Equilibrium Basic Concepts Reversible reactions do not go

Equilibrium Basic Concepts • Reversible reactions do not go to completion. – They can occur in either direction • Chemical equilibrium exists when two opposing reactions occur simultaneously at the same rate. – A chemical equilibrium is a reversible reaction that the forward reaction rate is equal to the reverse reaction rate. • Chemical equilibria are dynamic equilibria. – Molecules are continually reacting, even though the overall composition of the reaction mixture does not change. – Double arrow (⇋) indicates that both the forward and reverse reactions are occurring simultaneously and is read “is in equilibrium with” – At equilibrium, the composition of the system no longer changes with time – Composition of an equilibrium mixture is independent of the direction from which equilibrium is approached

Equilibrium Basic Concepts • One of the fundamental ideas of chemical equilibrium is that equilibrium can be established from either the forward or reverse direction. • Equilibrium state is achieved when the rate of the forward reaction equals the rate of the reverse reaction: Ratef = Rater • Under a given set of conditions, there must be a relationship between the composition of the system at equilibrium and the kinetics of a reaction represented by rate constants. The rates of the forward and reverse reactions can be represented as: • • • The ratio of the rate constants yields a new constant, the equilibrium constant (K), a unitless quantity and is defined as K = kf /kr. Equilibrium constants are uniless because they actually involve a thermodynamic quantity called activity. Activities are directly related to molarity This implies that equilibrium mixture is determined by the magnitudes of the rate constants for the forward and reverse reactions.

Equilibrium Basic Concepts Product Favored, ∆G˚ negative Systems can reach equilibrium when reactants have NOT converted completely to products. In the case ∆Grxn is < ∆Gorxn state with both reactants and products present is MORE STABLE than complete conversion.

Developing an Equilibrium Constant Expression • The ratio of the product of the equilibrium concentrations of the products (raised to their coefficients in the balanced equation) to the product of the equilibrium concentrations of the reactants (raised to their coefficients in the balanced equation) is always a constant under a given set of conditions: [C]c [D]d Kc = [A]a [B]b is called the equilibrium equation. • This relationship is known as the law of mass action where – Kc is the equilibrium constant for the reaction – Right side of the equation is called the equilibrium constant expression • Relationship is true for any pair of opposing reactions regardless of the mechanism of the reaction or of the number of steps in the mechanism.

The Equilibrium Constant • Write equilibrium constant expressions for the following reactions at 500 o. C. All reactants and products are gases at 500 o. C.

The Equilibrium Constant • One liter of equilibrium mixture from the following system at a high temperature was found to contain 0. 172 mole of phosphorus trichloride, 0. 086 mole of chlorine, and 0. 028 mole of phosphorus pentachloride. Calculate Kc for the reaction. • Equil []’s 0. 028 M 0. 172 M 0. 086 M

The Equilibrium Constant Products favored 103 > K > 10 -3 Reactants favored – Values of K greater than 103 indicate a strong tendency for reactants to form products, so equilibrium lies to the right, favoring the formation of products (kf >>kr) – Values of K less than 10– 3 indicate that the ratio of products to reactants at equilibrium is very small; reactants do not tend to form products readily, and equilibrium lies to the left, favoring the formation of reactants (kf<<kr) – Values of K between 103 and 10– 3 are not very large or small, so there is no strong tendency to form either products or reactants; at equilibrium, there are significant amounts of both products and reactants (kf kr)

The Equilibrium Constant • Equilibrium can be approached from either direction in a chemical reaction, so the equilibrium constant expression and the magnitude of the equilibrium constant depend on the form in which the chemical reaction is written. • When a reaction is written in the reverse direction, c. C + d. D ⇋ a. A + b. B K and the equilibrium constant expression are inverted: K´= [A]a [B]b [C]c [D]d so K´ = 1/K

Variation of Kc with the Form of the Balanced Equation • The value of Kc depends upon how the balanced equation is written. From the prior examples we have: PCl 5 Equil []’s 0. 028 M PCl 3 + Cl 2 0. 172 M 0. 086 M Kc = [PCl 3][Cl 2]/[PCl 5] = 0. 53 PCl 3 + Cl 2 Equil. []’s 0. 172 M 0. 086 M PCl 5 0. 028 M Kc’ = [PCl 5]/[PCl 3][Cl 2] = 1. 89 2 PCl 5 Equil []’s 0. 028 M 2 PCl 3 + 2 Cl 2 0. 172 M 0. 086 M Kc ’’ = [PCl 3] 2[Cl 2] 2/[PCl 5] 2 = 0. 28 2 PCl 3 + 2 Cl 2 Equil. []’s 0. 172 M 0. 086 M 2 PCl 5 0. 028 M Kc ‘“ = [PCl 5]2/[PCl 3]2[Cl 2]2 = 3. 56

The Equilibrium Constant • At a given temperature 0. 80 mole of N 2 and 0. 90 mole of H 2 were introduced in an evacuated 1. 00 -liter container. After equilibrium was established the equilibrium concentrations were determined to be 0. 20 mole of NH 3, 0. 70 moles N 2, and 0. 60 moles H 2. Calculate Kc for the reaction.

