© Houghton Mifflin Harcourt Publishing Company Preview Lesson Starter Objectives Reversible Reactions Equilibrium, a Dynamic State The Equilibrium Expression.

© Houghton Mifflin Harcourt Publishing Company Lesson Starter List two everyday processes that can easily be reversed and two that cannot. The freezing of water and the melting of ice can be reversed The cooking of an egg or the lighting of a match cannot be reversed. For the reversible processes, describe the conditions that favor the process going in a particular direction. Low temperature favors freezing, and high temperature favors melting. Section 1 The Nature of Chemical Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Objectives Define chemical equilibrium. Explain the nature of the equilibrium constant. Write chemical equilibrium expressions and carry out calculations involving them. Section 1 The Nature of Chemical Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Reversible Reactions Theoretically, every reaction can proceed in two directions, forward and reverse. Essentially all chemical reactions are considered to be reversible under suitable conditions. A chemical reaction in which the products can react to re-form the reactants is called a reversible reaction. Section 1 The Nature of Chemical Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Reversible Reactions, continued A reversible chemical reaction is in chemical equilibrium when the rate of its forward reaction equals the rate of its reverse reaction and the concentrations of its products and reactants remain unchanged. A a state of dynamic equilibrium has been reached when the amounts of products and reactants remain constant. Both reactions continue, but there is no net change in the composition of the system. Section 1 The Nature of Chemical Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Reversible Reactions, continued The chemical equation for the reaction at equilibrium is written using double arrows to indicate the overall reversibility of the reaction. Section 1 The Nature of Chemical Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Equilibrium, a Dynamic State Many chemical reactions are reversible under ordinary conditions of temperature and concentration. They will reach a state of equilibrium unless at least one of the substances involved escapes or is removed from the reaction system. When the products of the forward reaction are favored, there is a higher concentration of products than of reactants at equilibrium. Section 1 The Nature of Chemical Equilibrium Chapter 18 The equilibrium “lies to the right”

© Houghton Mifflin Harcourt Publishing Company Equilibrium, a Dynamic State, continued When the products of the reverse reaction are favored, there is a higher concentration of reactants than of products at equilibrium. Section 1 The Nature of Chemical Equilibrium Chapter 18 the equilibrium “lies to the left” In other cases, both forward and reverse reactions occur to nearly the same extent before chemical equilibrium is established. Neither reaction is favored, and considerable concentrations of both reactants and products are present at equilibrium.

© Houghton Mifflin Harcourt Publishing Company Equilibrium, a Dynamic State, continued products of the forward reaction favored, lies to the right Section 1 The Nature of Chemical Equilibrium Chapter 18 Neither reaction is favored products of the reverse reaction favored, lies to the left

© Houghton Mifflin Harcourt Publishing Company The Equilibrium Expression Initially, the concentrations of C and D are zero and those of A and B are maximum. Over time the rate of the forward reaction decreases as A and B are used up. The rate of the reverse reaction increases as C and D are formed. When these two reaction rates become equal, equilibrium is established. Section 1 The Nature of Chemical Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company The Equilibrium Expression, continued After equilibrium is reached, the individual concentrations of A, B, C, and D undergo no further change if conditions remain the same. A ratio of their concentrations should also remain constant. The equilibrium constant is designated by the letter K. Section 1 The Nature of Chemical Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company The Equilibrium Expression, continued The constant K is independent of the initial concentrations. K is dependent on the temperature of the system. The Equilibrium Constant The numerical value of K for a particular equilibrium system is obtained experimentally. If K is equal to 1 at equilibrium, there are roughly equal concentrations of reactants and products. Section 1 The Nature of Chemical Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company The Equilibrium Expression, continued The Equilibrium Constant, continued If the value of K is small, the reactants are favored. A large value of K indicates that the products are favored. Only the concentrations of substances that can actually change are included in K. Pure solids and liquids are omitted because their concentrations cannot change. Section 1 The Nature of Chemical Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company The Equilibrium Expression, continued The Equilibrium Constant, continued Section 1 The Nature of Chemical Equilibrium Chapter 18 The equilibrium constant, K, is the ratio of the mathematical product of the concentrations of substances formed at equilibrium to the mathematical product of the concentrations of reacting substances. Each concentration is raised to a power equal to the coefficient of that substance in the chemical equation. The equation for K is sometimes referred to as the chemical equilibrium expression.

