium nitrate.

10 a Name the precipitate formed when aqueous solutions of the following compounds are mixed.i Na2S and MgI2ii K2SO4 and Ba(NO3)2 iii NaOH and FeCl3iv NH4Cl and AgCH3COO

b Write a balanced chemical equation for each reaction.

c Write an ionic equation for the formation of each precipitate.

11 For each of the reactions in Question 10, identify the spectator ions.

12 Copy and complete the following table. Identify which reaction mixtures will produce precipitates and write their formulae.

MgSO4 KCl NaOH AgNO3 FeBr2

Pb(CH3COO)2

K2CO3

BaI

Na3PO4

NH4S

Page

Proo

fs


13 Species in which of the following states of matter are included in the equilibrium expression for a heterogeneous reaction?A gases and pure liquidsB gases and pure solidsC aqueous solutions, gases and pure liquidsD aqueous solutions, gases and mixtures of liquids

14 Write ionic equations for the dissolution of the following compounds:a FeCl2 b Al(OH)3c Ag2SO4 d Ba3(PO4)2

15 Write equilibrium expressions for the dissociation of the following compounds:a PbCl2 b Zn(OH)2c Ag2CrO4 d BaSO4

16 For each of the pairs of salts listed, state which is the more soluble of the two. a BaSO4 (Ksp = 1.08 × 10−10) and BaCO3

(Ksp = 2.58 × 10−9)b Zn(OH)2 (Ksp = 3.8 × 10−17) and Fe(OH)2

(Ksp = 4.87 × 10−15)c PbCl2 (Ksp = 1.70 × 10−5) and PbI2 (Ksp = 9.8 × 10−9)

17 Define the term ‘molar solubility’.

18 Equations can be used to represent what occurs when ionic salts are added to water. The equation for the dissolution of sodium chloride in water is:


The equation for silver chloride is:

�AgCl(s) Ag (aq) Cl (aq)++ −

Explain why an equilibrium arrow is used for the dissolution of silver chloride, but not for sodium chloride.

19 Explain why the change in concentration is always positive in an ICE table for the dissolution of sparingly soluble salts in water.

20 Calculate Ksp for the following substances:a lead(II) bromide, given a molar solubility of

1.18 × 10−2 mol L−1 at 25°Cb barium sulfate, given a molar solubility of

1.04 × 10−5 mol L−1 at 25°Cc silver chromate (Ag2CrO4), given a molar solubility

of 8.7 × 10−5 mol L−1 at 25°C.

21 Calculate the molar solubility of the following substances:a zinc sulfide, given a Ksp of 2.0 × 10−25 at 25°Cb silver bromide, given a Ksp of 5.35 × 10−13 at 25°Cc silver sulfate given a Ksp of 1.20 × 10−5 at 25°C.

22 Explain why you cannot compare the Ksp values of silver cyanide (AgCN) and aluminium hydroxide (Al(OH)3) to deduce their relative solubilities.

23 Calculate the masses of the following substances needed to form one litre of a saturated solution:a lead(II) sulfate, given a Ksp of 2.53 × 10−8

b silver sulfide, given a Ksp of 8.0 × 10−51.

24 a Predict whether a precipitate will form in the following solutions:i 0.0010 g of silver carbonate is added to 100 mL of

water. The Ksp of Ag2CO3 is 8.46 × 10−12 at 25°C.ii 0.20 g of copper(I) chloride is added to 1.0 L of

water. The Ksp of CuCl is 1.9 × 10−7 at 25°C.iii 3.0 × 10−5 g of strontium chromate is added to 1.0 L

of water. The Ksp of SrCrO4 is 2.0 × 10−5 at 25°C.b If a precipitate was not formed in the above solutions,

calculate the total mass of salt that would need to be added to the water to form a saturated solution.

25 Predict whether a precipitate will form in the following instances:a A 1.0 L solution contains 0.0030 mol L−1 NaI and

0.020 mol L−1 Pb(NO3)2b A 1.0 L solution contains 0.0040 mol L−1

(NH4)2CO3 and 3.0 × 10−5 mol L−1 Mg(CH3COO)2c A 1.0 L solution contains 2.0 × 10−4 mol L−1 KCl and

5.0 × 10−6 mol L−1 AgNO3.

26 Predict whether a precipitate will form if the following solutions are mixed:a 10 mL of 0.0010 mol L−1 AgCH3COO and 10 mL of

0.0050 mol L−1 KBrb 5.0 mL of 6.0 × 10−7 mol L−1 MgI2 and 5.0 mL

of 4.0 × 10−5 mol L−1 NaOHc 20 mL of 1.0 × 10−4 mol L−1 Pb(NO3)2 and 5.0 mL of

0.020 mol L−1 NaSO4.

27 Calculate the molar solubility of the following:a lead(II) iodide in 0.050 mol L−1 potassium iodide

solution (Ksp for lead(II) iodide is 9.8 × 10−9)b iron(III) hydroxide in 0.020 mol L−1 sodium hydroxide

solution (Ksp for iron(III) hydroxide is 2.79 × 10−39)c zinc carbonate in 0.20 mol L−1 ammonium

carbonate solution (Ksp for zinc carbonate is 1.4 × 10−11).

28 a The value of Ksp for silver iodide is 8.52 × 10−17. Calculate the molar solubility of silver iodide: i in pure waterii in 1.5 mol L−1 potassium iodide solution.

b Explain the difference in the calculated molar solubilities.

29 Explain how one food item was detoxified by Indigenous Australians.

30 Explain the purpose of crushing a cycad seed prior to leaching.

31 Reflect on the Inquiry activity on page 000. Using concepts from this chapter, explain why the dishwashing liquid did not produce suds in hard water.

CHAPTER REVIEW CONTINUED

Page

Proo

fs

chapter solubility and equilibria proofs

Documents