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How to Use This Presentation • To View the presentation as a slideshow with effects select “View” on the menu bar and click on “Slide Show.” • To advance through the presentation, click the right-arrow key or the space bar. • From the resources slide, click on any resource to see a presentation for that resource. • From the Chapter menu screen click on any lesson to go directly to that lesson’s presentation. • You may exit the slide show at any time by pressing the Esc key. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Resources Chapter Presentation Bellringer Transparencies Sample Problems Visual Concepts Standardized Test Prep Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Chemical Equilibrium Table of Contents Section 1 Reversible Reactions and Equilibrium Section 2 Systems at Equilibrium Section 3 Equilibrium Systems and Stress Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Bellringer • Describe what reversible means. • Find a synonym for reversible. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Objectives • Contrast reactions that go to completion with reversible ones. • Describe chemical equilibrium. • Give examples of chemical equilibria that involve complex ions. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Completion Reactions and Reversible Reactions • If enough oxygen gas is provided for the following reaction, almost all of the sulfur will react: S8(s) + 8O2(g) → 8SO2(g) • Reactions such as this one, in which almost all of the reactants react, are called completion reactions. • In other reactions, called reversible reactions, the products can re-form reactants. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Completion Reactions and Reversible Reactions, continued Reversible Reactions Reach Equilibrium • One reversible reaction occurs when you mix solutions of calcium chloride and sodium sulfate. CaCl2(aq) + Na2SO4(aq) → CaSO4(s) + 2NaCl(aq) • The net ionic equation best describes what happens. Ca2 (aq ) SO24 (aq ) CaSO 4 (s ) Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Completion Reactions and Reversible Reactions, continued Reversible Reactions Reach Equilibrium, continued • Solid calcium sulfate, the product, can break down to make calcium ions and sulfate ions in a reaction that is the reverse of the previous one. CaSO4 (s ) Ca2 (aq ) SO24 (aq ) • Use arrows that point in opposite directions when writing a chemical equation for a reversible reaction. CaSO4 (s ) Ca2 (aq ) SO2 4 (aq ) Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Completion Reactions and Reversible Reactions, continued Reversible Reactions Reach Equilibrium, continued • The reactions occur at the same rate after the initial mixing of CaCl2 and Na2SO4. • The amounts of the products and reactants do not change. • Chemical equilibrium is a state of balance in which the rate of a forward reaction equals the rate of the reverse reaction and the concentrations of products and reactants remain unchanged. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Completion Reactions and Reversible Reactions, continued Opposing Reaction Rates Are Equal at Equilibrium • The reaction of hydrogen, H2, and iodine, I2, to form hydrogen iodide, HI, reaches chemical equilibrium. 2HI(g ) H2 (g ) I2 (g ) • Only a very small fraction of the collisions between H2 and I2 result in the formation of HI. H2(g) + I2(g) → 2HI(g) Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Completion Reactions and Reversible Reactions, continued Opposing Reaction Rates Are Equal at Equilibrium, continued • After some time, the concentration of HI goes up. • As a result, fewer collisions occur between H2 and I2 molecules, and the rate of the forward reaction drops. • Similarly, in the beginning, few HI molecules exist in the system, so they rarely collide with each other. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Completion Reactions and Reversible Reactions, continued Opposing Reaction Rates Are Equal at Equilibrium, continued • As more HI molecules are made, they collide more often and form H2 and I2 by the reverse reaction. 2HI(g) → H2(g) + I2(g) • The greater the number of HI molecules that form, the more often the reverse reaction occurs. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Rate Comparison for H2(g) + I2(g) 2HI(g) Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Completion Reactions and Reversible Reactions, continued Opposing Reaction Rates Are Equal at Equilibrium, continued • When the forward rate and the reverse rate are equal, the system is at chemical equilibrium. • If you repeated this experiment at the same temperature, starting with a similar amount of pure HI instead of the H2 and I2, the reaction would reach chemical equilibrium again and produce the same concentrations of each substance. