Chemistry 40S Chemical Equilibrium (This unit has been adapted from

Chemistry 40S Chemical Equilibrium (This unit has been adapted from https://bblearn.merlin.mb.ca) Name: 1

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Lesson 1: Defining Equilibrium Goals: Describe physical and chemical equilibrium. Describe the conditions needed for equilibrium. Write the equilibrium law given the chemical equation for the equilibrium system. Chemical Equilibrium Closed system 3

Conditions for Equilibrium So far, we have assumed that reactions proceed completely from reactants to products. Many chemical processes are reversible and proceed in BOTH the forward and reverse directions. Let's consider the conversion of nitrogen dioxide gas, NO2, into dinitrogen tetraoxide, N2O4, by the reaction: 2NO2 (g) N2O4 (g). Nitrogen dioxide is produced in the exhaust of automobiles and is converted to dinitrogen tetraoxide by ultraviolet light form the sun. This is the major source of smog, which is prevalent in large urban centres. The brownish tint of the smog is caused by the presence of the nitrogen dioxide. Nitrogen dioxide is a brown gas while dinitrogen tetraoxide is colourless. If NO2 is placed in a sealed container, eventually the brown colour fades, but not completely. The light brown colour persists indefinitely, as long as the conditions remain constant. The fading is a result of the production of the colourless N2O4. The presence of a brownish tint indicates the presence of NO2 in the container. Therefore, the reaction does NOT go completely to products, but rests somewhere between. The equilibrium state is characterized by constant concentrations of both reactants (NO2) and products (N2O4). The equilibrium state is characterized by constant macroscopic properties. That is, the system appears to be static, or unchanging. As the concentration of N2O4 increases, it becomes converted back to NO2. According to the collision theory, the rate of conversion to reactants is increased as the product concentration increases. Eventually, the rate of N2O4 production, the forward reaction, equals the rate of NO2 production, the reverse reaction. This reaction is then written as 2NO2 (g) N2O4 (g). Notice how the number of each molecule remains constant even though the conversion to NO and its decomposition continue. The equilibrium state is dynamic, since BOTH forward and reverse reactions occur simultaneously. 4

Defining Equilibrium If we plot the concentration vs. time of the reactants and products in an equilibrium system it may appear as below: If we graphed the rate vs. time, it would appear as below. 5

Equilibrium Law Law of mass action or the equilibrium law They proposed, for the reaction aa + bb cc + dd where A, B, C and D are reactants and products and the lower case letters a, b, c and d are the molar coefficients in a balanced reaction. If the forward and reverse reactions were elementary reactions, the rate laws for both forward and reverse reactions would be: Homogeneous equilibria Example 1: Write the equilibrium law for the following equation: N2 (g) + 3H2 (g) 2NH3 (g) 6

Heterogeneous equilibria Example 2: Write the equilibrium law for the following equation: C (s) + H2O (g) CO (g) + H2 (g) Equilibrium Constant Equilibrium constant The equilibrium constant can indicate whether products or reactants are favoured at equilibrium. That is, whether there are more products or reactants at equilibrium. 7

If Keq = 1, [products] = [reactants]. Neither reactants nor products are favoured. If Keq > 1, the amount of product is greater than reactant. We say the products are favoured or the equilibrium position lies to the right. If Keq < 1, the product concentrations are less than reactant concentrations. We say the reactants are favoured or the position of equilibrium lies to the left. Practice: Defining Equilibrium 1. Write the equilibrium law for each of the following reactions: a) SO2 (g) + NO2 (g) SO3 (g) + NO (g) b) 2C (s) + 3H2 (g) C2H6 (g) c) 3O2 (g) 2O3 (g) d) MgCO3 (s) CO2 (g) + 2MgO (s) 8

4. Comment on the favorability of product formation in each of the reactions. a) H2 (g) + F2 (g) 2HF (g) Keq = 1.0 10 13 b) SO2 (g) + NO2 (g) NO (g) + SO3 (g) Keq = 1.0 10 2 c) 2H2O (g) 2H2 (g) + O2 (g) Keq = 6.0 10-28 5. Chemists have determined the equilibrium constants for several reactions. In which of these reactions are the products favoured over the reactants? Keq = 1.0 10 2 Keq = 3.5 Keq = 0.003 Keq = 6.0 10-4 9

