Unit 3: Solubility Equilibrium Topic B: Precipitation Reactions and Qualitative Analysis

Transcription

1 1 In this topic we will examine: Unit 3: Solubility Equilibrium Topic B: Precipitation Reactions and Qualitative Analysis how to predict if a precipitate forms when solutions are mixed how to selectively precipitate ions from solution qualitative analysis of solutions the common ion effect This topic will make extensive use of the table of Solubility of Common Compounds in Water found in the Chemistry 12 Data Booklet. 3.B.1: Predicting Precipitation Reactions When two salt solutions are mixed, a double displacement reaction may occur. Such a reaction may cause one or two precipitates to form. A precipitate is a solid that forms as a result of combining aqueous ions. Of course, not every combination of salt solutions will cause a precipitate to form. Because every salt is soluble to some degree, it is necessary that the ion concentrations after two solutions are mixed exceed those present in a saturated solution of some ionic solid. For example, the Table of Solubility tells us that Ag 2 SO 4 has low solubility (unable to form a solution with [Ag 2 SO 4 ] 0.1 M at 25 C). However, this compound has a solubility of M. This tells us that it is possible to mix a solution containing Ag + ions and SO 4 ions and NOT have a precipitate form. All that is necessary is that the concentration of each ion be low enough after mixing that the saturation point is not exceeded. Conversely, no salt is infinitely soluble in water. If the concentrations of two ions are sufficiently large after two solutions are mixed, we can cause the precipitation of a soluble substance. For example, the Table of Solubility tells us that Li 2 CO 3 is soluble (able to form a solution with [Li 2 CO 3 ] 0.1 M at 25 C). However, the solubility of Li 2 CO 3 is only M. It is possible to mix a solution containing a large concentration of Li + ions with a solution containing a large concentration of CO 3 ions and cause the precipitation of Li 2 CO 3. We will learn two different methods of predicting whether or not a precipitate forms after two solutions are mixed. The first method uses the table of Solubility of Common Compounds in Water. This method will be examined in this topic. The second method requires us to perform calculations similar to the "Trial K eq " calculations that we learned in the previous unit. This method will be examined in Topic C of this unit.

2 2 3.B.1.1: Predicting Precipitation Reactions Using the Table of Solubility of Common Compounds in Water The table of Solubility of Common Compounds in Water may be used to predict the formation of a precipitate ONLY if the concentrations of all ions is greater than or equal to 0.1 M immediately after the mixing of two solutions. If equal volumes of two solutions are mixed, the limitation of the table requires that the original concentrations of all ions be greater than or equal to 0.2 M. Applying what we learned in the previous topic, if an ion concentration in one of the two solutions is equal to 0.2 M, then the mixing of equal volumes of the solutions will dilute the concentration of that ion to one-half of its initial value, or 0.1 M. Example: Predict the result of mixing equal volumes of 0.20 M Na 2 CO 3 and 0.20 M CaCl 2. Solution: When the two solutions are mixed, the ion concentrations immediately after mixing are: [Na + ] = [Ca 2+ ] = [CO 3 ] = [Cl - ] = When we consider the possible recombination of anions and cations, the combination of Ca 2+ ions and CO 3 ions present in the solution with concentrations equal to 0.1 M will cause the precipitation of CaCO 3 because it has low solubility. The "Table" tells us this. THINK: What other possible recombination of anions and cations is possible? Why does that combination not form a precipitate? Should any other recombination of ions be considered? (e.g., a combination of Cl - and CO 3 ) 3.B.2: Formula Equations, Complete Ionic Equations and Net Ionic Equations Often we are asked to illustrate the result of a precipitation reaction by writing an appropriate chemical reaction equation. There are three types of reaction equations used to illustrate precipitation reactions: formula equations, complete ionic equations, and net ionic equations. 3.B.2.1: Formula Equation for a Precipitation Reaction A formula equation (FE) gets its name from the fact that all reactants and products appear in the reaction equation as non-dissociated salts. Since the reactants are dissolved in water, they are assigned the phase (aq). We indicate which product is the precipitate by indicating that it is in the solid phase (s). Any ions that remain dissolved in solution are represented as an aqueous ionic compound.