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Visual Concepts Chemical Equilibrium Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Chemical Equilibria Are Dynamic • If you drop a ball into a bowl, it will bounce. • When the ball comes to a stop it has reached static equilibrium, a state in which nothing changes. • Chemical equilibrium is different from static equilibrium because it is dynamic. • In a dynamic equilibrium, there is no net change in the system. • Two opposite changes occur at the same time. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Chemical Equilibria Are Dynamic, continued • In equilibrium, an atom may change from being part of the products to part of the reactants many times. • But the overall concentrations of products and reactants stay the same. • For chemical equilibrium to be maintained, the rates of the forward and reverse reactions must be equal. • Arrows of equal length also show equilibrium. products reac tants Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium Chemical Equilibria Are Dynamic, continued • In some cases, the equilibrium has a higher concentration of products than reactants. • This type of equilibrium is also shown by using two arrows. reac tants products • The forward reaction has a longer arrow to show that the products are favored. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium More Examples of Equilibria • Even when systems are not in equilibrium, they are continuously changing in order to reach equilibrium. • For example, combustion produces carbon dioxide, CO2, and poisonous carbon monoxide, CO. After combustion, a reversible reaction produces soot. C(s ) CO2 (g ) 2CO(g ) • This reaction of gases and a solid will reach chemical equilibrium. • Equilibria can involve any state of matter, including aqueous solutions. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium More Examples of Equilibria, continued Equilibria Involving Complex Ions • Complex ion, or coordination compound, is the name given to any metal atom or ion that is bonded to more than one atom or molecule. • Some ions have a metal ion surrounded by ligands, molecules or anions that readily bond to metal ions. • Complex ions may be positively charged cations or negatively charged anions. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued • In this complex ion, [Cu(NH3)4]2+, ammonia molecules bond to the central copper(II) ion. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued Complex ions formed from transition metals are often deeply colored. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued • The charge on a complex ion is a sum of the charges on the species from which the complex ion forms. • For example, when the cobalt ion, Co2+, bonds with four Cl− ligands, the total charge is (+2) + 4(−1) = −2. • Metal ions and ligands can form complexes that have no charge. These are not complex ions. • Complex ions often form in systems that reach equilibrium. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued • Consider zinc nitrate dissolving in water: Zn2 (aq ) 2NO3- (aq ) Zn(NO3 )2 (s ) • In the absence of other ligands, water molecules bond with zinc ions. So, this reaction can be written: [Zn(H2O)4 ]2 (aq ) 2NO3- (aq ) Zn(NO3 )2 (s ) 4H2O(l ) • If another ligand, such as CN−, is added, the new system will again reach chemical equilibrium. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued • Both water molecules and cyanide ions “compete” to bond with zinc ions, as shown in the equation below. [Zn(CN)4 ]2– (aq ) 4H2O( l ) [Zn(H2O)4 ]2 (aq ) 4CN– (aq ) • All of these ions are colorless, so you cannot see which complex ion has the greater concentration. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 1 Reversible Reactions and Equilibrium More Examples of Equilibria, continued Equilibria Involving Complex Ions, continued • In the chemical equilibrium of nickel ions, ammonia, and water, the complex ions have different colors. • You can see which ion has the greater concentration. 2 [Ni(H2O)6 ]2 (aq ) 6NH3 (aq ) [Ni(NH ) ] (aq ) 6H2O( l ) 3 6 green blue-violet • The starting concentration of NH3 will determine which one will have the greater concentration. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium Bellringer • Make a list of numbers that are “constants” under constant conditions. • An example is the speed of light. • What do these constants have in common? • Answer: Each constant is always the same number for a certain and constant set of conditions. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium Objectives • Write Keq expressions for reactions in equilibrium, and perform calculations with them. • Write Ksp expressions for the solubility of slightly soluble salts, and perform calculations with them. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Equilibrium Constant, Keq • Limestone caverns form as rainwater, slightly acidified by H3O+, dissolves calcium carbonate. • The reverse reaction also takes place, depositing calcium carbonate and forming stalactites and stalagmites. Ca2 (aq ) CO2 (g ) 3H2O(l ) CaCO3 (s ) 2H3O (aq ) • When the rates of the forward and reverse reactions become equal, the reaction reaches chemical equilibrium. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Equilibrium Constant, Keq, continued • There is a mathematical relationship between product and reactant concentrations at equilibrium. • For limestone reacting with acidified water at 25°C: [Ca2 ][CO2 ] -9 Keq 1.4 10 [H3O ]2 • Keq is the equilibrium constant of the reaction. • Keq for a reaction is unitless, applies only to systems in equilibrium, and depends on temperature and must be found experimentally or from tables. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Visual Concepts Equilibrium Constant Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Equilibrium Constant, Keq, continued Determining Keq for Reactions at Chemical Equilibrium 1. Write a balanced chemical equation. • Make sure that the reaction is at equilibrium before you write a chemical equation. 2. Write an equilibrium expression. • To write the expression, place the product concentrations in the numerator and the reactant concentrations in the denominator. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Equilibrium Constant, Keq, continued Determining Keq for Reactions at Chemical Equilibrium, continued • The concentration of any solid or a pure liquid that takes part in the reaction is left out. • For a reaction occurring in aqueous solution, water is omitted. 3. Complete the equilibrium expression. • Finally, raise each substance’s concentration to the power equal to the substance’s coefficient in the balanced chemical equation. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Visual Concepts Equilibrium Constant Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium Calculating Keq from Concentrations of Reactants and Products Sample Problem A An aqueous solution of carbonic acid reacts to reach equilibrium as described below. – H2CO3 (aq ) H2O(l ) HCO ( aq ) H O (aq ) 3 3 The solution contains the following solution concentrations: carbonic acid, 3.3 × 10−2 mol/L; bicarbonate ion, 1.19 × 10−4 mol/L; and hydronium ion, 1.19 × 10−4 mol/L. Determine the Keq. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Section 2 Systems at Equilibrium Chapter 14 Calculating Keq from Concentrations of Reactants and Products Sample Problem A Solution [H2CO3 ] 3.3 10-2 [HCO3– ] [H3O ] 1.19 10-4 For this reaction, the equilibrium constant expression is K eq [HCO3– ][H3O ] [H2CO3 ] Substitute the concentrations into the expression. Keq [HCO3– ][H3O ] (1.19 10-4 ) (1.19 10-4 ) -7 4.3 10 [H2CO3 ] (3.3 10-2 ) Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Equilibrium Constant, Keq, continued Keq Shows If the Reaction Is Favorable • When Keq is large, the numerator of the equilibrium constant expression is larger than the denominator. • Thus, the concentrations of the products will usually be greater than those of the reactants. • In other words, when a reaction that has a large Keq reaches equilibrium, there will be mostly products. • Reactions in which more products form than reactants form are said to be “favorable.” Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Equilibrium Constant, Keq, continued Keq Shows If the Reaction Is Favorable, continued • The synthesis of ammonia is very favorable at 25°C and has a large Keq value. 2NH3 (g ) N2 (g ) 3H2 (g ) K eq [NH3 ]2 8 3.3 10 [N2 ][H2 ]3 Chapter menu at 25 C Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Equilibrium Constant, Keq, continued Keq Shows If the Reaction Is Favorable, continued • However, the reaction of oxygen and nitrogen to give nitrogen monoxide is not favorable at 25°C. 2NO(g ) N2 (g ) O2 (g ) K eq [NO2 ]2 4.5 10 –31 [N2 ][O2 ] Chapter menu at 25 C Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Equilibrium Constant, Keq, continued Keq Shows If the Reaction Is Favorable, continued • When Keq is small, the denominator of the equilibrium constant expression is larger than the numerator. • The larger denominator shows that the concentrations of reactants at chemical equilibrium may be greater than those of products. • A reaction that has larger concentrations of reactants than concentrations of products is an “unfavorable” reaction. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Equilibrium Constant, Keq, continued Keq Shows If the Reaction Is Favorable, continued These pie charts show the relative amounts of reactants and products for three Keq values of a reaction. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium Calculating Concentrations of Products from Keq and Concentrations of Reactants Sample Problem B Keq for the equilibrium below is 1.8 × 10−5 at a temperature of 25°C. Calculate [NH4 ] when [NH3] = 6.82 × 10−3. – NH3 (aq ) H2O(l ) NH ( aq ) OH (aq ) 4 Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium Calculating Concentrations of Products from Keq and Concentrations of Reactants, continued Sample Problem B Solution The equilibrium expression is NH4and OH− ions are produced in equal numbers, so [OH– ] [NH4 ]. So, the numerator can be written as x2. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium Calculating Concentrations of Products from Keq and Concentrations of Reactants, continued Sample Problem B Solution, continued Keq and [NH3] are known and can be put into the expression. 