Partial Pressure and Kc • For gas phase reactions the equilibrium constants can be expressed in partial pressures rather than concentrations. • For gases, the pressure is proportional to the concentration. • We can see this by looking at the ideal gas law. – – PV = n. RT P = n. RT/V n/V = M P= MRT and M = P/RT (PC)c (PD)d {[C](RT)}c{[D](RT)}d Kp = = a b (PA) (PB) {[A](RT)}a{[B](RT)}b [C]c(RT)c[D]d(RT)d [C]c[D]d (RT)c(RT)d = [A]a(RT)a[B]b(RT)b = [A]a[B]b x(RT)a(RT)b = Kc {(RT)(c+d)-(a+b)} (nprod-nreact ) = Kc(RT) n = Kc(RT)

Partial Pressures and the Equilibrium Constant • Consider this system at equilibrium at 5000 C. 2 SO 2(g) + O 2(g) 2 SO 3(g) • Kp is unitless. • Partial pressures are expressed in atmospheres or mm. Hg, so the molar concentration of a gas and its partial pressure do not have the same numerical value but are related by the ideal gas constant R and the temperature Kp = Kc(RT) n Kc = Kp(RT)- n where K is the equilibrium constant expressed in units of concentration and n is the difference between the number of moles of gaseous products and gaseous reactants; temperature is expressed in Kelvins. • If n = 0, Kp = Kc

Relationship Between Kp and Kc • Kc is 49 for the following reaction at 450 o. C. If 1. 0 mole of H 2 and 1. 0 mole of I 2 are allowed to reach equilibrium in a 3. 0 -liter vessel, (a) How many moles of I 2 remain unreacted at equilibrium? (b) What are the equilibrium partial pressures of H 2, I 2 and HI? (c) What is the total pressure in the reaction vessel?

Relationship Between Kp and Kc • Nitrosyl bromide, NOBr, is 34% dissociated by the following reaction at 25 o. C, in a vessel in which the total pressure is 0. 25 atmosphere. What is the value of Kp?

Homogeneous and Heterogeneous Equilibria • Homogeneous equilibrium – When the products and reactants of an equilibrium reaction form a single phase, whether gas or liquid – Concentrations of the reactants and products can vary over a wide range • Heterogeneous equilibrium – A system whose reactants, products, or both are in more than one phase – An example is the reaction of a gas with a solid or liquid • Molar concentrations of pure liquids and solids do not vary with temperature, so they are treated as constants, this simplifies their equilibrium constant expressions

Heterogeneous Equlibria • Heterogeneous equilibria have more than one phase present. – For example, a gas and a solid or a liquid and a gas. • How does the equilibrium constant differ for heterogeneous equilibria? – Pure solids and liquids have activities of unity. – Solvents in very dilute solutions have activities that are essentially unity. – The Kc and Kp for the reaction shown above are:

Heterogeneous Equlibria

Heterogeneous Equlibria • What are Kc and Kp for this reaction?

Heterogeneous Equlibria • What are Kc and Kp for this reaction?

Equilibrium Constant Expressions for the Sums of Reactions An aside: Similar to Hess’ Law we can add several reactions together to get to an overall reaction not listed in a table. To determine K for the sum of the reactions simply multiply all the values for each intermediate step to get the K for the overall reaction.

Solving Equilibrium Problems • Two fundamental kinds of equilibrium problems 1. Those in which the concentrations of the reactants and products at equilibrium are given and the equilibrium constant for the reaction needs to be calculated 2. Those in which the equilibrium constant and the initial concentrations of reactants are known and the concentration of one or more substances at equilibrium needs to be calculated

Calculating an Equilibrium Constant from Equilibrium Concentrations • can be calculated when equilibrium concentrations, or partial pressures are substituted into the equilibrium constant expression for the reaction. • When equilibrium concentrations are not given the equilibrium concentrations can be obtained from the initial concentrations of the reactants and the balanced equation for the reaction, as long as the equilibrium concentration of one of the substances is known. • Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium.

Uses of the Equilibrium Constant, Kc • The equilibrium constant, Kc, is 3. 00 for the following reaction at a given temperature. If 1. 00 mole of SO 2 and 1. 00 mole of NO 2 are put into an evacuated 2. 00 L container and allowed to reach equilibrium, what will be the concentration of each compound at equilibrium?

Uses of the Equilibrium Constant, Kc • The equilibrium constant is 49 for the following reaction at 450 o. C. If 1. 00 mole of HI is put into an evacuated 1. 00 -liter container and allowed to reach equilibrium, what will be the equilibrium concentration of each substance?

The Reaction Quotient • The mass action expression or reaction quotient has the symbol Q. – Q has the same form as Kc • The major difference between Q and Kc is that the concentrations used in Q are not necessarily equilibrium values. • Why do we need another “equilibrium constant” that does not use equilibrium concentrations? • Q will help us predict how the equilibrium will respond to an applied stress. • Q may be derived from a set of values measured at any time during the reaction of any mixture of the reactants and products, regardless of whether the system is at equilibrium • To make this prediction we compare Q with Kc.

The Reaction Quotient (Q) • Comparing the magnitudes of Q and K allows the determination of whether a reaction mixture is already at equilibrium and, if it is not, how to predict whether its composition will change with time (whether the reaction will proceed to the right or to the left) 1. If Q = K, the system is at equilibrium, no further change in the composition of the system will occur unless the conditions are changed 2. If Q < K, then the ratio of the concentrations of products to the concentration of reactants is less than the ratio at equilibrium; reaction will proceed to the right, forming products at the expense of reactants 3. If Q > K, then the ratio of the concentrations of products to the concentrations of reactants is greater than at equilibrium; reaction will proceed to the left, forming reactants at the expense of products

The Reaction Quotient • The equilibrium constant for the following reaction is 49 at 450 o. C. If 0. 22 mole of I 2, 0. 22 mole of H 2, and 0. 66 mole of HI were put into an evacuated 1. 00 -liter container, would the system be at equilibrium? If not, what must occur to establish equilibrium?