© Houghton Mifflin Harcourt Publishing Company The Equilibrium Expression, continued The H 2, I 2, HI Equilibrium System The rate of the reaction between H 2 and I 2 vapor in a sealed flask at an elevated temperature can be followed by observing the rate at which the violet color of the iodine vapor diminishes. The color fades to a constant intensity but does not disappear completely because the reaction is reversible. Hydrogen iodide decomposes to re-form hydrogen and iodine. Section 1 The Nature of Chemical Equilibrium Chapter 18 The constant color achieved indicates that equilibrium exists among hydrogen, iodine, and hydrogen iodide.

© Houghton Mifflin Harcourt Publishing Company The Equilibrium Expression, continued The H 2, I 2, HI Equilibrium System, continued The net chemical equation for the reaction is Section 1 The Nature of Chemical Equilibrium Chapter 18 The value for K is constant for any system of H 2, I 2, and HI at equilibrium at a given temperature. The following chemical equilibrium expression is

© Houghton Mifflin Harcourt Publishing Company The Equilibrium Expression, continued The H 2, I 2, HI Equilibrium System, continued At 425°C, the equilibrium constant for this equilibrium reaction system has the average value of 54.34. Section 1 The Nature of Chemical Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company The Equilibrium Expression, continued The H 2, I 2, HI Equilibrium System, continued The balanced chemical equation for an equilibrium system is necessary to write the expression for the equilibrium constant. Once the value of the equilibrium constant is known, the equilibrium constant expression can be used to calculate concentrations of reactants or products at equilibrium. Section 1 The Nature of Chemical Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company The Equilibrium Expression, continued Sample Problem A An equilibrium mixture of N 2, O 2, and NO gases at 1500 K is determined to consist of 6.4  10 –3 mol/L of N 2, 1.7  10 –3 mol/L of O 2, and 1.1  10 –5 mol/L of NO. What is the equilibrium constant for the system at this temperature? Section 1 The Nature of Chemical Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company The Equilibrium Expression, continued Sample Problem A Solution Section 1 The Nature of Chemical Equilibrium Chapter 18 Given: [N 2 ] = 6.4  10 –3 mol/L [O 2 ] = 1.7  10 –3 mol/L [NO] = 1.1  10 –5 mol/L Solution: The balanced chemical equation is Unknown: K The chemical equilibrium expression is

© Houghton Mifflin Harcourt Publishing Company Preview Lesson Starter Objectives Predicting the Direction of Shift Reactions That Go to Completion Common-Ion Effect Chapter 18 Section 2 Shifting Equilibrium

© Houghton Mifflin Harcourt Publishing Company Imagine children playing on a seesaw. Five boys are sitting on one side and five girls on the other, and the seesaw is just balanced. Then, one girl gets off, and the system is no longer at equilibrium. One way to get the seesaw in balance again is for one of the boys to move toward the girls’ side. Section 2 Shifting Equilibrium Chapter 18 Lesson Starter

© Houghton Mifflin Harcourt Publishing Company When he gets to the middle, the seesaw is again at equilibrium. The stress of one girl getting off is relieved by having one of the boys move his position. How would a chemical system in equilibrium respond to removing one of the products? Section 2 Shifting Equilibrium Chapter 18 Lesson Starter, continued

© Houghton Mifflin Harcourt Publishing Company Objectives Discuss the factors that disturb equilibrium. Discuss conditions under which reactions go to completion. Describe the common-ion effect. Section 2 Shifting Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Predicting the Direction of Shift Changes in pressure, concentration, or temperature can alter the equilibrium position and thereby change the relative amounts of reactants and products. Section 2 Shifting Equilibrium Chapter 18 Le Châtelier’s principle states that if a system at equilibrium is subjected to a stress, the equilibrium is shifted in the direction that tends to relieve the stress. This principle is true for all dynamic equilibria, chemical as well as physical. Changes in pressure, concentration, and temperature illustrate Le Châtelier’s principle.