3 3 For example, when 0.2 M Na 2 CO 3 and 0.2 M CaCl 2 are mixed, we know that a precipitate of CaCO 3 forms while the Na + and Cl - ions remain in solution. The formula equation for the precipitation reaction is: Na 2 CO 3(aq) + CaCl 2(aq) 2NaCl (aq) + CaCO 3(s) A correctly written formula equation has the following properties: correct chemical formulae are written for each substance involved in the reaction (we reviewed the procedure for doing this in Unit 1 Topic R) ALL phases/states of the substances involved are indicated [(aq) or (s)] a one-way arrow is used to indicate the non-reversibility of the precipitation reaction it is BALANCED 3.B.2.2: Complete Ionic Equation for a Precipitation Reaction The difference between a complete ionic equation (CIE) and a formula equation is that all dissolved ionic compounds are represented in dissociated form. Any precipitate that forms is represented in the same manner as in a formula equation. The most common error made when writing CIE's is to not properly balance all the ions. As a first step, it is very helpful to write dissociation equations for each compound that appears in the formula equation. The formula equation above includes three compounds in aqueous form. The dissociation equations for these are: Na 2 CO 3(aq) 2Na + (aq) + CO 3 (aq) CaCl 2(aq) Ca 2+ (aq) + 2Cl - (aq) NaCl (aq) Na + (aq) + Cl - (aq) When writing the CIE for the reaction such as the one between Na 2 CO 3 and CaCl 2, the common error is to not include two Na + ions and two Cl - ions on the product side. The correct CIE for the reaction is: 2Na + (aq) + CO 3 (aq) + Ca 2+ (aq) + 2Cl - (aq) 2Na + (aq) + 2Cl - (aq) + CaCO 3(s) 3.B.2.3: Net Ionic Equation for a Precipitation Reaction In most precipitation reactions, there are some ionic species that do not take part in the chemical reaction that occurs. These species appear on both the reactant side and the product side of the CIE in dissociated aqueous form. These ions are referred to as spectator ions. When writing the net ionic equation (NIE) for a precipitation reaction, we do not include the spectator ions. We only include the ions that take part in the reaction. For the reaction we have been examining, the net ionic equation is:

4 4 Ca 2+ (aq) + CO 3 (aq) CaCO 3(s) Common errors made when writing NIE's include failing to write the proper chemical formula for the compound produced and not reducing molar coefficients to lowest terms (i.e., having a "2" in front of each species). Examples: Equal volumes of each of the following pairs of 0.2 M solutions are mixed at 25 C. For each pair where one or more precipitates form, write the formula equation, the complete ionic equation and the net ionic equation to illustrate the reactions that occur. a) AgNO 3(aq) and CaCl 2(aq) b) Aqueous magnesium sulphate and aqueous strontium hydroxide

5 5 Provincial Exam Examples: Predicting Precipitation Reactions Using the Table of Solubility of Common Compounds in Water Note: Unless told otherwise, a temperature of 25 C must always be assumed. 1. When equal volumes of 0.2 M solutions are mixed, which of the following combinations forms a precipitate? A. CaS and Sr(OH) 2 C. H 2 SO 4 and MgCl 2 B. (NH 4 ) 2 SO 4 and K 2 CO 3 D. H 2 SO 3 and NaCH 3 COO 2. Which of the following would be true when equal volumes of 0.2 M NaBr and 0.2 M AgNO 3 are combined? A. No precipitate forms. B. A precipitate of AgBr forms. C. A precipitate of NaNO 3 forms. D. Precipitates of both NaNO 3 and AgBr form. 3. What happens when 10.0 ml of 0.2 M KOH is added to 10.0 ml of 0.2 M CuSO 4? A. No precipitate forms. B. A precipitate of K 2 SO 4 forms. C. A precipitate of Cu(OH) 2 forms. D. Precipitates of both K 2 SO 4 and Cu(OH) 2 form. 4. What will happen when equal volumes of 0.20 M (NH 4 ) 2 S and 0.20 M Sr(OH) 2 are mixed? A. SrS precipitates. C. NH 4 OH precipitates. B. Both NH 4 OH and SrS precipitate. D. No precipitate forms. 5. Which of the following precipitates may form when equal volumes of 0.3 M AgNO 3, 0.3 M SrCl 2 and 0.3 M Na 2 CO 3 are mixed together? A. SrCO 3 and AgCl C. Ag 2 CO 3 and AgCl B. SrCO 3 and Ag 2 CO 3 D. SrCO 3, Ag 2 CO 3 and AgCl 6. What is the net ionic equation for the reaction between BaS (aq) and Sr(OH) 2(aq)? A. Sr 2+ (aq) + S (aq) SrS (s) B. Ba 2+ (aq) + 2OH - (aq) Ba(OH) 2(s) C. Ba 2+ (aq) + S (aq) + Sr 2+ (aq) + 2OH - (aq) Ba(OH) 2(s) + SrS (s) D. Ba 2+ (aq) + S (aq) + Sr 2+ (aq) + 2OH - (aq) Ba(OH) 2(s) + Sr 2+ (aq) + S (aq)