1.8 10–5 Keq [NH4 ][OH– ] x2 [NH3 ] 6.82 10-3 x2 = (1.8 10−5) (6.82 10−3) = 1.2 × 10−7 Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium Calculating Concentrations of Products from Keq and Concentrations of Reactants, continued Sample Problem B Solution, continued Take the square root of x2. [NH4 ] 1.2 10-7 3.5 10-4 Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Solubility Product Constant, Ksp • The maximum concentration of a salt in an aqueous solution is called the solubility of the salt in water. • Solubilities can be expressed in moles of solute per liter of solution (mol/L or M). • For example, the solubility of calcium fluoride in water is 3.4 × 10−4 mol/L. • So, 0.00034 mol of CaF2 will dissolve in 1 L of water to give a saturated solution. • If you try to dissolve 0.00100 mol of CaF2 in 1 L of water, 0.00066 mol of CaF2 will remain undissolved. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Visual Concepts Solution Equilibrium Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Solubility Product Constant, Ksp, continued • Calcium fluoride is one of a large class of salts that are said to be slightly soluble in water. • The ions in solution and any solid salt are at equilibrium. 2 – CaF2 (s ) Ca ( aq ) 2F (aq ) • Solids are not a part of equilibrium constant expressions, so Keq for this reaction is the product of [Ca2+] and [F−]2, which is equal to a constant. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Solubility Product Constant, Ksp, continued • Equilibrium constants for the dissolution of slightly soluble salts are called solubility product constants, Ksp, and have no units. • The Ksp for calcium fluoride at 25°C is 1.6 10−10. Ksp = [Ca2+][F−]2 = 1.6 10−10 • This relationship is true whenever calcium ions and fluoride ions are in equilibrium with calcium fluoride, not just when the salt dissolves. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Solubility Product Constant, Ksp, continued • For example, if you mix solutions of calcium nitrate and sodium fluoride, calcium fluoride precipitates. • The net ionic equation for this precipitation is the reverse of the dissolution. CaF2 (s ) Ca2 (aq ) 2F– (aq ) • This equation is the same equilibrium. So, the Ksp for the dissolution of CaF2 in this system is the same and is 1.6 × 10−10. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Visual Concepts Solubility Product Constant Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Visual Concepts Solubility Product Constant Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Solubility Product Constant, Ksp, continued Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Solubility Product Constant, Ksp, continued Determining Ksp for Reactions at Chemical Equilibrium 1. Write a balanced chemical equation. • Solubility product is only for salts that have low solubility. Soluble salts do not have Ksp values. • Make sure that the reaction is at equilibrium. • Equations are always written so that the solid salt is the reactant and the ions are products. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Solubility Product Constant, Ksp, continued Determining Ksp for Reactions at Chemical Equilibrium 2. Write a solubility product expression. • Write the product of the ion concentrations. • Concentrations of solids or liquids are omitted. 3. Complete the solubility product expression. • Raise each concentration to a power equal to the substance’s coefficient in the balanced chemical equation. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Visual Concepts Using Solubility Product Constants Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium Calculating Ksp from Solubility Sample Problem C Most parts of the oceans are nearly saturated with CaF2.The mineral fluorite, CaF2, may precipitate when ocean water evaporates. A saturated solution of CaF2 at 25°C has a solubility of 3.4 × 10−4 M. Calculate the solubility product constant for CaF2. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium Calculating Ksp from Solubility, continued Sample Problem C Solution 2 – CaF2 (s ) Ca ( aq ) 2F (aq ) [CaF2] = 3.4 10–4, [F−] = 2[Ca2+] Ksp = [Ca2+][F−]2 Because 3.4 × 10−4 mol CaF2 dissolves in each liter of solution, you know from the balanced equation that every liter of solution will contain 3.4 × 10−4 mol Ca2+ and 6.8 × 10−4 mol F−. Thus, the Ksp is: [Ca2+][F−]2 = (3.4 10−4)(6.8 10−4)2 = 1.6 × 10−10 Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium Calculating Ionic Concentrations Using Ksp, Sample Problem D Copper(I) chloride has a solubility product constant of 1.2 × 10−6 and dissolves according to the equation below. Calculate the solubility of this salt in ocean water in which the [Cl−] = 0.55. – CuCl(s ) Cu ( aq ) Cl (aq ) Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium Calculating Ionic Concentrations Using Ksp, continued Sample Problem D Solution The product of [Cu+][Cl−] must equal Ksp = 1.2 × 10−6. [Cl−] = 0.55 Ksp = [Cu+][Cl−] = 1.2 × 10−6 –6 K 1.2 10 [Cu ] sp– 2.2 10–6 [Cl ] 0.55 This is the solubility of copper(I) chloride because the dissolution of 1 mol of CuCl produces 1 mol of Cu+. Therefore, the solubility of CuCl is 2.2 × 10−6 mol/L. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Solubility Product Constant, Ksp, continued Using Ksp to Make Magnesium • Though slightly soluble hydroxides are not salts, they have solubility product constants. • Magnesium hydroxide is an example. 2 – Mg(OH)2 (s ) Mg ( aq ) 2OH (aq ) [Mg2+][OH−]2 = Ksp = 1.8 × 10−11 • This equilibrium is the basis for obtaining magnesium. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Solubility Product Constant, Ksp, continued Using Ksp to Make Magnesium, continued • The table at right lists the most abundant ions in ocean water and their concentrations. • Mg2+ is the third most abundant ion in the ocean. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Solubility Product Constant, Ksp, continued Using Ksp to Make Magnesium, continued • To get magnesium, calcium hydroxide is added to sea water, which raises the hydroxide ion concentration to a large value so that [Mg2+][OH−]2 would be greater than 1.8 × 10−11. • Magnesium hydroxide precipitates. • Magnesium hydroxide is treated with hydrochloric acid to make magnesium chloride, MgCl2. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 2 Systems at Equilibrium The Solubility Product Constant, Ksp, continued Using Ksp to Make Magnesium, continued • Finally, magnesium is obtained by the electrolysis of MgCl2 in the molten state. • One cubic meter of sea water yields 1 kg of magnesium metal. • Because of magnesium’s low density and rigidity, alloys of magnesium are used when light weight and strength are needed. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Bellringer • List examples of everyday adjustments that are made to relieve stress on a system. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Objectives • State Le Châtelier’s principle. • Apply Le Châtelier’s principle to determine whether the forward or reverse reaction is favored when a stress such as concentration, temperature, or pressure is applied to an equilibrium system. • Discuss the common-ion effect in the context of Le Châtelier’s principle. • Discuss practical uses of Le Châtelier’s principle. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle • Stress is another word for something that causes a change in a system at equilibrium. • Chemical equilibrium can be disturbed by a stress, but the system soon reaches a new equilibrium. • Le Châtelier’s principle states that when a system at equilibrium is disturbed, the system adjusts in a way to reduce the change. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued • Chemical equilibria respond to three kinds of stress: • changes in the concentrations of reactants or products • changes in temperature • changes in pressure • When a stress is first applied to a system, equilibrium is disturbed and the rates of the forward and backward reactions are no longer equal. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued • The system responds to the stress by forming more products or by forming more reactants. • A new chemical equilibrium is reached when enough reactants or products form. • At this point, the rates of the forward and backward reactions are equal again. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Visual Concepts Le Châtelier's Principle Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Changes in Concentration Alter Equilibrium Composition • If you increase a reactant’s concentration, the system will respond to decrease the concentration of the reactant by changing some of it into product. • Therefore, the rate of the forward reaction must be greater than the rate of the reverse reaction. • The equilibrium is said to shift right, and the reactant concentration drops until the reaction reaches equilibrium. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Changes in Concentration Alter Equilibrium Composition, continued • In a reaction of two colored complex ions: 2 [Cu(H2O)4 ]2 (aq) 4NH3 (aq ) [Cu(NH ) ] (aq ) 4H2O( l ) 3 4 pale blue blue-purple • When the reaction mixture in a beaker is pale blue, we know that chemical equilibrium favors the formation of reactants. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Changes in Concentration Alter Equilibrium Composition, continued If ammonia is added, the system responds by forming more product and the solution becomes blue-purple. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Changes in Concentration Alter Equilibrium Composition, continued • The equilibrium below occurs in a bottle of soda. 2H2O(l ) CO2 (aq ) H3O (aq ) HCO3– (aq ) • After you uncap the bottle, the dissolved carbon dioxide leaves the solution and enters the air. • The forward reaction rate of this system will increase to produce more CO2. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Changes in Concentration Alter Equilibrium Composition, continued • This increase in the rate of the forward reaction decreases the concentration of H3O+. • As a result, the drink gets “flat.” • If you could increase the concentration of CO2 in the bottle, the reverse reaction rate would increase, and [H3O ] and [HCO3– ] would increase. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Temperature Affects Equilibrium Systems • The effect of temperature on the gas-phase equilibrium of nitrogen dioxide, NO2, and dinitrogen tetroxide, N2O4, can be seen because of the difference in color of NO2 and N2O4. • The intense brown NO2 gas is the pollution that is responsible for the colored haze that you sometimes see on smoggy days. N2O4 (g ) 2NO2 (g ) brown colorless Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Temperature Affects Equilibrium Systems, continued • Recall that endothermic reactions absorb energy and have positive ∆H values. • Exothermic reactions release energy and have negative ∆H values. • The forward reaction is an exothermic process, as the equation below shows. 2NO2(g) → N2O4(g) ∆H = −55.3 kJ Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Temperature Changes Affect an Equilibrium System Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Temperature Affects Equilibrium Systems, continued • Now suppose that you heat the flask to 100°C. • The mixture becomes dark brown because the reverse reaction rate increased to remove some of the energy that you added to the system. • The equilibrium shifts to the left, toward the formation of NO2. • Because this reaction is endothermic, the temperature of the flask drops as energy is absorbed. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Temperature Affects Equilibrium Systems, continued • This equilibrium shift is true for all exothermic forward reactions: • Increasing the temperature of an equilibrium mixture usually leads to a shift in favor of the reactants. • The opposite statement is true for endothermic forward reactions: • Increasing the temperature of an equilibrium mixture usually leads to a shift in favor of the products. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Temperature Affects Equilibrium Systems, continued • The following is an endothermic reaction involving two colored cobalt complex ions: 2– [Co(H2O)6 ]2 (aq ) 4Cl– (aq) [CoCl ] (aq ) 6H2O(l ) 4 pink blue • The forward reaction is endothermic, so the forward reaction is favored at high temperatures. • The reverse reaction is favored at low temperatures. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Temperature Affects Equilibrium Systems, continued • Equilibrium changes with changes in temperature they affect the value of equilibrium constants. • Consider Keq for the ammonia synthesis equilibrium: 2NH3 (g ) N2 (g ) 3H2 (g ) • The forward reaction is exothermic (∆H = −91.8 kJ), so the equilibrium constant decreases a lot as temperature increases. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Visual Concepts Factors Affecting Equilibrium Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Pressure Changes May Alter Systems in Equilibrium • Changes in pressure can affect gases in equilibrium. • The NO2 and N2O4 equilibrium can show the effect of a pressure stress on a chemical equilibrium. 2NO2(g) → N2O4(g) • When the gas mixture has a larger concentration of N2O4 than of NO2, it has a pale color. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Pressure Changes May Alter Systems in Equilibrium When the gas is suddenly compressed to about half its former volume, the pressure doubles. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Pressure Changes May Alter Systems in Equilibrium, continued • Before the system has time to adjust to the pressure stress, the concentration of each gas doubles, which results in a darker color. • Le Châtelier’s principle predicts that the system will adjust in an attempt to reduce the pressure. • According to the equation, 2 mol of NO2 produce 1 mol of N2O4. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Pressure Changes May Alter Systems in Equilibrium, continued • At constant volume and temperature, pressure is proportional to the number of moles. • So, the pressure reduces when there are fewer moles of gas. • Thus, the equilibrium shifts to the right, and more N2O4 is produced. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Le Châtelier’s Principle, continued Pressure Changes May Alter Systems in Equilibrium, continued • In an equilibrium, a pressure increase favors the reaction that produces fewer gas molecules, which for the following equilibrium is the reverse reaction. 2NO(g ) Cl2 (g ) 2NOCl(g ) • When there is no change in the number of molecules, a change in pressure will not affect equilibrium. H2 (g ) CO2 (g ) H2O(g ) CO(g ) Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Visual Concepts Equilibrium Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress The Common-Ion Effect • The solubility of CuCl in water is 1.1 10−3 mol/L. • The solubility of CuCl in sea water is 2.2 10−6 mol/L • CuCl is 500 times less soluble in sea water. • The dramatic reduction in solubility of CuCl shows Le Châtelier’s principle and is described below. – CuCl(s ) Cu ( aq ) Cl (aq ) [Cu+][Cl−] = Ksp = 1.2 × 10−6 Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress The Common-Ion Effect, continued • If you add chloride-rich ocean water to a saturated solution of copper(I) chloride, [Cl−] increases. • However, the Ksp, or [Cl−] [Cu+], remains constant. • [Cu+] must decrease. • This decrease can occur only by the precipitation of the CuCl salt. • The ion Cl− is the common-ion in this case. • The reduction of the solubility of a salt in the solution due to the addition of a common ion is called the common-ion effect. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Visual Concepts Common–Ion Effect Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress The Common-Ion Effect, continued • Doctors use solutions of barium sulfate, BaSO4, to diagnose problems in the digestive tract. • BaSO4 is in equilibrium with a small concentration of Ba2+(aq), a poison. • To reduce the Ba2+ concentration to a safe level, a common ion is added. • Doctors add sodium sulfate, Na2SO4, into the BaSO4 solution that a patient must swallow. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress The Common-Ion Effect, continued • Another example of the common-ion effect emerges when solutions of potassium sulfate and calcium sulfate are mixed and either solution is saturated. • Immediately after the two solutions are mixed, the 2 2– [Ca ] and [SO product of the 4 ] is greater than the Ksp of calcium sulfate. • So, CaSO4 precipitates to establish equilibrium: 2 2– CaSO4 (s ) Ca ( aq ) SO 4 (aq ) Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Practical Uses of Le Châtelier’s Principle • The chemical industry makes use of Le Châtelier’s principle in the synthesis of ammonia by the Haber Process. • High pressure is used to drive the following equilibrium to the right. 2NH3 (g ) N2 (g ) 3H2 (g ) • The forward reaction converts 4 mol of gas into 2 mol of another gas, so it is favored at high pressures. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Section 3 Equilibrium Systems and Stress Practical Uses of Le Châtelier’s Principle, continued • The ammonia synthesis is an exothermic reaction, so the forward reaction is favored at low temperatures. 0°C 250°C 500°C Keq = 6.5 × 108 Keq = 52 Keq = 5.8 × 10−2 • The Haber Process is operated at temperatures of 500°C even though the Keq is small at that temperature. • The reaction proceeds too slowly at lower temperatures. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Understanding Concepts 1. In which of these reactions is the formation of the products favored by an increase in pressure? A. 2O3(g) 3O2(g) B. C(s) + O2(g) CO2(g) C. 2NO(g) + O2(g) 2NO2(g) D. CO2 (aq ) + 2H2O(l ) H3O (aq ) + HCO3– (aq ) Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Understanding Concepts 1. In which of these reactions is the formation of the products favored by an increase in pressure? A. 2O3(g) 3O2(g) B. C(s) + O2(g) CO2(g) C. 2NO(g) + O2(g) 2NO2(g) D. CO2 (aq ) + 2H2O(l ) H3O (aq ) + HCO3– (aq ) Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Understanding Concepts 2. What is the effect of an increase in temperature on an exothermic reaction at equilibrium? F. It has no effect on the equilibrium. G. It shifts the equilibrium in favor of the forward reaction. H. It shifts the equilibrium in favor of the reverse reaction. I. It shifts the equilibrium in favor of both the forward and reverse reactions. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Understanding Concepts 2. What is the effect of an increase in temperature on an exothermic reaction at equilibrium? F. It has no effect on the equilibrium. G. It shifts the equilibrium in favor of the forward reaction. H. It shifts the equilibrium in favor of the reverse reaction. I. It shifts the equilibrium in favor of both the forward and reverse reactions. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Understanding Concepts 3. Which of the following properties of a reactant is included in a Keq equation? A. charge B. concentration C. mass D. volume Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Understanding Concepts 3. Which of the following properties of a reactant is included in a Keq equation? A. charge B. concentration C. mass D. volume Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Understanding Concepts 4. Explain how the common ion effect is involved in the addition of NaIO3 to a solution of Cd(IO3)2, which is much less soluble than NaIO3. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Understanding Concepts 4. Explain how the common ion effect is involved in the addition of NaIO3 to a solution of Cd(IO3)2, which is much less soluble than NaIO3. Answer: Some of the Cd(IO3)2 would precipitate out of the solution, reducing the cadmium ion concentration. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Understanding Concepts 5. Explain how pressure can be used to maximize the production of carbon dioxide in the reaction 2CO2 (g )? 2CO(g ) O2 (g ) Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Understanding Concepts 5. Explain how pressure can be used to maximize the production of carbon dioxide in the reaction 2CO2 (g )? 