© Houghton Mifflin Harcourt Publishing Company Predicting the Direction of Shift, continued Changes in Pressure A change in pressure affects only equilibrium systems in which gases are involved. For changes in pressure to affect the system, the total number of moles of gas on the left side of the equation must be different from the total number of moles of gas on the right side of the equation. An increase in pressure is an applied stress. Section 2 Shifting Equilibrium Chapter 18 It causes an increase in the concentrations of all species. The system can reduce the total pressure by reducing the number of molecules.

© Houghton Mifflin Harcourt Publishing Company Predicting the Direction of Shift, continued Changes in Pressure, continued the Haber process for the synthesis of ammonia Section 2 Shifting Equilibrium Chapter 18 4 molecules of gas 2 molecules of gas When pressure is applied, the equilibrium will shift to the right, and produce more NH 3. By shifting to the right, the system can reduce the total number of molecules. This leads to a decrease in pressure.

© Houghton Mifflin Harcourt Publishing Company Predicting the Direction of Shift, continued Changes in Pressure, continued Even though changes in pressure may shift the equilibrium position, they do not affect the value of the equilibrium constant. The introduction of an inert gas, such as helium, into the reaction vessel increases the total pressure in the vessel. But it does not change the partial pressures of the reaction gases present. Section 2 Shifting Equilibrium Chapter 18 Increasing pressure by adding a gas that is not a reactant or a product cannot affect the equilibrium position of the reaction system.

© Houghton Mifflin Harcourt Publishing Company Predicting the Direction of Shift, continued Changes in Concentration An increase in the concentration of a reactant is a stress on the equilibrium system. Section 2 Shifting Equilibrium Chapter 18 An increase in the concentration of A creates a stress. To relieve the stress, some of the added A reacts with B to form products C and D. The equilibrium is reestablished with a higher concentration of A than before the addition and a lower concentration of B.

© Houghton Mifflin Harcourt Publishing Company Predicting the Direction of Shift, continued Changes in Concentration, continued Changes in concentration have no effect on the value of the equilibrium constant. Section 2 Shifting Equilibrium Chapter 18 Such changes have an equal effect on the numerator and the denominator of the chemical equilibrium expression. The concentrations of pure solids and liquids do not change, and are not written in the equilibrium expression. When a solvent, such as water, in a system involving acids and bases, is in an equilibrium equation, it is not included in the equilibrium expression.

© Houghton Mifflin Harcourt Publishing Company Predicting the Direction of Shift, continued Changes in Concentration, continued High pressure favors the reverse reaction. Low pressure favors the formation of CO 2. Because both CaO and CaCO 3 are solids, changing their amounts will not change the equilibrium concentration of CO 2. Section 2 Shifting Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Predicting the Direction of Shift, continued Changes in Temperature Reversible reactions are exothermic in one direction and endothermic in the other. The effect of changing the temperature of an equilibrium mixture depends on which of the opposing reactions is endothermic and which is exothermic. The addition of energy in the form of heat shifts the equilibrium so that energy is absorbed. This favors the endothermic reaction. The removal of energy favors the exothermic reaction. Section 2 Shifting Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Predicting the Direction of Shift, continued Changes in Temperature, continued A rise in temperature increases the rate of any reaction. In an equilibrium system, the rates of the opposing reactions are raised unequally. The value of the equilibrium constant for a given system is affected by the temperature. Section 2 Shifting Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Predicting the Direction of Shift, continued Changes in Temperature, continued The synthesis of ammonia by the Haber process is exothermic. Section 2 Shifting Equilibrium Chapter 18 A high temperature favors the decomposition of ammonia, the endothermic reaction. At low temperatures, the forward reaction is too slow to be commercially useful. The temperature used represents a compromise between kinetic and equilibrium requirements.