6 6 7. Which net ionic equation best describes the reaction that exists in a solution prepared by mixing equal volumes of 0.20 M Ca(NO 3 ) 2 and 0.20 M Na 2 CO 3? A. Ca 2+ (aq) + CO 3 (aq) CaCO 3(s) B. Na + (aq) + NO 3 - (aq) NaNO 3(s) C. Ca(NO 3 ) 2(aq) + Na 2 CO 3(aq) 2NaNO 3(s) + CaCO 3(s) D. Ca(NO 3 ) 2(aq) + Na 2 CO 3(aq) 2NaNO 3(aq) + CaCO 3(s) 8. When equal volumes of 0.20 M Pb(NO 3 ) 2 and 0.20 M KCl are mixed, a precipitate of PbCl 2 forms. a) Write the formula equation for the above reaction. b) Write the complete ionic equation for the above reaction. c) Write the net ionic equation for the above reaction. Hebden Reference: Sections III.3 and III.4. Practice Exercises: Complete the following exercises from the Hebden text: Pages 83 to 87 Exercises 22 and B.3: Removing Ions From Solution With Help From the Table of Solubility of Common Compounds in Water Knowing how to determine which combinations of ions in solution will react to form a precipitate allows us to devise strategies to remove one or more ions from an aqueous solution by precipitation. The first step is to be able to readily identify what anions or cations we can introduce into the solution the cause the precipitation the desired ion(s).

7 7 Examples: 1. A solution contains SO 4 ions. Which ions can we introduce into the solution to cause the SO 4 ions to precipitate? 2. A solution contains I - ions. Which ions can we introduce into the solution to cause the I - ions to precipitate? 3. A solution contains Sr 2+ ions. Which ions can we introduce into the solution to cause the Sr 2+ ions to precipitate? In laboratory practice, causing the precipitation of an ion that is present in a solution is not as straightforward as it seems. For example, we know that the addition of Pb 2+ ions to a solution containing SO 4 ions may initiate the precipitation of PbSO 4(s). However, we need to consider the following factors: How much Pb 2+ must be added before precipitation starts, How much Pb 2+ must we add to reduce the [SO 4 ] to nearly zero (as we will learn later, we can never get rid of all the SO 4 by precipitation), and How, exactly, do we go about adding Pb 2+ ions into the solution? Since precipitation of PbSO 4 will not begin until the solution reaches the saturation point, the amount of Pb 2+ that we must add to the solution depends on the [SO 4 ] ions initially present in the solution and the solubility of PbSO 4. In the next topic we will learn how to mathematically calculate the minimum required [Pb 2+ ] that will initiate the precipitation of SO 4 in a solution. Once the precipitation of PbSO 4 starts, if we continue to add more and more Pb 2+ to a solution containing SO 4 ions, precipitation will continue until nearly all the SO 4 ions have been removed. At some point, however, precipitation of PbSO 4 will stop, leaving the solution with a nearly immeasurable [SO 4 ]. Later, we will learn how to calculate the exact value of the [SO 4 ] ions that can exist in a solution containing a known [Pb 2+ ]. As for adding Pb 2+ to a solution, it is important to be aware that it is impossible to add a single cation (or anion, for that matter) to a solution. Anions and cations must be introduced into a solution in the form of a soluble salt, either in solid form or as a fairly concentrated solution of the selected soluble salt. It is common practice to always add nitrate salts of any cation that we want to introduce into a solution. This means that if we want to add Pb 2+ ions to a solution containing SO 4 ions we would add solid