Lesson 2: Equilibrium Law Calculations Goals: Given equilibrium concentrations, calculate the value of K. Given K and all but one equilibrium concentration, calculate the missing equilibrium concentration. Given K and initial concentrations, calculate equilibrium concentrations. Given initial concentrations and an equilibrium concentration, calculate the value of K. You can calculate the value Keq if you are given the concentrations of reactants and products at equilibrium. The values are then substituted into the equilibrium law. Example 1: For the reaction N2 (g) + 3H2 (g) 2NH3 (g) at 225 C, a 2.0 L container holds 0.040 mole of N2, 0.15 mole of H2 and 0.50 mole of NH3. If the system is at equilibrium, calculate Keq. 10

Example 2: For the reaction N2 (g) + O2 (g) 2NO (g) at 210 C, the Keq is 64.0. The equilibrium concentrations of N2 and O2 are 0.40 mol/l and 0.60 mol/l, respectively. Calculate the equilibrium concentration of NO. 11

Another problem type is when the initial concentrations of some species, usually reactants are given and the equilibrium concentration of one of the products is given, calculate the equilibrium constant. Example 3: For the reaction H2 (g) + F2 (g) 2HF (g), 1.00 mole of hydrogen and 1.00 mole of fluorine are sealed in a 1.00 L flask at 150 C and allowed to react. At equilibrium, 1.32 moles of HF are present. Calculate the equilibrium constant. 12

If we are given initial concentrations and the equilibrium constant, we can calculate the equilibrium concentrations of all reactants and products for many equilibrium systems. Example 4: For the reaction N2 (g) + O2 (g) 2NO (g), the equilibrium constant is 6.76. If 6.0 moles of nitrogen and oxygen gases are placed in a 1.0 L container, what are the concentrations of all reactants and products at equilibrium? 13

Example 5: A 2.0 L container contains 6.00 moles of NO2 (g), 3.0 moles of NO (g) and 0.20 mole of O2 (g) at equilibrium. What is Keq for 2NO (g) + O2 (g) 2NO2 (g)? 14

Example 6: 4.00 moles of NO2 (g) is introduced to into a 2.00 L container. After a while equilibrium is attained according to the equation 2NO (g) + O2 (g) 2NO2 (g). At equilibrium 0.500 mole of NO (g) is found. What is the Keq value? 15

Example 7: A certain amount of NO2 (g) was introduced into a 5.00 L container. When equilibrium was attained according to the equation 2NO (g) + O2 (g) 2NO2 (g), the concentration of NO (g) was 0.800 mol/l. If Keq has a value of 24.0, how many moles of NO2 were originally put into the container? 16

Example 8: Keq = 3.5 for SO2 (g) + NO2 (g) SO3 (g) + NO (g). If 4.0 moles of SO2 (g) and 4.0 moles of NO2 (g) are placed in a 5.0 L container and allowed to come to equilibrium, what concentration of all species will exist at equilibrium? 17

Example 9: A 1.0 L reaction vessel contained 1.0 mole of SO2, 4 moles of SO3, and 4.0 moles of NO at equilibrium according to SO2 (g) + NO2 (g) SO3 (g) + NO (g). If 3.0 moles of SO2 is added to the mixture, what will be the new concentration of NO when equilibrium is reattained? 18

Practice: Equilibrium Law Calculations 1. A mixture at equilibrium at 827 C contains 0.552 moles of CO2, 0.552 moles H2, 0.448 moles CO, and 0.448 moles of H2O in a 1.00 L container. What is the value of the equilibrium constant for CO2 (g) + H2 (g) CO (g) + H2O (g)? 2. The equilibrium constant for the reaction 4H2 (g) + CS2 (g) CH4 (g) + 2H2S (g) at 755 C is 0.256. What is the equilibrium concentration of H2S if at equilibrium [CH4] = 0.00108 mol/l, [H2] = 0.316 mol/l, and [CS2] = 0.0898 mol/l? 19