Le Châtelier’s Principle • • When a system at equilibrium is perturbed in some way, the effects of the perturbation can be predicted qualitatively using Le Châtelier’s principle. This principle states that if a stress is applied to a system at equilibrium, the composition of the system will change to relieve the applied stress. Stress occurs when any change in the system affects the magnitude of Q or K. Three types of stresses can change the composition of an equilibrium mixture: 1. A change in the concentrations (or partial pressures) of the components by the addition or removal of reactants or products 2. A change in the total pressure or volume 3. A change in the temperature of the system

Disturbing a System at Equlibrium: Predictions 1 Changes in Concentration of Reactants and/or Products • Also true for changes in pressure for reactions involving gases. • An equilibrium is disturbed by adding or removing a reactant or product 1. Stress of an added reactant or product is relieved by reaction in the direction that consumes the added substance a. Add reactant—reaction shifts right toward product b. Add product—reaction shifts left toward reactant 2. Stress of removing reactant or product is relieved by reaction in the direction that replenishes the removed substance a. Remove reactant—reaction shifts left b. Remove product—reaction shifts right

Disturbing a System at Equlibrium: Predictions 2 Changes in Volume (and pressure for reactions involving gases) • Increase in pressure (due to decrease in volume) results in a reaction in the direction of a fewer number of moles of gas • Decrease in pressure (due to increase in volume) results in a reaction in the direction of a greater number of moles of gas 1. Decrease volume—molarity increases 2. If reactant side has more moles of gas a. Increase in denominator is greater than increase in numerator and Qc < Kc b. To return to equilibrium, Qc must increase; the numerator of the Qc expression must increase and denominator must decrease—it shifts toward fewer moles of gas (reactants to products) 3. If product side has more moles of gas a. Increase in numerator is greater than increase in denominator and Qc > Kc b. b. To return to equilibrium, Qc must decrease; the denominator of the Qc expression must decrease and the numerator must increase—it shifts toward fewer moles of gas (products to reactants) – Predict what will happen if the volume of this system at equilibrium is changed by changing the pressure at constant temperature:

Disturbing a System at Equlibrium: Predictions 3 Changing the Reaction Temperature Changes in temperature can change the value of K without affecting Q (Q K) • Predictions on the response of a system to a change in requires knowledge regarding Hrxn. Remember: 1. Exothermic (heat is lost by the system, H<0): reactants ⇋ products + heat 2. Endothermic (heat is gained by the system, H>0): reactants + heat ⇋ products • Le Châtelier’s principle predicts 1. that an exothermic reaction will shift to the left (toward reactants) if the temperature of the system is increased (heat is added); 2. that an endothermic reaction will shift to the right (toward the products) if the temperature of the system is increased; 3. that if Hrxn = 0, then a change in temperature has no affect on composition. • Increasing the temperature increases the magnitude of the equilibrium constant for an endothermic reaction • Increasing the temperature decreases the equilibrium constant for an exothermic reaction

Disturbing a System at Equlibrium: Predictions • Introduction of a Catalyst – Catalysts decrease the activation energy of both the forward and reverse reaction equally. • Catalysts do not affect the position of equilibrium. – The concentrations of the products and reactants will be the same whether a catalyst is introduced or not. – Equilibrium will be established faster with a catalyst.

Disturbing a System at Equlibrium: Predictions • How will an increase in pressure (caused by decreasing the volume) affect the equilibrium in each of the following reactions?

Disturbing a System at Equlibrium: Predictions • How will an increase in temperature affect each of the following reactions?

Disturbing a System at Equlibrium: Predictions • A 2. 00 liter vessel in which the following system is in equilibrium contains 1. 20 moles of COCl 2, 0. 60 moles of CO and 0. 20 mole of Cl 2. Calculate the equilibrium constant.

Disturbing a System at Equlibrium: Predictions • An additional 0. 80 mole of Cl 2 is added to the vessel at the same temperature. Calculate the molar concentrations of CO, Cl 2, and COCl 2 when the new equilibrium is established.

Controlling the Products of Reactions • One of the primary goals of modern chemistry is to control the identity and quantity of the products of chemical reactions. • Two approaches 1. To get a high yield of a desired compound, make the rate of the desired reaction much faster than the rate of any of the other possible reactions that might occur in the system; altering reaction conditions to control reaction rates, thereby obtaining a single product or set of products is called kinetic control. 2. Thermodynamic control—consists of adjusting conditions so that at equilibrium only the desired products are present in significant quantities.

Disturbing a System at Equlibrium: Predictions • Given the reaction below at equilibrium in a closed container at 500 o. C. How would the equilibrium be influenced by the following?

• An Application of Equilibrium: The Haber Process The Haber process is used for the commercial production of ammonia. – This is an enormous industrial process in the US and many other countries. – Ammonia is the starting material for fertilizer production. Fritz Haber 1868 -1934 Nobel Prize, 1918 Carl Bosch 1874 -1940 Nobel Prize, 1931

o G rxn Relationship Between and the Equilibrium Constant • The relationship for K at conditions other than thermodynamic standard state conditions is derived from this equation. When Q < K or Q > K, reaction is spontaneous. When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K

Acids and bases Autoionization reaction of liquid water 2. p. H, p. OH, and p. Kw 3. conjugate acid-base pairs 1. 4. acid or base strength and the magnitude of Ka, Kb, p. Ka, and p. Kb 5. leveling effect 6. To be able to predict whether reactants or products are favored in an acid-base equilibrium 7. use molecular structure and acid and base strengths 8. use Ka and Kb values to calculate the percent ionization and p. H of a solution of an acid or a base 9. calculate the p. H at any point in an acid-base titration 10. common ion effects and the position of an acid-base equilibrium 11. how a buffer works and how to use the Henderson. Hasselbalch equation to calculate the p. H of a buffer

Acids and bases • There are three classes of strong electrolytes. 1 Strong Water Soluble Acids Remember the list of strong acids from Chapter 4. 2 Strong Water Soluble Bases The entire list of these bases was also introduced in Chapter 4. 3 Most Water Soluble Salts The solubility guidelines from Chapter 4 will help you remember these salts. • Weak acids and bases ionize or dissociate partially, much less than 100%, and is often less than 10%.