8 8 Pb(NO 3 ) 2 or a fairly concentrated solution of Pb(NO 3 ) 2. In practice, we add Pb(NO 3 ) 2 "in excess", meaning that we add it beyond the point where it appears that the precipitation of PbSO 4 has stopped. If we wish to introduce an anion into a solution to cause the precipitation of some known cation, we also need to add it in the form of a soluble salt (in solid form or as a fairly concentrated solution). In such cases it is common to add sodium salts of the desired anion. Sodium salts, like nitrate salts, have high solubility, are readily obtained, and are rather inexpensive. For example, if we wish to remove Sr 2+ ions from a solution by precipitation it in the form of SrSO 4, we would add Na 2 SO 4 "in excess", either in solid form or as a fairly concentrated solution. In pretty much all cases, when an ion is removed from a solution by means of precipitation it is normal to filter out any precipitates that form. After all, if the goal is to remove a substance from a solution one should take out any solids containing that substance as a final step. This idea will be emphasized later on in this topic. Examples 1. A solution contains Sr 2+ ions. Give the formulae of three soluble salts that may be added to the solution to cause the Sr 2+ ions to precipitate. 2. A solution contains Cl - ions. Give the formulae of three soluble salts that may be added to the solution to cause the Cl - ions to precipitate. 3. A solution contains PO 4 3- ions. Give the formulae of three soluble salts that may be added to the solution to cause the PO 4 3- ions to precipitate. Hebden Reference: Sections III.3 and III.4. Practice Exercises: Complete the following Exercises from the Hebden text: Page 84 Exercise B.3.1: Applications of This Wonderful New Skill Now that we know how to select salts that will cause the precipitation of selected anions and cations, how shall we apply this knowledge? We will examine three applications of this knowledge.

9 9 1. How to "soften" hard water. 2. How to devise schemes to sequentially precipitate and remove ions from a solution containing a mixture of anions and/or cations. 3. How to determine what ions are or are not present in a solution by the selective addition of soluble salts. The last two applications may seem similar at times. The important difference is that in one case, we know what ions are present in a solution, and our task is to remove them one at a time. In the other case, we are uncertain which ions are actually present in a sample of a solution, and our task is to determine what ions are and are not present in the solution. 3.B.3.1.1: Softening Hard Water In many areas, the water provided to homes and businesses contain rather high concentrations of dissolved Ca 2+, Mg 2+, and Fe 2+ /Fe 3+ ions (you must remember this). Such water is commonly referred to as hard water. Many soaps contain soluble compounds of the stearate ion (C 17 H 35 O 2 - ). This ion is the cleaning agent and is responsible for the "lathering" of soaps and detergents. The metal ions present in hard water form stearate compounds of low solubility, preventing soaps and detergents from lathering and reduce their effectiveness. These lowly soluble stearate compounds are the primary component of "soap scum". The same metal ions present in hard water cause the buildup of "scale" in devices that boil water. Scale is a hard deposit of carbonates that do not conduct heat very well. When scale builds up on heating elements, the heat is not efficiently transferred to the water and can cause the heating elements to overheat and "burn out". In industrial applications, the buildup of scale inside boilers can cause large explosions that often result in fatalities. In areas where the water is very hard, it is necessary to "soften" the water before it is used. The procedure used for softening water involves the addition of soluble salts to the water to cause the metal ions to precipitate, usually in the form of carbonates, which are either filtered out or allowed to settle out before the water goes on to where it is used. On the household level, many people add "washing soda" (Na 2 CO 3 H 2 O) to their laundry before adding detergent. This causes the metal ions to form carbonates of low solubility preventing the metal ions from combining with the stearate ions. In earlier days, soluble phosphate salts were added to soaps and detergents to soften hard water. This practice has been all but abandoned because of environmental reasons.