3. Find the value of Keq if there is 25.0 moles of P4, 10.0 moles of H2 and 5.00 moles of PH3, in a 5.00 L container. The equation is P4 (g) + 6H2 (g) 4PH3 (g). 4. Find the value of Keq for ZnO (s) + CO (g) Zn (s) + CO2 (g) if at equilibrium there are 3.0 moles of CO, 4.0 moles of Zn and 4.0 moles of CO2 in a 500.0 ml container. 20

5. If Keq = 46.0 for H2(g) + I2(g) 2 HI(g) what [I2] would be in equilibrium with 0.50 mol/l HI and 0.10 mol/l H2? 6. If Keq = 10.0 for N2 (g) + 3H2 (g) 2NH3 (g) how many moles of NH3, at equilibrium, will be in a 2.00 L container if [H2] is 0.600 mol/l and [N2] is 0.100 mol/l? 21

7. The formation of ammonia from hydrogen and nitrogen occurs by the reaction 3H2 (g) + N2 (g) 2NH3 (g). Analysis of an equilibrium mixture of nitrogen, hydrogen, and ammonia contained in a 1.0 L flask at 300 C gives the following results: hydrogen 0.15 mole, nitrogen 0.25 mole, ammonia 0.10 mole. Calculate Keq for the reaction. 8. Bromine chloride, BrCl, decomposes to form bromine and chlorine. 2 BrCl(g) Cl2(g) + Br2(g) At a certain temperature the equilibrium constant for the reaction is 11.1, and the equilibrium mixture contains 4.00 mol of Cl2. How many moles of Br2 and BrCl are present in the equilibrium mixture? 22

9. The decomposition of hydrogen iodide to hydrogen and iodine occurs by the reaction 2HI (g) H2 (g) + I2 (g). Hydrogen iodide is placed in a container at 450 C an equilibrium mixture contains 0.50 mole of hydrogen iodide. The equilibrium constant is 0.020 for the reaction. How many moles of iodine and hydrogen iodide are present in the equilibrium mixture? 10. For the reaction of H2 (g) + Cl2 (g) 2HCl (g), a student places 2.00 moles of H2 and 2.00 moles of Cl2 into a 0.500 L container and the reaction is allowed to go to equilibrium at 516 C. If Keq is 76.0, what are the equilibrium concentrations of H2, Cl2 and HCl? 23

11. If Keq = 78.0 for the reaction A (s) + 2B (g) 2C (g) and initially there are 5.00 moles of A and 4.84 moles of B in a 2.00 L container, how many moles of B are left at equilibrium? 24

12. For the reaction of C (s) + O2 (g) CO2 (g), Keq = 25.0. Find the moles of CO2 at equilibrium, if initially there are 100.0 moles of C, 50.0 moles of O2, and 2.0 moles of CO2 in a 2.00 L container. 25

13. For the reaction of NH4Cl (s) NH3 (g) + HCl (g), Keq = 3.50 10-4. Find the concentration of NH3 in a 1.00 L container at equilibrium if initially there were 0.200 mole of NH3 added to 0.200 mole of HCl. 14. Initially the concentrations of N2 and O2 are 1.8 mol/l each and there is no NO. If at equilibrium the [NO] is 2.0 mol/l, find Keq for the reaction of N2 (g) + O2 (g) 2NO (g). 26

15. Find Keq for the reaction 2CO (g) + O2 (g) 2CO2 (g) if initially, there is 5.0 moles of CO, 10.0 moles of O2 and 1.0 mole of CO2 in a 2.0 L container and at equilibrium CO2 has a concentration of 2.5 mol/l. 27