© Houghton Mifflin Harcourt Publishing Company Predicting the Direction of Shift, continued Changes in Temperature, continued Catalysts have no effect on relative equilibrium amounts. They only affect the rates at which equilibrium is reached. Catalysts increase the rates of forward and reverse reactions in a system by equal factors. Therefore, they do not affect K. Section 2 Shifting Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Reactions That Go to Completion Some reactions involving compounds formed by the chemical interaction of ions in solutions appear to go to completion in the sense that the ions are almost completely removed from solution. The extent to which reacting ions are removed from solution depends on the solubility of the compound formed and, if the compound is soluble, on the degree of ionization. Section 2 Shifting Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Reactions That Go to Completion, continued Formation of a Gas Section 2 Shifting Equilibrium Chapter 18 H 2 CO 3 (aq) H 2 O(l) + CO 2 (g) This reaction goes practically to completion because one of the products, CO 2, escapes as a gas if the container is open to the air.

© Houghton Mifflin Harcourt Publishing Company Reactions That Go to Completion, continued Formation of a Precipitate Section 2 Shifting Equilibrium Chapter 18 If chemically equivalent amounts of the two solutes are mixed, almost all of the Ag + ions and Cl − ions combine and separate from the solution as a precipitate of AgCl. AgCl is only very sparingly soluble in water. The reaction thus effectively goes to completion because an essentially insoluble product is formed.

© Houghton Mifflin Harcourt Publishing Company Reactions That Go to Completion, continued Formation of a Slightly Ionized Product Neutralization reactions between H 3 O + ions from aqueous acids and OH − ions from aqueous bases result in the formation of water molecules, which are only slightly ionized. Section 2 Shifting Equilibrium Chapter 18 Hydronium ions and hydroxide ions are almost entirely removed from the solution. The reaction effectively runs to completion because the product is only slightly ionized.

© Houghton Mifflin Harcourt Publishing Company Common-Ion Effect The phenomenon in which the addition of an ion common to two solutes brings about precipitation or reduced ionization is an example of the common-ion effect. Section 2 Shifting Equilibrium Chapter 18 example: hydrogen chloride gas is bubbled into a saturated solution of sodium chloride.

© Houghton Mifflin Harcourt Publishing Company Common-Ion Effect, continued As the hydrogen chloride dissolves in sufficient quantity, it increases the concentration of Cl − ions in the solution, which is a stress on the equilibrium system. The system can compensate by forming some solid NaCl. The NaCl precipitates out, relieving the stress of added chloride. The new equilibrium has a greater concentration of Cl − ions but a decreased concentration of Na + ions. Section 2 Shifting Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company The common-ion effect is also observed when one ion species of a weak electrolyte is added in excess to a solution. Small additions of sodium acetate,NaCH 3 COO, to a solution containing acetic acid increase the acetate ion concentration. The equilibrium then shifts in the direction that uses up some of the acetate ions. More molecules of acetic acid are formed, and the concentration of hydronium ions is reduced. Section 2 Shifting Equilibrium Chapter 18 Common-Ion Effect, continued

© Houghton Mifflin Harcourt Publishing Company In general, the addition of a salt with an ion common to the solution of a weak electrolyte reduces the ionization of the electrolyte. Section 2 Shifting Equilibrium Chapter 18 Common-Ion Effect, continued

© Houghton Mifflin Harcourt Publishing Company Preview Lesson Starter Objectives Ionization Constant of a Weak Acid Buffers Ionization Constant of Water Hydrolysis of Salts Chapter 18 Section 3 Equilibria of Acids, Bases, and Salts

© Houghton Mifflin Harcourt Publishing Company 10G define acids and bases and distinguish between Arrhenius and Bronsted-Lowry definitions and predict products in acid base reactions that form water 10H understand and differentiate among acid-base reactions, precipitation reactions, and oxidation- reduction reactions 10J distinguish between degrees of dissociation for strong and weak acids and bases TEKS Section 3 Equilibria of Acids, Bases, and Salts The student is expected to: Chapter 18

© Houghton Mifflin Harcourt Publishing Company pH of 0.10 M HCl _________________ pH of 0.10 M CH 3 COOH ____________ Why is the pH of the 0.10 M HCl different from the pH of the 0.10 M CH 3 COOH? Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 Lesson Starter