10 10 Hebden Reference: Sections III.9. Hebden goes much farther in depth than is necessary for you to know. All you need to know about hard water is what ions cause it and what compounds can be used to soften it. You don't need to worry about how "permanent hard water" and "temporary hard water" differ. Practice Exercises: None that you need to worry about. Provincial Exam Example: Softening Hard Water 1. Which of the following would be most effective in removing the cations responsible for hard water? A. S C. Cl - 3- B. PO 4 D. SO 4 3.B.3.1.2: Sequential Precipitation of Ions in a Mixture of Aqueous Ions At times, chemists are confronted with an aqueous solution containing one or more dissolved ions that must be removed from the solution. Reasons for having to remove the ions vary, but typically it is to prevent these ions from forming undesired precipitates with other substances when the solution goes on to be part of another chemical process. Our knowledge of the "solubility rules" permits us to devise schemes to remove each undesired ion (in the form of a precipitate). Very often it is necessary to remove each kind of ion individually so that we do not wind up with a mixture of different kinds of precipitates. The usual reason for this is that each precipitate may present different safety hazards requiring them to be disposed of in different ways. In some cases, the precipitates formed are the desired product of a chemical reaction. Things to consider in devising schemes to sequentially precipitate ions from a solution containing a mixture of ions are: A. We must add soluble salts that cause the precipitation of only ONE ion at a time. This will require us to carefully think about the order in which the ions are precipitated. B. In each precipitation step, we must add the selected salt "in excess" to ensure that each precipitated ion is as thoroughly removed as possible. As mentioned previously, it is impossible to completely remove an ion from solution, but its concentration can be reduced to an immeasurable level. C. After each precipitate is formed it must be filtered out of the solution before the next precipitation step occurs. This includes the last precipitate formed in the procedure.

11 11 D. There is nearly always more than one possible method of obtaining the desired results. Not only is there more than one choice of salt to use for each precipitation step, but the sequence in which the ions are precipitated may also vary. Examples: Devise a scheme to sequentially precipitate each ion separately from a mixture of the following set of ions. For each precipitation step, write the net ionic equation to show the reaction that occurs. a) SO 4 and PO 4 3- b) Fe 2+, Pb 2+, and Mg 2+ c) Br -, CO 3, and SO 4

12 12 3.B.3.1.3: Qualitative Analysis Qualitative analysis is any process that permits us to determine: whether or not one or more specific ions are present in a solution, or which unknown ion(s) is/are present in a solution. The first application mentioned above is very similar to sequential precipitation of ions discussed earlier. The difference is that we are uncertain whether or not each ion is actually present or not present in a solution. Example: A solution may contain Ba 2+ and/or Pb 2+ ions. Devise a scheme to determine whether or not each ion is present in the solution. The second application of qualitative analysis is somewhat different. In this application a number of different salts or solutions of salts are added to a number of samples of a solution containing one or more unknown ions. By observing the formation or non-formation of precipitates after the addition of each salt, we can deduce which ion or ions are present in the solution. Example: Na 2 S (s), Na 2 SO 4(s) and NaOH (s) are added to samples of a solution containing an unknown ion. The table below shows the results of each addition. Added Salt Na 2 S Na 2 SO 4 NaOH Did a Precipitate Form? No Yes No Determine the unknown ion in the solution.

13 13 Hebden Reference: Section III.5. Practice Exercises: Complete the following Exercises from the Hebden text: Page 90 Exercises 26 to 36. Provincial Exam Examples: Removing Ions From Solution and Qualitative Analysis 1. Which cation can be used to separate SO 4 from S ions by precipitation? A. Sr 2+ C. Pb 2+ B. Cs + D. Be Using the solubility table, determine which of the following ions could not be used to separate S from SO 4 by precipitation. A. Be 2+ C. Ca 2+ B. Ba 2+ D. Sr An experiment is conducted to identify an unknown cation that is present in each of four beakers. Which of the following could be the unknown cation? A. Ag + B. Fe 3+ C. Ba 2+ D. Be 2+