Lesson 3: Predicting Equilibrium Goals: Determine the reaction quotient for a system. Determine if a system is at equilibrium and, if not, which reaction is favoured. Reaction Quotient To determine which reaction is favoured and in which direction the system is moving, Q is compared to Keq: If Q = Keq, the system is at equilibrium. The forward and reverse rates are equal and the reactant and product concentrations remain constant. If Q > Keq, the system is NOT at equilibrium. There is too much product, so the reverse reaction is favoured to bring the reactant-product ratio to equal Keq by increasing reactant concentration. If Q < Keq, the system is NOT at equilibrium. The concentration of reactants is too large, so the forward reaction is favoured. This results in decreasing reactant concentrations and increasing product concentrations, bringing their ratio to a value equal to Keq. Example 1: For the reaction N2 (g) + O2 (g) 2NO (g), it was found that 8.50 moles of nitrogen, 11.0 moles of oxygen and 2.20 moles of nitrogen monoxide were in a 5.00 L container. If the equilibrium constant is 0.035, are the following concentrations at equilibrium? If not, which reaction is favoured and which concentrations are increasing and which are decreasing? 28

Example 2: For the following imaginary equilibrium system 2A (g) + B (g) 3C (g), the value of Keq is 0.22. 1.50 moles of A and 3.20 moles of B are placed into a 1.0 L container and allowed to react. After several minutes, a sample is taken and found to contain 1.00 mole of A. Is the system at equilibrium? Which reaction rate is fastest? Which concentrations are increasing? 29

Example 3: Keq = 49 for 2NO (g) + O2 (g) 2NO2 (g). If 2.0 moles of NO (g), 0.20 mole of O2 (g) and 0.40 mole of NO2 (g) are put into a 2.0 L container, which way will the reaction shift in order to reach equilibrium? 30

Practice: Predicting Equilibrium 1. For the reaction A (s) + 2B (g) 2C (g), Keq = 25.0. If [A] = 12.0 mol/l, [B] = 2.0 mol/l, and [C] = 30.0 mol/l: a) Is the system is at equilibrium? b) Which reaction is favoured, forward or reverse? c) Which concentrations are increasing or decreasing? 2. There exists an equilibrium if 5.0 moles of CO2, 5.0 moles of CO and 0.20 moles of O2 are in a 2.0 L container at 562 C. Find Keq for the reaction 2CO (g) + O2 (g) 2CO2 (g). Would the system be at equilibrium if [CO2] = 15.8 mol/l, [CO] = 10.0 mol/l and [O2] = 0.25 mol/l? If not, which reaction is favoured? 31

3. For the reaction 2SO2 (g) + O2 (g) 2SO3 (g), Keq = 16.0. Initially, [SO2] = 5.0 mol/l, [O2] = 10.0 mol/l, and [SO3] = 0.0 mol/l. After two hours [O2] = 7.9 mol/l. Is the system at equilibrium? If not, which substances are increasing and which are decreasing? 4. The reaction 4NH3 (g) + 5O2 (g) 4NO2 (g) + 6H2O (g) is at equilibrium when [H2O] = 0.100 mol/l, [O2] = 2.00 mol/l, [NO] = 0.200 mol/l and [NH3] = 0.500 mol/l. If 0.75 mole of H2O, 12.0 moles of NO, 30.0 moles of O2, and 0.30 mole of NH3 are in a 3.0 L container at the same temperature, is equilibrium achieved? If not, which reaction is favoured? 32

5. Keq = 46.0 for the reaction H2 (g) + I2 (g) 2HI (g). Initially there are 6.90 moles of H2 and 2.40 moles of I2 in a 1.00 L container. After 5 hours there is still 1.00 moles of I2 left. Is the system at equilibrium? If not, which substances are increasing and which are decreasing? 33

Lesson 4: Le Chatelier s Principle Goal: Use Le Chatelier s Principle to explain the effects of changing concentration, changing pressure, changing temperature and adding a catalyst on the position of a system at equilibrium. Concentration Changes In a system at equilibrium, a change in the concentration of products or reactants present at equilibrium creates a stress. At equilibrium, the ratio of product to reactant concentrations is constant. 34