2CO(g ) O2 (g ) Answer: Increasing the pressure will cause the reaction to favor the production of carbon dioxide because there are three gas molecules in the reactants and only two in the products. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Reading Skills Read the passage below. Then answer the questions. Coral reefs are made by tiny organisms known as polyps. They attach themselves permanently in one place and survive by eating tiny marine animals that swim past. The polyps secrete calcium carbonate to make their shells or skeletons. When the polyps die, the calcium carbonate structures remain and accumulate over time to form a reef. This reef building is possible because calcium carbonate is only slightly soluble in water. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Reading Skills 6. What is the Ksp expression for calcium carbonate? F. Ksp [Ca2 ][CO32– ] G. Ksp [Ca2 ][CO32– ]2 2 2 2– H. Ksp [Ca ] [CO3 ] I. Ksp [Ca2 ]2 [CO32– ]2 Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Reading Skills 6. What is the Ksp expression for calcium carbonate? F. Ksp [Ca2 ][CO32– ] G. Ksp [Ca2 ][CO32– ]2 2 2 2– H. Ksp [Ca ] [CO3 ] I. Ksp [Ca2 ]2 [CO32– ]2 Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Reading Skills 7. What is the most likely source of calcium used by the polyps to build their shells? A. calcium atoms in solution in sea water B. calcium ions in solution in sea water C. calcium particles that reach the water in acid rain D. calcium containing rocks gathered from the ocean floor Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Reading Skills 7. What is the most likely source of calcium used by the polyps to build their shells? A. calcium atoms in solution in sea water B. calcium ions in solution in sea water C. calcium particles that reach the water in acid rain D. calcium containing rocks gathered from the ocean floor Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Reading Skills 8. The Ksp for calcium carbonate is 2.8 10-9 and the Ksp value for calcium sulfate is 9.1 10-6. If coral polyps secreted calcium sulfate rather than calcium carbonate, how would this affect the formation of the coral reef? Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Reading Skills 8. The Ksp for calcium carbonate is 2.8 10-9 and the Ksp value for calcium sulfate is 9.1 10-6. If coral polyps secreted calcium sulfate rather than calcium carbonate, how would this affect the formation of the coral reef? Answer: A reef would not be built as quickly if polyps secreted calcium sulfate, because calcium sulfate has a higher solubility in water. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Interpreting Graphics Use the information from the table below to answer questions 9 through 12. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Interpreting Graphics 9. What is the concentration of magnesium carbonate in a saturated aqueous solution? F. 0.0000068 M G. 0.0026 M H. 0.84 M I. 1.31 M Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Interpreting Graphics 9. What is the concentration of magnesium carbonate in a saturated aqueous solution? F. 0.0000068 M G. 0.0026 M H. 0.84 M I. 1.31 M Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Interpreting Graphics 10. If a saturated solution of silver carbonate is mixed with a saturated solution of zinc sulfide, a precipitate forms. What compound precipitates? A. Ag2CO3 B. Ag2S C. ZnCO3 D. ZnS Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Interpreting Graphics 10. If a saturated solution of silver carbonate is mixed with a saturated solution of zinc sulfide, a precipitate forms. What compound precipitates? A. Ag2CO3 B. Ag2S C. ZnCO3 D. ZnS Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Interpreting Graphics 11. Calculate the concentration of S2– ions in a saturated solution of FeS that contains 0.010 M Fe2+ ions. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Interpreting Graphics 11. Calculate the concentration of S2– ions in a saturated solution of FeS that contains 0.010 M Fe2+ ions. Answer: 1.6 10–17 M Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Interpreting Graphics 12. What will happen if a solution containing 1 10–10 M Na2CO3 is mixed with an equal volume of a solution containing 1 10–10 M MgCl2 and 1 10–10 M ZnCl2? F. No precipitate will form. G. MgCO3 will precipitate out of the solution. H. ZnCO3 will precipitate out of the solution. I. MgCO3 and ZnCO3 will precipitate out of the solution. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved. Chapter 14 Standardized Test Preparation Interpreting Graphics 12. What will happen if a solution containing 1 10–10 M Na2CO3 is mixed with an equal volume of a solution containing 1 10–10 M MgCl2 and 1 10–10 M ZnCl2? F. No precipitate will form. G. MgCO3 will precipitate out of the solution. H. ZnCO3 will precipitate out of the solution. I. MgCO3 and ZnCO3 will precipitate out of the solution. Chapter menu Resources Copyright © by Holt, Rinehart and Winston. All rights reserved.