14 14 4. A solution contains both 0.2 M Mg 2+ and 0.2 M Sr 2+. These ions can be removed separately through precipitation by adding equal volumes of 0.2 M solutions of A. OH - and then S C. Cl - and then OH - B. CO 3 and then SO 3 D. SO 3-4 and then PO 4 5. A solution contains 0.2 M Zn 2+ and 0.2 M Sr 2+. An equal volume of a second solution is added, forming a precipitate with Sr 2+ but not with Zn 2+. What is present in the second solution? A. 0.2 M Cl - C. 0.2 M OH - B. 0.2 M SO 4 3- D. 0.2 M PO 4 3.B.4: Factors Affecting the Position of Solubility Equilibrium Many of the general factors affecting the position of equilibrium that we studied in Unit 2 affect the position of solubility equilibrium, but some of those factors do not. A. Changes in Pressure/Volume Since no gases are present in a solubility equilibrium involving ionic substances, changes in total system pressure or volume do not or cannot affect the position of solubility equilibrium (in fact, one cannot actually compress such systems!). B. Changes in Temperature The dissociation of a salt may be an endothermic or an exothermic process. Examples of each are: i) MgCl 2(s) + heat Mg 2+ (aq) + 2Cl - (aq) ii) Ca(CH 3 COO) 2(s) Ca 2+ (aq) + 2CH 3 COO - (aq) + heat In i) above, an increase in T shifts the equilibrium to the. A decrease in T shifts the equilibrium to the. In ii) above, an increase in T shifts the equilibrium to the. A decrease in T shifts the equilibrium to the. When the position of solubility equilibrium shifts to the products, some of the undissolved solid present in the system will dissociate into ions. Therefore, we can say that any stress causing solubility equilibrium to shift to the right increases the solubility of the salt. When the position of solubility equilibrium shifts to the reactants, some of the ions in solution will precipitate, increasing the amount of undissolved solid present in the system. Therefore, we can say that any stress causing solubility equilibrium to shift to the left decreases the solubility of the salt.

15 15 Therefore, an increase in temperature may cause the solubility of a salt to either increase or decrease, depending on whether the dissociation of the salt is exothermic or endothermic. C. Addition/Removal of the Reactant The only reactant in solubility equilibrium is the undissolved solid present. Changing the amount of solid present in the system does not affect the position of equilibrium. However, the amount of undissolved solid present does affect the rate of both the forward and reverse reaction. In order for the forward reaction to occur, some amount of solid must be present. The greater the amount of solid present (or the greater the surface area of the solid present), the faster the rate at which the solid dissolves, as discussed in Unit 1. However, the rate of the reverse reaction (the rate of precipitation) increases in an equal amount so no shift in the position of equilibrium occurs. This result is explained by the fact that when solubility equilibrium exists in a saturated solution, the processes of dissociation and crystallization occur at the surface of the solid present in the container. Therefore, a greater quantity of undissolved solid provides a greater number of locations for the forward and reverse reactions to occur. D. Addition/Removal of a Product Consider the following solubility equilibrium: AgCl (s) Ag + (aq) + Cl - (aq) How can we add either one of the product ions to the equilibrium? As we discussed earlier, one cannot simply add Ag + or Cl - by itself. To add either ion, one must add a soluble salt containing an ion "in common" with AgCl (e.g., AgNO 3 or NaCl). If we dissolve solid AgNO 3 into a saturated solution of AgCl, the AgNO 3 will dissociate and release Ag + ions into the solution. According to LeChatelier's Principle, this will shift the equilibrium to the left, causing the precipitation of some solid AgCl. Since less AgCl is now dissolved in the solution, we can say that the addition of any soluble silver salt to a saturated solution of AgCl reduces the solubility of the AgCl. Similarly, since the addition of NaCl will also shift the equilibrium to the left, the addition of any soluble chloride salt to a saturated solution of AgCl reduces the solubility of the AgCl. The Common Ion Effect is the name given to the lowering of the solubility of a salt by adding a second salt having an ion in common with the first. Of course, the added salt must be more soluble than the first salt. The above discussion leads to the conclusion that the solubility of any salt is greater in pure water than it is in a solution already containing one of the ions in that salt. In other words, compared to its solubility in pure water, the solubility of AgCl is lower in a solution that already contains some dissolved Ag + ions (such as a solution of AgNO 3 ) or some dissolved Cl - ions (such as a solution of NaCl).