Table of Common Ions Common Negative Ions (Anions) Monovalent Hydride Fluoride Chloride Bromide Iodide Hydroxide Permangante Cyanide Thiocynate Acetate Nitrate Bisulfite Bisulfate Bicarbonate Dihydrogen phosphate Nitrite Amide Hypochlorite Chlorate Perchlorate HFl. Cl. Br. IOHMn. O 4 CNSCNCH 3 COONO 3 HSO 4 HCO 3 H 2 PO 4 NO 2 NH 2 Cl. O 3 Cl. O 4 - Divalent Oxide Peroxide Sulfide Selenide Oxalate Chromate Dichromate Tungstate Molybdate Tetrathionate Thiosulfate Sulfite Sulfate Carbonate Hydrogen phosphate O 2 O 22 S 2 Se 2 C 2 O 42 Cr 2 O 72 WO 42 Mo. O 42 S 4 O 62 S 2 O 32 SO 42 CO 32 HPO 42 - Trivalent Nitride N 3 - Phosphate PO 43 -

Table of Common Ions Common Positive Ions (Cations) Monovalent Hydronium (aqueous) Hydrogen (proton) Lithium Sodium Potassium Rubidium Cesium Francium Silver Ammonium Thalium Copper I H 3 O+ H+ Li+ Na+ K+ Rb+ Cs+ Fr+ Ag+ NH 4+ Tl+ Cu+ Divalent Magnesium Calcium Strontium Beryllium Manganese II Barium Zinc Cadmium Nickel II Palladium II Platinum II Copper II Mercury I Iron II Cobalt II Chromium II Lead II Tin II Mg 2+ Ca 2+ Sr 2+ Be 2+ Mn 2+ Ba 2+ Zn 2+ Cd 2+ Ni 2+ Pd 2+ Pt 2+ Cu 2+ Hg 22+ Fe 2+ Co 2+ Cr 2+ Pb 2+ Sn 2+ Trivalent Aluminium Antimony III Bismuth III Al 3+ Sb 3+ Bi 3+ Iron III Cobalt III Chromium III Fe 3+ Co 3+ Cr 3+

Water Solubility of Ionic Compounds If one ion from the “Soluble Compound” list is present in a compound, the compound is water soluble.

Acids and bases • Most salts of strong or weak electrolytes can dissolve in water to produce a neutral, basic, or acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion (A–) or the conjugate acid of a weak base as the cation (BH+), or possibly both. • Salts that contain small, highly charged metal ions produce acidic solutions in H 2 O. • The most important parameter for predicting the effect of a metal ion on the acidity of coordinated water molecules is the charge-to-radius ratio of the metal ion. • The reaction of a salt with water to produce an acidic or basic solution is called a hydrolysis reaction, reaction which is just an acid-base reaction in which the acid is a cation or the base is an anion.

Acids and bases Acid (HCl) Base (Na. OH) Arrhenius Brönsted-Lowery Lewis Two species that differ by only a proton constitute a conjugate acid-base pair. 1. Conjugate base has one less proton than its acid; A– is the conjugate base of HA 2. Conjugate acid has one more proton than its base; BH+ is the conjugate acid of B 3. Conjugates are weaker than strong parents and stronger than weak parents. 4. All acid-base reactions involve two conjugate acid-base pairs. • HCl (aq) + H 2 O (l) H 3 O+ (aq) + Cl– (aq) parent acid parent base conjugate acid conjugate base Water is amphiprotic: amphiprotic it can act as an acid by donating a proton to a base to form the hydroxide ion, or as a base by accepting a proton from an acid to form the hydronium ion, H 3 O+. Substances that can behave as both an acid and a base are said to be amphoteric

Acids and bases can be defined in different ways: 1. Arrhenius definition: An acid is a substance that dissociates in water to produce H+ ions (protons), and a base is a substance that dissociates in water to produce OH– ions (hydroxide); an acid-base reaction involves the reaction of a proton with the hydroxide ion to form water. – Three limitations 1. Definition applied only to substances in aqueous solutions. 2. Definition restricted to substances that produce H+ and OH– ions 3. Definition does not explain why some compounds containing hydrogen such as CH 4 dissolve in water and do not give acidic solutions 2. Brønsted–Lowry definition: definition An acid is any substance that can donate a proton, and a base is any substance that can accept a proton; acid-base reactions involve two conjugate acid-base pairs and the transfer of a proton from one substance (the acid) to another (the base). Not restricted to aqueous solutions, expanding to include other solvent systems and acid-base reactions for gases and solids. Not restricted to bases that only produce OH– ions. Acids still restricted to substances that produce H+ ions. Limitation 3 not dealt with. 3. Lewis definition: definition A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor.

Acids and bases CO 2 + H 2 O Baking Soda Na. HCO 3 Soda Pop CO 2 Atmosphere. 8317 CO 2+3 H 2 O KH 2. 5× 10− 4 H 2 CO 3+2 H 2 O K 1 H 2 CO 3 + H 2 O HCO 3 - + H+ HCO 3 - + H 2 O CO 32 - + H+ 5. 61× 10− 11 K 2 HCO 3 -+H 3 O++H 2 O CO 32 -+2 H 3 O+ CO 2 + Ca. CO 3 + H 2 O 2 HCO 3 - + Ca 2 Biological Calcification (not a reversible reaction) unneeded critter's shells fall to ocean floor H 2 CO 3