Pressure Changes Changing the pressure of a system only affects those equilibria with gaseous reactants and/or products. Example 1: For the reaction N2 (g) + 3H2 (g) 2NH3 (g): a) What is the effect on the equilibrium if the size of the container is cut in half, but the number of particles and temperature remain unchanged? 35

b) What is the effect on equilibrium if the reaction chamber is increased in volume, while keeping temperature and total number of particles constant? For the reaction H2 (g) + I2 (g) 2HI (g), pressure changes would have NO effect on the equilibrium position. Each side of the reaction has two moles of molecules. There is no way to either increase or reduce the number of particles. Therefore, in response to pressure changes, the equilibrium position remains unchanged. Temperature Changes Recall from Kinetics that increasing temperature always increases the rate of a reaction. However, increasing temperature always increases the rate of an endothermic reaction more than the rate of an exothermic reaction. 36

Example 2: The conversion of dinitrogen tetroxide to nitrogen dioxide is reversible and temperature dependent. N2O4 (g) 2NO2 (g) ΔH = +58 kj colourless brown Temperature and Keq 37

Effect of a Catalyst Practice: Le Chatelier s Principle 1. For the reaction PCl3 (g) + Cl2 (g) PCl5 (g), ΔH = -92.5 kj, predict the effect on the position of the equilibrium that results from a) Increasing the total pressure by decreasing volume. b) Injecting more Cl2 gas without changing the volume. c) Increasing the temperature. d) Increasing the volume of the container. e) Adding a catalyst. 38

2. For CH4 (g) + H2O (g) + 49.3 kj CO (g) + 3H2 (g), predict the effect on the position of the equilibrium that results from a) Increasing temperature. b) Decreasing temperature. c) Decreasing the pressure. d) Decreasing the volume of the container. e) Adding a solid drying agent such as CaCl2, which reacts with H2O (g). 3. For the reaction 9.4 kj + 2HI (g) H2 (g) + I2 (g). a) What is the effect on [HI] if a small amount of H2 is added? b) What is the effect on [HI] if the pressure of the system is increased? c) What is the effect on [HI] if the temperature is increased? d) What is the effect on [HI] if a catalyst is added? 39

4. For the reaction CO (g) + 2H2 (g) CH3OH (g) + energy, predict the effect of the following changes on the equilibrium concentration of CH3OH (g). a) A decrease in temperature. b) An increase in pressure. c) Addition of H2 (g). d) Addition of a catalyst. 5. In the reaction 2NO (g) + O2 (g) 2NO2 (g) + 114.6 kj. What will be the change in the equilibrium of [NO2] under each of the following conditions? a) O2 is added. b) NO is removed. c) Energy is added. 6. For the reaction N2O4 (g) 2NO2 (g), ΔH = +58.9 kj, how will the equilibrium of [NO2] be affected by the following? a) An increase in pressure. b) An increase in temperature. c) The addition of a catalyst. 40

Lesson 5: Graphs Goals: Interpret rate vs. time graphs for equilibrium systems. Interpret concentration vs. time graphs for equilibrium systems. Rate vs. Time Graphs for Changing Reactant Concentrations Let s analysis the Rate vs. Time graphs for the equilibrium system of 2NO2 (g) N2O4 (g), ΔH = -58.0 kj/mol N2O4 and its instantaneous change after adding/removing reactant to/from the system. Adding more NO2 to the system. Removing some NO2 from the system. 41

Rate vs. Time Graphs for Changing Product Concentrations Let s analysis the Rate vs. Time graphs for the equilibrium system of 2NO2 (g) N2O4 (g) and its instantaneous change after adding/removing product to/from the system. Adding more N2O4 to the system. Removing some N2O4 from the system. 42

Concentration vs. Time Graphs for Changing Reactant Concentrations Let s analysis the Concentration vs. Time graphs for the equilibrium system of 2NO2 (g) N2O4 (g), ΔH = -58.0 kj/mol N2O4 and its instantaneous change after adding/removing reactant to/from the system. Adding more NO2 to the system. Removing some NO2 from the system. 43

Concentration vs. Time Graphs for Changing Product Concentrations Let s analysis the Rate vs. Time graphs for the equilibrium system of 2NO2 (g) N2O4 (g) and its instantaneous change after adding/removing product to/from the system. Adding more N2O4 to the system. Removing some N2O4 from the system. 44