16 16 One of the product ions of a solubility equilibrium may be removed by introducing another ion to cause it to precipitate. If we add solid NaBr to a saturated solution of AgCl, Br - ions will be released into the solution because of the dissociation of the NaBr: NaBr (s) Na + (aq) + Br - (aq) When the [Br - ] is high enough, the precipitation of AgBr (s) will begin: Ag + (aq) + Br - (aq) AgBr (s) This will cause the [Ag + ] ions to decrease and shift the equilibrium to the right. A shift to the right will cause some of the AgCl (s) that is present to dissolve and dissociate. Since a greater amount of AgCl is now dissolved in the solution, the addition of NaBr (s) to a saturated solution of AgCl effectively increases the solubility of AgCl. Similarly, the addition of a soluble salt that causes the precipitation of the Cl - ions (e.g., Pb(NO 3 ) 2 ) will also cause the equilibrium to shift to the right and some of the AgCl (s) that is present to dissolve and dissociate. From these last discussions, we can conclude that it is possible to dissolve a greater quantity of solid AgCl in a solution of NaBr or Pb(NO 3 ) 2 than we can dissolve in pure water. In other words, AgCl has a greater solubility in solutions containing dissolved Br - or Pb 2+ ions (or any other ion that precipitates Ag + or Cl - ) than it has in pure water. (Personally, I don't like this last interpretation of the previous discussions) Examples: The following equilibrium: BaS (s) Ba 2+ (aq) + S (aq) exists with a significant amount of undissolved solid present. Describe the effect that the addition of each of the following salts has on the solubility of BaS. a) Na 2 S (this example is continued on the next page)

17 17 b) Ba(NO 3 ) 2 c) Cu(NO 3 ) 2 Hebden Reference: Section III.10. Practice Exercises: Complete the following Exercises from the Hebden text: Page 79 Exercise 17. Page 108 Exercises 81, 82 and 85. Provincial Exam Examples: Factors Affecting the Position of Solubility Equilibrium and the Common Ion Effect 1. Consider the following equilibrium: CaSO 4(s) Ca 2+ (aq) + SO 4 (aq) Which of the following would shift the equilibrium to the left? A. Adding CaSO 4(s) C. Adding MgSO 4(s) B. Removing some Ca 2+ (aq) D. Removing some SO 4 (aq) 2. Consider the following equilibrium: AgCl (s) + energy Ag + (aq) + Cl - (aq) Addition of which of the following will increase the solubility of AgCl? A. heat C. HCl B. AgNO 3 D. a catalyst

18 18 3. A saturated solution of PbI 2 was subjected to a stress and the following graph was obtained. Which stress was applied at time t 1? A. The addition of PbI 2 B. A temperature change C. An increase in volume D. The evaporation of water. 4. Which of the following is true when solid Na 2 S is added to a saturated solution of CuS and equilibrium is reestablished? A. [S ] increases. C. [Cu 2+ ] increases. B. [S ] does not change. D. [Cu 2+ ] does not change. 5. A saturated solution is prepared by dissolving a salt in water. Which of the following graphs could represent the ion concentrations as the temperature is changed?

19 19 6. Solid NaCl is added to a saturated AgCl solution. How have the [Ag + ] and [Cl - ] changed when equilibrium has been reestablished? [Ag + ] [Cl - ] A. increased increased B. decreased increased C. increased decreased D. decreased decreased 7. Consider the following equilibrium: MgCO 3(s) Mg 2+ (aq) + CO 3 (aq) Adding which of the following would cause the solid to dissolve? A. HCl C. K 2 CO 3 B. MgCO 3 D. Mg(NO 3 ) 2 8. Consider the following equilibrium: CaS (s) Ca 2+ (aq) + S (aq) When Ca(NO 3 ) 2(aq) is added to this solution, the equilibrium shifts to the A. left and [S ] increases. C. left and [S ] decreases. B. right and [S ] increases. D. right and [S ] decreases. 9. In which would PbCl 2 be least soluble? A. 1 M HCl C. 1 M BaCl 2 B. 1 M K 2 SO 4 D. 1 M Pb(NO 3 ) When a solution of Na 2 CO 3(aq) is mixed with a solution of Ca(NO 3 ) 2(aq) a precipitate forms. a) Write the net ionic equation for the precipitation reaction. b) Explain what happens to the precipitate when HCl is added. It is now time to complete the Unit 3 Topic B Hand-in Assignment.