© Houghton Mifflin Harcourt Publishing Company pH of 0.10 M NaOH __________________ pH 0.10 M NH 3 _____________________ Why is the pH of the 0.10 M NaOH different from the pH of the 0.10 M NH 3 ? Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 Lesson Starter, continued

© Houghton Mifflin Harcourt Publishing Company Objectives Explain the concept of acid ionization constants, and write acid ionization equilibrium expressions. Review the ionization constant of water. Explain buffering. Compare cation and anion hydrolysis. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18

© Houghton Mifflin Harcourt Publishing Company Ionization Constant of a Weak Acid The term K a is called the acid ionization constant. The acid ionization constant, K a, is constant for a specified temperature but has a new value for each new temperature. Acetic acid, CH 3 COOH, is a weak acid and most of the CH 3 COOH molecules remain unionized. An acetic solution contains CH 3 COOH molecules, H 3 O + ions, and acetate ions, CH 3 COO . Section 3 Equilibria of Acids, Bases, and Salts Chapter 18

© Houghton Mifflin Harcourt Publishing Company Ionization Constant of a Weak Acid, continued The equilibrium equation for the ionization of acetic acid is Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 Because water is the solvent, one can assume that the molar concentration of H 2 O molecules remains constant. The concentration of water is not included in the equilibrium expression. The equation for K a is

© Houghton Mifflin Harcourt Publishing Company Ionization Constant of a Weak Acid, continued Ionization data and constants for some dilute acetic acid solutions at 25°C are given below. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 The numerical value of K a is almost identical for each solution molarity shown.

© Houghton Mifflin Harcourt Publishing Company Ionization Constant of a Weak Acid, continued At constant temperature, an increase in the concentration of CH 3 COO − ions through the addition of sodium acetate, NaCH 3 COO, disturbs the equilibrium. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 This disturbance causes a decrease in [H 3 O + ] and an increase in [CH 3 COOH]. The equilibrium is reestablished with the same value of K a. But there is a higher concentration of nonionized acetic acid molecules and a lower concentration of H 3 O + ions. Changes in the hydronium ion concentration affect pH.

© Houghton Mifflin Harcourt Publishing Company Buffers Buffered solutions resist changes in pH. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 Buffered solutions contains both a weak acid and a salt of the weak acid example: CH 3 COOH and NaCH 3 COO – Buffered solution can react with either an acid or a base. When small amounts of acids or bases are added, the pH of the solution remains nearly constant.

© Houghton Mifflin Harcourt Publishing Company Buffers, continued If a small amount of acid is added to the acetic acid– sodium acetate solution, acetate ions react with most of the added hydronium ions to form nonionized acetic acid molecules. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 The hydronium ion concentration and the pH of the solution remain practically unchanged.

© Houghton Mifflin Harcourt Publishing Company Buffers, continued If a small amount of a base is added, the OH − ions of the base react with and remove hydronium ions to form nonionized water molecules. Acetic acid molecules then ionize and mostly replace the hydronium ions neutralized by the added OH − ions. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 The hydronium ion concentration and the pH of the solution remain practically unchanged.

© Houghton Mifflin Harcourt Publishing Company Buffers, continued A solution of a weak base containing a salt of the base also behaves as a buffered solution. Buffer action has many important applications in chemistry and physiology. Human blood is naturally buffered to maintain a pH of between 7.3 and 7.5. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18

© Houghton Mifflin Harcourt Publishing Company Ionization Constant of Water The self-ionization of water is an equilibrium reaction. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 Equilibrium is established with a very low concentration of H 3 O + and OH − ions. K w =[H 3 O + ][OH – ] = 1.0  10 -14

© Houghton Mifflin Harcourt Publishing Company Hydrolysis of Salts Salts are formed during the neutralization reaction between a Brønsted acid and a Brønsted base. When a salt dissolves in water, it produces positive ions (cations) of the base from which it was formed negative ions (anions) of the acid from which it was formed If the ions formed are from weak acids or bases, they react chemically with the water molecules, and the pH of the solution will have a value other than 7. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18