You should know the strong acids & bases

p. H, a Concentration Scale p. H: a way to express acidity -- the concentration of H+ in solution. Low p. H: high [H+] High p. H: low [H+] p. H = log (1/ [H+]) = - log [H+] Acid Formula p. H at half equivalence point Acetic CH 3 COOH 4. 7 Nitrous HNO 2 3. 3 Hydrofluoric HF 3. 1 Hypochlorous HCl. O 7. 4 Hydrocyanic HCN 9 Acidic solution Neutral Basic solution p. H < 7 p. H = 7 p. H > 7

p. H, a Concentration Scale • A convenient way to express the acidity and basicity of a solution is the p. H and p. OH scales. • The p. H of an aqueous solution is defined as: Acid Formula p. H Acetic CH 3 COOH 4. 7 Nitrous HNO 2 3. 3 Hydrofluoric HF 3. 1 Hypochlorous HCl. O 7. 4 Hydrocyanic HCN 9

Titration Titrant Equivalence Point Primary Standard End Point Secondary Standard Titration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. 4. Net ionic equation H+ + OH- --> H 2 O 5. At equivalence point moles H+ = moles OH-

http: //www. chem 1. com/acad/webtext/abcon-2. html

Autoionization of Water • Because water is amphiprotic, one water molecule can react with another to form an OH– ion and an H 3 O+ ion in an autoionization process: 2 H 2 O(l)⇋H 3 O+ (aq) + OH– (aq) • Equilibrium constant K for this reaction can be written as Kc = [H 3 O+] [OH–] [H 2 O]2 • 1 L of water contains 55. 5 moles of water. In dilute aqueous solutions: • The water concentration is many orders of magnitude greater than the ion concentrations. The concentration is essentially that of pure water. Recall that the activity of pure water is 1. • When pure liquid water is in equilibrium with hydronium and hydroxide ions at 25ºC, the concentrations of hydronium ion and hydroxide ion are equal: [H 3 O+]=[OH–] = 1. 0 x 10– 7 M [H 3 O+][OH–] = 1. 0 x 10– 14 M = Kw p. H = p. OH = 7 p. H + p. OH = p. Kw = 14 Kc [H 2 O]2 = Kw = [H 3 O+][OH–] = 1. 0 x 10– 14

Leveling Effect It’s all because of Gibbs Free Energy No acid stronger than H 3 O+ and no base stronger than OH– can exist in aqueous solution, leading to the phenomenon known as the leveling effect. • Any species that is a stronger acid than the conjugate acid of water (H 3 O+) is leveled to the strength of H 3 O+ in aqueous solution because H 3 O+ is the strongest acid that can exist in equilibrium with water. • In aqueous solution, any base stronger than OH– is leveled to the strength of OH– because OH– is the strongest base that can exist in equilibrium with water • Any substance whose anion is the conjugate base of a compound that is a weaker acid than water is a strong base that reacts quantitatively with water to form hydroxide ion http: //www. chem 1. com/acad/webtext/abcon-2. html

Acid-Base Equilibrium Constants: Ka, Kb, p. Ka, and p. Kb • The magnitude of the equilibrium constant for an ionization reaction can determine the relative strengths of acids and bases • The general equation for the ionization of a weak acid in water, where HA is the parent acid and A– is its conjugate base, is HA(aq) + H 2 O(l) ⇋ H 3 O+(aq) + A–(aq) • The equilibrium constant for this dissociation is K = [H 3 O+] [A–] [H 2 O] [HA] • The concentration of water is constant for all reactions in aqueous solution, so [H 2 O] can be incorporated into a new quantity, the acid ionization constant (Ka): Ka = K[H 2 O] = [H 3 O+] [A–] [HA]

Ionization Constants for Weak Monoprotic Acids and Bases • Strong acids and bases ionize essentially completely in water; the percent ionization is always approximately 100%, regardless of the concentration • The percent ionization in solutions of weak acids and bases is small and depends on the analytical concentration of the weak acid or base; percent ionization of a weak acid or a weak base actually increases as its analytical concentration decreases and percent ionization increases as the magnitude of the ionization constants Ka and Kb increases

Ionization Constants for Weak Acids and Bases When is a 1 M solution not a 1 M solution? Ø A 1 M solution is prepared by dissolving 1 mol of acid or base in water and adding enough water to give a final volume of exactly 1 L. Ø If the actual concentrations of all species present in the solution were listed, it would be determined that none of the values is exactly 1 M because a weak acid or a weak base always reacts with water to some extent. Ø Only the total concentration of both the ionized and unionized species is equal to 1 M. Ø The analytical concentration (C) is defined as the total concentration of all forms of an acid or base that are present in solution, regardless of their state of protonation. Ø Thus; a 1 M solution has an analytical concentration of 1 M, which is the sum of the actual concentrations of unionized acid or base and the ionized form.

Ionization Constants for Weak Monoprotic Acids and Bases • Let’s look at the dissolution of acetic acid, a weak acid, in water as an example. • The equation for the ionization of acetic acid is: • The equilibrium constant for this ionization is expressed as:

Ionization Constants for Weak Monoprotic Acids and Bases • The water concentration in dilute aqueous solutions is very high. • 1 L of water contains 55. 5 moles of water. • Thus in dilute aqueous solutions: • The water concentration is many orders of magnitude greater than the ion concentrations. • Thus the water concentration is essentially that of pure water. – Recall that the activity of pure water is 1.

Ionization Constants for Weak Monoprotic Acids and Bases • We can define a new equilibrium constant for weak acid equilibria that uses the previous definition. – This equilibrium constant is called the acid ionization constant. – The symbol for the ionization constant is Ka.