Concentration vs. Time Graphs for Changing Temperature We will now look at the graphs that illustrate the effect of changing temperature on reactant and product concentrations. A forward reaction is exothermic. Increasing temperature. Decreasing Temperature. 45

Practice: Graphs The graph below shows concentration versus time for a system containing carbon monoxide (CO), dichlorine (Cl2), and phosgene (COCl2). (Bodenstein and Plaut studied this system (Z. physik. Chemie, 1924, 110, 399 416).) 1. Write a balanced equation to represent the reaction studied. 2. How much time was required for the system to reach equilibrium? 3. Calculate an approximate value for the equilibrium constant, K, using the concentrations at time t = 60 s. 4. Explain the changes 70 s after the initiation of the reaction. 46

5. What changes in conditions might have been imposed on the system 120 s after the initiation of the reaction? 6. Are any events taking place between the interval 50 s and 70 s? Between 280 s and 300 s? Explain your answers. 47

Lesson 6: Solubility Equilibrium Goals: Describe and write a balanced chemical equation to represent the equilibrium in a saturated aqueous solution of an ionic compound. Write a solubility product expression, given a balanced chemical equation for a solubility reaction. Distinguish between solubility and solubility product constant, (Ksp). Calculate the solubility product, given the solubility of a compound in water, and vice versa. Saturated solution Solubility Product When a sparingly soluble ionic solid is dissolved in water to form a saturated solution the general equilibrium equation is AaBb (s) aa + (aq) + bb - (aq). 48

However, the term Keq[AaBb] can be replaced by a new constant, Ksp, called the solubility product. The equilibrium law for the dissolved compound is Ksp = [A + ] a [B - ] b. The solubility product constant is the product of ion concentrations in a saturated solution. The solubility product constant takes into account the presence of the solid. Example 1: Write the dissociation equation and the expression for the solubility product constant for calcium hydroxide. Example 2: Write the solubility product expression for Pb3(PO4)2. Example 3: If at equilibrium, the concentration of silver ions is 1.3 10-5 mol/l and the concentration of chloride ions is 1.3 10-5 mol/l, what is the Ksp of silver chloride? 49

Solubility Solubility product Example 4: The solubility of PbF2 is 0.466 g/l. What is the value of the solubility product? 50

Example 5: The Ksp of magnesium hydroxide is 8.9 10-12. What will be the equilibrium concentrations of the dissolved ions in a saturated solution of Mg(OH)2? 51

Example 6: What is the solubility, in grams per Litre for the magnesium hydroxide when the Ksp = 8.9 10-12. Practice: Solubility Equilibrium 1. Write the dissociation equation and the solubility product expression for each of the solid below. Assume that all the solid that dissolves exists as ions. a) Ba(OH)2 b) CuCl c) Ag2CO3 d) Fe2(SO4)3 52

2. Given the following compounds Ksp, calculate their solubilities in mol/l and g/l. a) PbI2, Ksp = 1.39 10-8 b) SrC2O4, Ksp = 1.58 10-7 c) Al(OH)3, Ksp = 1.26 10-33 53

3. For the following solubilities below, calculate the Ksp. a) Pb(OH)2, 4.20 10-6 mol/l b) AgI, 2.88 10-6 g/l c) Ca3(PO4)2, 7.15 10-7 mol/l d) CaF2, 1.70 10-2 g/l 54

4. If 6.7 10-5 g of AgBr is all that can be dissolved at 25 C in 500.0 ml, calculate the solubility product of AgBr. 5. A saturated solution of calcium hydroxide has an hydroxide ion concentration of 3.0 10-3 mol/l. Calculate the Ksp of calcium hydroxide. 6. What are the equilibrium concentrations of all the ions in a saturated solution of AgCN at 25 C, if the Ksp is 1.6 10-14? 55

7. At 25 C, a saturated solution of iron (III) hydroxide has an iron (III) ion concentration of 1.3 10-13 mol/l. Calculate the Ksp of iron (III) hydroxide. 8. What are the equilibrium concentrations of all the ions in a saturated solution of Cu(OH)2 at 25 C, if the Ksp is 1.6 10-19. 56