© Houghton Mifflin Harcourt Publishing Company Hydrolysis of Salts, continued A reaction between water molecules and ions of a dissolved salt is hydrolysis. If the anions react with water, the process is anion hydrolysis and results in a more basic solution. If the cations react with water molecules, the process is cation hydrolysis and results in a more acidic solution. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18

© Houghton Mifflin Harcourt Publishing Company Hydrolysis of Salts, continued Anion Hydrolysis In the Brønsted sense, the anion of the salt is the conjugate base of the acid from which it was formed. It is also a proton acceptor. If the acid is weak, its conjugate base (the anion) will be strong enough to remove protons from some of the water molecules. An equilibrium is established in which the net effect of the anion hydrolysis is an increase in the hydroxide ion concentration, [OH − ], of the solution. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18

© Houghton Mifflin Harcourt Publishing Company Hydrolysis of Salts, continued Anion Hydrolysis, continued The equilibrium equation for a typical weak acid in water, HA, is Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 The general equilibrium equation is

© Houghton Mifflin Harcourt Publishing Company Hydrolysis of Salts, continued Anion Hydrolysis, continued The hydrolysis reaction between water and the anion, A −, that is produced by the dissociation of the weak acid, HA, is Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 The extent of OH − ion formation and the position of the equilibrium depends on the relative strength of the anion, A −. The lower the K a value of HA, the weaker the acid, HA, the stronger its conjugate base, A −

© Houghton Mifflin Harcourt Publishing Company Hydrolysis of Salts, continued Anion Hydrolysis, continued Aqueous solutions of sodium carbonate are strongly basic. The carbonate ions react as a Brønsted base. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 The OH − ion concentration increases until equilibrium is established.

© Houghton Mifflin Harcourt Publishing Company Hydrolysis of Salts, continued Cation Hydrolysis In the Brønsted sense, the cation of the salt is the conjugate acid of the base from which it was formed. It is also a proton donor. If the base is weak, the cation is an acid strong enough to donate a proton to a water molecule to form H 3 O + ions. An equilibrium is established in which the net effect of the cation hydrolysis is an increase in the [H 3 O + ] of the solution. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18

© Houghton Mifflin Harcourt Publishing Company Hydrolysis of Salts, continued Cation Hydrolysis, continued The equilibrium equation for a typical weak base, B, is Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 K b, the base dissociation constant is

© Houghton Mifflin Harcourt Publishing Company Hydrolysis of Salts, continued Cation Hydrolysis, continued The hydrolysis reaction between water and the cation, BH +, produced by the dissociation of the weak base, B, is Section 3 Equilibria of Acids, Bases, and Salts Chapter 18 The extent of H 3 O + ion formation and the position of the equilibrium depend on the relative strength of the cation, BH +. The lower the K b value of B, the weaker the base, the stronger its conjugate acid will be.

© Houghton Mifflin Harcourt Publishing Company Hydrolysis of Salts, continued Hydrolysis in Acid-Base Reactions Hydrolysis can help explain why the end point of a neutralization reaction can occur at a pH other than 7. Salts can be placed in four general categories, depending on their hydrolysis properties: strong acid–strong base strong acid–weak base weak acid–strong base weak acid–weak base Section 3 Equilibria of Acids, Bases, and Salts Chapter 18

© Houghton Mifflin Harcourt Publishing Company Hydrolysis of Salts, continued Hydrolysis in Acid-Base Reactions, continued Salts of strong acids and strong bases produce neutral solutions. Neither the cation of a strong base nor the anion of a strong acid hydrolyzes appreciably in aqueous solutions. The aqueous solutions of salts formed from reactions between weak acids and strong bases are basic. Anions of the dissolved salt are hydrolyzed by the water molecules, and the hydroxide-ion concentration increases. This raises the pH of the solution. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18