Ionization Constants for Weak Monoprotic Acids and Bases • The ionization constant values for several acids are given below. – Which acid is the strongest? – Are all of these acids weak acids? – What is the relationship between Ka and strength? – What is the relationship between p. H and strength? Acid Formula Ka value p. Ka value -log Ka p. H of 1 M analytical [HA] Acetic CH 3 COOH 1. 8 x 10 -5 4. 7 2. 4 Nitrous HNO 2 4. 5 x 10 -4 3. 3 1. 7 Hydrofluoric HF 7. 2 x 10 -4 3. 1 1. 6 Hypochlorous HCl. O 3. 5 x 10 -8 7. 5 3. 7 Hydrocyanic HCN 4. 0 x 10 -10 9. 4 4. 7

Determining Ka and Kb • The ionization constants Ka and Kb are equilibrium constants that are calculated from experimentally measured concentrations. • What does the concentration of an aqueous solution of a weak acid or base exactly mean? – A 1 M solution is prepared by dissolving 1 mol of acid or base in water and adding enough water to give a final volume of exactly 1 L. – If the actual concentrations of all species present in the solution were listed, it would be determined that none of the values is exactly 1 M – Only the total concentration of both the ionized and unionized species is equal to 1 M. – The analytical concentration (C) is defined as the total concentration of all forms of an acid or base that are present in solution, regardless of their state of protonation. – 1 M solution has an analytical concentration of 1 M, the sum of the actual concentrations of unionized acid or base and the ionized form. Two common ways to obtain the concentrations 1. By measuring the electrical conductivity of the solution, which is related to the total concentration of ions present 2. By measuring the p. H of the solution, which gives [H+] or [OH–]

Determining Ka and Kb • RICE Procedure for determining Ka for a weak acid and Kb for a weak base 1. The analytical concentration of the acid or base ionization Reaction is the Initial concentration 2. The stoichiometry of the reaction with water determines the Change in concentrations 3. The final (Equilibrium) concentrations of all species are calculated from the initial concentrations and the changes in the concentrations 4. Inserting the final concentration into the equilibrium constant expression enables the value of Ka or Kb to be calculated

Ionization Constants for Weak Monoprotic Acids: the math In a 0. 12 M solution of a weak monoprotic acid, HY, the acid is 5. 0% ionized. Calculate the ionization constant for the weak acid. R HY H 3 O + + Y- I C E • Use the concentrations that were just determined in the ionization constant expression to get the value of Ka.

Ionization Constants for Weak Monoprotic Acids the MATH • The p. H of a 0. 10 M solution of a weak monoprotic acid, HA, is found to be 2. 97. What is the value for its ionization constant? • Use the [H 3 O+] and the stoichiometry of the ionization reaction to determine concentrations of all species. R Simplifying Assumption: Is the change significant? Later we will find that in general, if the Ka/[] is < 1 x 10 -3 you can apply the simplifying assumption. C HA I E • Calculate the ionization constant from this information. H 3 O+ + A-

Ionization Constants for Weak Monoprotic Acids the MATH Calculate the concentrations of the various species in 0. 15 M acetic acid, CH 3 COOH, solution. 1. It is always a good idea to write down the ionization reaction and the ionization constant expression. 2. Next we combine the basic chemical concepts with some algebra to solve the problem + - R CH 3 COOH I C E H 3 O + CH 3 COO

Ionization Constants for Weak Monoprotic Acids and Bases • Substitute these algebraic quantities into the ionization expression. • Solve the algebraic equation, using a simplifying assumption that is appropriate for all weak acid and base ionizations.

Ionization Constants for Weak Monoprotic Acids and Bases • Complete the algebra and solve for the concentrations of the species. • Note that the properly applied simplifying assumption gives the same result as solving the quadratic equation does.

Ionization Constants for Weak Monoprotic Acids: the MATH • Calculate the concentrations of the species in 0. 15 M hydrocyanic acid, HCN, solution. Ka= 4. 0 x 10 -10 for HCN R HCN H 3 O+ + CNI C E • Substitute these algebraic quantities into the ionization expression. • Solve the algebraic equation, using the simplifying assumption that is appropriate for all weak acid and base ionizations.

Ionization Constants for Weak Monoprotic Acids and Bases • Let’s look at the percent ionization of two weak acids as a function of their ionization constants. Solution Ka [H+] p. H % ionization 0. 15 M acetic acid 0. 15 M HCN 1. 8 x 10 -5 1. 6 x 10 -3 2. 80 1. 1 4. 0 x 10 -10 7. 7 x 10 -6 5. 11 0. 0051 • Note that the [H+] in 0. 15 M acetic acid is 200 times greater than for 0. 15 M HCN. [ionized HY] % ionization = x 100% [unionized HY]

Ionization Constants for Weak Monoprotic Acids and Bases • All of the calculations and understanding we have at present can be applied to weak acids and weak bases. Calculate the concentrations of the various species in 0. 15 M aqueous ammonia.

Ionization Constants for Weak Monoprotic Bases: the MATH • All of the calculations and understanding we have at present can be applied to weak acids and weak bases. Calculate the concentrations of the various species in 0. 15 M aqueous ammonia. Kb = 1. 8 E-5 R I C E NH 3 NH 4+ + OH-

Ionization Constants for Weak Monoprotic Bases: the MATH • The p. H of an aqueous ammonia solution is 11. 37. Calculate the molarity (original concentration) of the aqueous ammonia solution R NH 3 NH 4+ + OH- I C E • Examination of the last equation suggests that our simplifying assumption can be applied. In other words (x-2. 3 x 10 -3) x. – Making this assumption simplifies the calculation.

Acid-Base Properties of Solutions of Salts • A salt can dissolve in water to produce a neutral, basic, or acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion (A–) or the conjugate acid of a weak base as the cation (BH+), or both. • Salts that contain small, highly charged metal ions produce acidic solutions in water. • The most important parameter for predicting the effect of a metal ion on the acidity of coordinated water molecules is the charge-to-radius ratio of the metal ion. • The reaction of a salt with water to produce an acidic or basic solution is called a hydrolysis reaction, which is just an acid-base reaction in which the acid is a cation or the base is an anion.