Lesson 7: Precipitates Goals: Use the reaction quotient to predict whether a solution will be saturated or unsaturated. Use the reaction quotient to predict whether a precipitate will form when two solutions are mixed. Saturated Solutions Not all reactions produce a precipitate. If the volume of solution is large enough, and the amount of solute is small enough no precipitate will form. Example 1: The Ksp of lead (II) chloride is 1.6 10-5. If 0.57 g of lead (II) chloride is added to 1500 ml of water, is the solution saturated? Assume no volume change. 57

Example 2: If 20.0 ml of a 0.0010 mol/l silver nitrate solution is mixed with 20.0 ml of a 3.0 10-5 mol/l potassium bromide solution, does silver bromide precipitate? The Ksp of silver bromide is 5.0 10-13. Assume the volumes are additive. Practice: Precipitates 1. To 1.0 L of 1.0 M H2SO4 is added 0.0020 mole of solid Pb(NO3)2. As the lead nitrate dissolves, will lead sulfate precipitate if the Ksp is 1.3 10-8? 58

2. The Ksp of CaF2 at 25 C is 1.7 10-10. If 0.75 g of CaF2 are dissolved in 25.0 L of hot water then cooled to 25 C, will a precipitate form? Assume no volume change. 3. If 2.5 10-5 moles of aluminum hydroxide are added to 10.0 L of water, will all the solid dissolve if the Ksp is 5.0 10-33? 4. The Ksp of PbSO4 is 1.3 10-8. If 0.20 g of solid lead (II) sulfate is added to 7.5 L of water, will all the solid dissolve? 59

5. If equal volumes of 0.020 mol/l TlNO3 and 0.0040 mol/l NaCl are mixed, will a precipitate form if the Ksp is 1.9 10-4? 6. 50.0 ml of 0.040 mol/l calcium nitrate solution is added to 150.0 ml of 0.0080 mol/l ammonium sulphate solution. Does a precipitate form if the Ksp is 2.6 10-4? Justify your answer. 60

7. Does a precipitate form when 2.0 10-3 moles of strontium nitrate are added to 50.0 ml of 4.2 10-6 mol/l ammonium sulphate? Justify your answer. The Ksp is 7.6 10-7. 8. If 2.5 ml of 0.30 mol/l AgNO3 is mixed with 7.5 ml of 0.015 mol/l Na2CrO4 will a precipitate form if the Ksp is 9.2 10-12? 61

Lesson 8: Common Ions Goal: Calculate the solubility of an ionic compound in the presence of an ion in common with the ionic compound. The Common Ion Effect Example 1: Determine the solubility of silver chloride in pure water and in a solution of 0.10 mol/l sodium chloride. The Ksp of AgCl is 1.7 10-10. 62

Example 2: The Ksp of lead (II) chloride, PbCl2, is 1.6 10-5. What is the solubility of lead (II) chloride in a 0.10 mol/l solution of magnesium chloride, MgCl2? 63

Example 3: The Ksp of lead (II) chloride is 1.6 10-5. What is the solubility of lead (II) chloride in a 0.10 mol/l solution of lead (II) nitrate, Pb(NO3)2? 64

Practice: Common Ions 1. Silver iodide, AgI, has a solubility product of 8.5 10-17. What is the solubility, in mol/l, of AgI in: a) pure water b) 0.010 mol/l HI 65

c) 0.010 mol/l MgI2 d) 0.010 mol/l AgNO3 66

2. Magnesium fluoride, MgF2, has a solubility product of 8.0 10-8. Calculate the solubility, in mol/l, of magnesium fluoride in: a) pure water b) 0.50 mol/l NaF c) 0.50 mol/l MgCl2 67

3. Gold (III) chloride, AuCl3, has a Ksp of 3.2 10 25. Calculate its solubility, in mol/l, in: a) pure water b) 0.20 mol/l HCl 68

c) 0.20 mol/l MgCl2 d) 0.20 mol/l Au(NO3)3 69

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