© Houghton Mifflin Harcourt Publishing Company Hydrolysis of Salts, continued Hydrolysis in Acid-Base Reactions, continued Salts of strong acids and weak bases produce acidic aqueous solutions. Cations of the dissolved salt are hydrolyzed in the water solvent, and hydronium ion concentration increases. The pH of the solution is lowered. Salts of weak acids and weak bases can produce either acidic, neutral, or basic aqueous solutions, depending on the salt dissolved. Both ions of the dissolved salt are hydrolyzed extensively. Section 3 Equilibria of Acids, Bases, and Salts Chapter 18

© Houghton Mifflin Harcourt Publishing Company Preview Objectives Solubility Product Calculating Solubilities Precipitation Calculations Limitations on the Use of Ksp Chapter 18 Section 4 Solubility Equilibrium

© Houghton Mifflin Harcourt Publishing Company 10H understand and differentiate among acid-base reactions, precipitation reactions, and oxidation- reduction reactions TEKS Section 4 Solubility Equilibrium The student is expected to: Chapter 18

© Houghton Mifflin Harcourt Publishing Company Objectives Explain what is meant by solubility product constants, and calculate their values. Calculate solubilities using solubility product constants. Carry out calculations to predict whether precipitates will form when solutions are combined. Section 4 Solubility Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Solubility Product A saturated solution contains the maximum amount of solute possible at a given temperature in equilibrium with an undissolved excess of the substance. A saturated solution is not necessarily a concentrated solution. The equilibrium principles developed in this chapter apply to all saturated solutions of sparingly soluble salts. Section 4 Solubility Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Solubility Product, continued The heterogeneous equilibrium system in a saturated solution of silver chloride containing an excess of the solid salt is represented by Section 4 Solubility Equilibrium Chapter 18 The solubility product constant, K sp, of a substance is the product of the molar concentrations of its ions in a saturated solution, each raised to the power that is the coefficient of that ion in the balanced chemical equation.

© Houghton Mifflin Harcourt Publishing Company Solubility Product, continued The equation for the solubility equilibrium expression for the dissolution reaction of AgCl is Section 4 Solubility Equilibrium Chapter 18 The equilibrium expression is written without including the solid species. The numerical value of K sp can be determined from solubility data.

© Houghton Mifflin Harcourt Publishing Company Solubility Product, continued For a saturated solution of CaF 2, the equilibrium equation is Section 4 Solubility Equilibrium Chapter 18 The expression for the solubility product constant is The solubility of CaF 2 is is 8.6  10 −3 /100 g of water at 25°C. Expressed in moles per liter this concentration becomes 1.1  10 −3 mol/L.

© Houghton Mifflin Harcourt Publishing Company Solubility Product, continued CaF 2 dissociates to yield twice as many F − ions as Ca 2+ ions. [Ca 2+ ] = 1.1  10 −3 mol/L [F − ] = 2.2  10−3 mol/L Section 4 Solubility Equilibrium Chapter 18 K sp = 5.3  10 -9 Calculations of K sp ordinarily should be limited to two significant figures.

© Houghton Mifflin Harcourt Publishing Company Solubility Product, continued The solubility product constant is an equilibrium constant representing the product of the molar concentrations of its ions in a saturated solution. Section 4 Solubility Equilibrium Chapter 18 The solubility of a solid is an equilibrium position that represents the amount of the solid required to form a saturated solution with a specific amount of solvent. It has only one value for a given solid at a given temperature. It has an infinite number of possible values at a given temperature and is dependent on other conditions, such as the presence of a common ion.

© Houghton Mifflin Harcourt Publishing Company Solubility Product, continued Sample Problem B Calculate the solubility product constant, K sp,for copper(I) chloride, CuCl, given that the solubility of this compound at 25°C is 1.08  10 –2 g/100. g H 2 O. Section 4 Solubility Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Solubility Product, continued Sample Problem B Solution K sp =[Cu + ][Cl – ] Section 4 Solubility Equilibrium Chapter 18 Unknown: K sp Given: solubility of CuCl = 1.08  10 −2 g CuCl/100. g H 2 O Solution: [Cu + ] = [Cl – ] = solubility in mol/L