Acids, Bases, and ionization constants • Acid and Base strengths can be compared using Ka and Kb values. The larger the Ka or Kb value the more product favored the dissociation. • An acid-base equilibrium always favors the side with the weaker acid and base. stronger acid + stronger base weaker acid + weaker base • In an acid-base reaction the proton always reacts with the strongest base until totally consumed before reacting with any weaker bases. • Any substance whose anion is the conjugate base of a weak acid weaker than OH- reacts quantitatively with water to form more hydroxide ions. H 2 O Step 1. Na. CH 3 COO → Na+ + CH 3 COOAcetate ion is the conjugate base of acetic acid, a weak acid. Step 2. CH 3 COO- + H 2 O CH 3 COOH + OH- • Hydrolysis: Aqueous solutions of salts that dissociate into both: 1. A strong conjugate acid and a strong conjugate base are neutral (KNO 3). 2. A strong conjugate acid and a weak conjugate base are acidic (HCl). 3. A strong conjugate base and a weak conjugate acid are basic (Na. OH). 4. A weak conjugate base and a weak conjugate acid can be neutral, basic or acidic: • The comparison of the values of Ka and Kb determine the p. H of these solutions. a. Kbase = Kacid make neutral solutions (NH 4 CH 3 OO) b. Kbase > Kacid make basic solutions (NH 4 Cl. O) c. Kbase < Kacid make acidic solutions (CH 3)3 NHF

Polyprotic Acids and Bases • Polyprotic acids contain more than one ionizable proton, and the protons are lost in a stepwise manner. • The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion; the fully deprotonated species is the strongest base. • Acid strength decreases with the loss of subsequent protons, and the p. Ka increases. • The strengths of the conjugate acids and bases are related by p. Ka + p. Kb = p. Kw, and equilibrium favors formation of the weaker acid-base pair.

• • • 1 2 3 Polyprotic Acids Many weak acids contain two or more acidic hydrogens. – Examples include H 3 PO 4 and H 3 As. O 4. The calculation of equilibria for polyprotic acids is done in a stepwise fashion. – There is an ionization constant for each step. Consider arsenic acid, H 3 As. O 4, which has three ionization constants. Ka 1 = 2. 5 x 10 -4 Ka 2 = 5. 6 x 10 -8 Ka 3 = 3. 0 x 10 -13 • Notice that the ionization constants vary in the following fashion: • This is a general relationship. – For weak polyprotic acids the Ka 1 is always > Ka 2, etc.

Polyprotic Acids • The first ionization step for arsenic acid is: • The second ionization step for arsenic acid is: • The third ionization step for arsenic acid is:

Polyprotic Acids The MATH • Calculate the concentration of all species in 0. 100 M arsenic acid, H 3 As. O 4, solution. 1 Write the first ionization step and represent the concentrations. Approach this problem exactly as previously done. R I C E The simplifying assumption cannot be used. Using the quadratic equation x = H 3 As. O 4 H 3 O+ + H 2 As. O 4 -

Polyprotic Acids The MATH 2. Next, write the equation for the second step ionization and represent the concentrations and work as before. R I C E The simplifying assumption can be used. H 2 As. O 4 - H 3 O+ + HAs. O 4 -2

Polyprotic Acids The MATH 3. Finally, repeat the entire procedure for the third ionization step. R HAs. O 4 -2 H 3 O+ + As. O 4 -3 I C E The simplifying assumption can be used. 3. 0 x 10 -13/5. 6 x 10 -8 = 5. 4 x 10 -6<1. 0 x 10 -3

Polyprotic Acids • A comparison of the various species in 0. 100 M H 3 As. O 4 solution follows. Species Concentration H 3 As. O 4 0. 095 M H+ 4. 9 x 10 -3 M H 2 As. O 4 - 4. 9 x 10 -3 M HAs. O 42 - 5. 6 x 10 -8 M As. O 43 - 3. 4 x 10 -18 M OH- 2. 0 x 10 -12 M When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. 1. The most acidic group is titrated first, followed by the next most acidic, and so forth 2. If the p. Ka values are separated by at least three p. Ka units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton

Polyprotic Acids

The Common Ion Effect and Buffers The ionization equilibrium of a weak acid (HA) is affected by the addition of either the conjugate base of the acid (A–) or a strong acid (a source of H+); Le. Châtelier’s principle is used to predict the effect on the equilibrium position of the solution • Common-ion effect—the shift in the position of an equilibrium on addition effect of a substance that provides an ion in common with one of the ions already involved in the equilibrium; equilibrium is shifted in the direction that reduces the concentration of the common ion Buffers are characterized by the following: 1. the p. H range over which they can maintain a constant p. H—depends strongly on the chemical properties of the weak acid or base used to prepare the buffer (on K) 2. their buffer capacity, capacity the amount of strong acid or base that can be absorbed before the p. H changes significantly—depends solely on the concentration of the species in the buffered solution (the more concentrated the buffer solution, the greater its buffer capacity) 3. observed change in the p. H of the buffer is inversely proportional to the concentration of the buffer

Buffers are characterized by the following: 1. the p. H range over which they can maintain a constant p. H—depends strongly on the chemical properties of the weak acid or base used to prepare the buffer (on K) 2. buffer capacity, capacity is the number of moles of strong acid or strong base needed to change the p. H of 1 Liter of buffer solution by 1 p. H unit. a. depends solely on the concentration of the species in the buffered solution (the more concentrated the buffer solution, the greater its buffer capacity) b. A general estimate of the buffer capacity is 40% of the sum of the molarities of the conjugate acid and conjugate base 3. observed change in the p. H of the buffer is inversely proportional to the concentration of the buffer

The Common Ion Effect and Buffers • There are two common kinds of buffer solutions: 1 Solutions made from a weak acid plus a soluble ionic salt of the weak acid. 2 Solutions made from a weak base plus a soluble ionic salt of the weak base 1. • Solutions made of weak acids plus a soluble ionic salt of the weak acid One example of this type of buffer system is: – The weak acid - acetic acid CH 3 COOH – The soluble ionic salt - sodium acetate Na. CH 3 COO