© Houghton Mifflin Harcourt Publishing Company Solubility Product, continued Sample Problem B Solution, continued 1.09  10 -3 mol/L CuCl Section 4 Solubility Equilibrium Chapter 18 K sp = (1.09  10 -3 )(1.09  10 -3 ) = 1.19  10 -6 [Cu + ] = [Cl – ]=1.09  10 -3 mol/L

© Houghton Mifflin Harcourt Publishing Company Calculating Solubilities The solubility product constant can be used to determine the solubility of a sparingly soluble salt. Section 4 Solubility Equilibrium Chapter 18 The molar solubility of BaCO 3 is 7.1  10 −5 mol/L. How many moles of barium carbonate, BaCO 3, can be dissolved in 1 L of water at 25°C?

© Houghton Mifflin Harcourt Publishing Company Calculating Solubilities, continued Sample Problem C Calculate the solubility of silver bromide, AgBr, in mol/L, using the K sp value for this compound. Section 4 Solubility Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Calculating Solubilities, continued Sample Problem C Solution Given: K sp = 5.0  10 −13 Unknown: solubility of AgBr Solution: [Ag + ] = [Br − ], so let [Ag + ] = x and [Br − ] = x Section 4 Solubility Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Precipitation Calculations The equilibrium condition does not require that the two ion concentrations be equal. Equilibrium will still be established so that the ion product does not exceed the value of K sp for the system. If the ion product is less than the value of K sp at a particular temperature, the solution is unsaturated. If the ion product is greater than the value for K sp, solid precipitates. Section 4 Solubility Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Precipitation Calculations, continued Unequal quantities of BaCl 2 and Na 2 CO 3 are dissolved in water and the solutions are mixed. If the ion product exceeds the K sp of BaCO 3, a precipitate of BaCO 3 forms. Precipitation continues until the ion concentrations decrease to the point at which equals the K sp. Section 4 Solubility Equilibrium Chapter 18 The solubility product can be used to predict whether a precipitate forms when two solutions are mixed.

© Houghton Mifflin Harcourt Publishing Company Precipitation Calculations, continued Sample Problem D Will a precipitate form if 20.0 mL of 0.010 M BaCl 2 is mixed with 20.0 mL of 0.0050 M Na 2 SO 4 ? Section 4 Solubility Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Precipitation Calculations, continued Sample Problem D Solution Given: concentration of BaCl 2 = 0.010 M volume of BaCl 2 = 20.0 mL concentration of Na 2 SO 4 = 0.0050 M volume of Na 2 SO 4 = 20.0 mL Unknown: whether a precipitate forms Solution: The two possible new pairings of ions are NaCl and BaSO 4. BaSO 4 is a sparingly soluble salt. Section 4 Solubility Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Precipitation Calculations, continued Sample Problem D Solution, continued total volume of solution: 0.020 L + 0.020 L = 0.040 L concentration Ba 2+ ion in combined solution: Section 4 Solubility Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Precipitation Calculations, continued Sample Problem D Solution, continued The ion product: Section 4 Solubility Equilibrium Chapter 18 Precipitation occurs.

© Houghton Mifflin Harcourt Publishing Company Limitations on the Use of K sp The solubility product principle can be very useful when applied to solutions of sparingly soluble substances. It cannot be applied very successfully to solutions of moderately soluble or very soluble substances. The positive and negative ions attract each other, and this attraction becomes appreciable when the ions are close together. Sometimes it is necessary to consider two equilibria simultaneously. Section 4 Solubility Equilibrium Chapter 18

© Houghton Mifflin Harcourt Publishing Company Preview Lesson Starter Objectives Reversible Reactions Equilibrium, a Dynamic State The Equilibrium Expression.

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© Houghton Mifflin Harcourt Publishing Company Preview Lesson Starter Objectives Reversible Reactions Equilibrium, a Dynamic State The Equilibrium Expression.

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Presentation on theme: "© Houghton Mifflin Harcourt Publishing Company Preview Lesson Starter Objectives Reversible Reactions Equilibrium, a Dynamic State The Equilibrium Expression."— Presentation transcript:

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