Buffer Solutions Weak Acids Plus Salts of Their Conjugate Bases One example of the type I of buffer system is: The weak acid - acetic acid CH 3 COOH H 3 O+ + CH 3 COO~100% Na. CH 3 COO →Na+ + CH 3 COOThe soluble ionic salt - sodium acetate Na. CH 3 COO • This is an equilibrium problem with a starting concentration for both the cation and anion. After calculating the concentration of H+ and the p. H of a solution that is 0. 15 M in both acetic acid sodium acetate yields: R CH 3 COOH I C E H 3 O+ + CH 3 COO-

Buffer Solutions Weak Acids Plus Salts of Their Conjugate Bases Alternatively you might have noticed: [base]/[acid] = 100 log 100 = 0 p. H = p. Ka + 0 = p. Ka = 4. 74 [H+] = Ka = 1. 8 E-5 Solution [H+] p. H 0. 15 M CH 3 COOH 1. 6 x 10 -3 2. 80 0. 15 M CH 3 COOH & 0. 15 M Na. CH 3 COO buffer 1. 8 x 10 -5 4. 74 n [H+] is ~90 times greater in pure acetic acid than in buffer solution. n. Note that the p. H of the buffer equals the p. Ka of the buffering acid.

The Common Ion Effect and Buffers • The general expression for the ionization of a weak monoprotic acid is: • The generalized ionization constant expression for a weak acid is:

The Common Ion Effect and Buffers • If we solve the expression for [H+], this relationship results: • By making the assumption that the concentrations of the weak acid and the salt are reasonable, the expression reduces to:

The Common Ion Effect and Buffers • The relationship developed in the previous slide is valid for buffers containing a weak monoprotic acid and a soluble, ionic salt. • If the salt’s cation is not univalent the relationship changes to:

The Common Ion Effect and Buffers • Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation. The Henderson-Hasselbach equation is one method to calculate the p. H of a buffer given the concentrations of the salt and acid.

Henderson-Hasselbalch - Caveats and Advantages • Henderson-Hasselbalch equation is valid for solutions whose concentrations are at least 100 times greater than the value of their Ka’s • The Henderson-Hasselbach equation is one method to calculate the p. H of a buffer given the concentrations of the salt and acid. • A special case exists for the Henderson-Hasselbalch equation when [base]/[acid] = some power of 10, regardless of the actual concentrations of the acid and base, where the Henderson. Hasselbalch equation can be interpreted without the need for calculations: [base]/[acid] = 10 x log 10 x = x in general p. H = p. Ka + x Examples: 1. when [base] = [acid], [base]/[acid] = 1 or 100, log 1 = 0, p. H = p. Ka, (corresponds to the midpoint in the titration of a weak acid or base) 2. when [base]/[acid] = 10 or 101, log 10 = 1 then p. H = p. Ka + 1 3. when [base]/[acid] =. 001 or 10 -2, log 10 = -2 then p. H = p. Ka -2

Buffer Solutions There are two common kinds of buffer solutions: I. II. Commonly, solutions made from a weak acid plus a soluble ionic salt of the conjugate base of the weak acid. Less common, solutions made from a weak base plus a soluble ionic salt of the conjugate acid of the weak base. Both of the above may also be prepared by starting with a weak acid (or weak base) and add half as many moles of strong base (acid)

Buffer Solutions: Weak Bases Plus Salts of Their Conjugate Acids One example of the type II of buffer system is: The weak base – ammonia NH 3 The soluble ionic salt – ammonium nitrate NH 4 NO 3 NH 4 ++ OH ~100% → NH 4 ++ NO 3 • This is an equilibrium problem with a starting concentration for both the cation and anion. After calculating the concentration of OH- and the p. OH of the solution that is 0. 15 M in aqueous ammonia, NH 3, and 0. 30 M in ammonium nitrate, NH 4 NO 3 yeilds: R NH 3 NH 4+ + OH- I C E • Substitute the quantities determined in the previous relationship into the ionization expression for ammonia.

The Common Ion Effect and Buffers • We can derive a general relationship for buffer solutions that contain a weak base plus a salt of a weak base similar to the acid buffer relationship. – The general ionization equation for weak bases is: • Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation.

The Common Ion Effect and Buffers • A comparison of the aqueous ammonia concentration to that of the buffer described above shows the buffering effect. Solution [OH-] p. H 0. 15 M NH 3 1. 6 x 10 -3 M 11. 20 0. 15 M NH 3 & 0. 15 M NH 4 NO 3 buffer 9. 0 x 10 -6 M 8. 95 n The [OH-] in aqueous ammonia is 180 times greater than in the buffer.

Buffering Action • If 0. 020 mole of gaseous HCl is added to 1. 00 liter of a buffer solution that is 0. 100 M in aqueous ammonia and 0. 200 M in ammonium chloride, how much does the p. H change? Assume no volume change due to addition of the HCl. 1 Calculate the p. H of the original buffer solution. NH 4 NO 3 NH 3 R NH 3 → NH ~100% 4 ++ NO 3 NH 4 ++ OH NH 4+ + OH- I C E • Substitute the quantities determined in the previous relationship into the ionization expression for ammonia.

Buffering Action 2 Next, calculate the concentration of all species after the addition of the gaseous HCl. – The HCl will react with some of the ammonia and change the concentrations of the species. – This is another limiting reactant problem. ~100% R I C E HCl → H + Cl NH 3 + H+ NH 4 + NH 3 + NH 4+ - + OH-

Buffering Action 3 Using the concentrations of the salt and base and the Henderson. Hassselbach equation, the p. H can be calculated. 4 Finally, calculate the change in p. H.

Buffering Action 1. If 0. 020 mole of Na. OH is added to 1. 00 liter of solution that is 0. 100 M in aqueous ammonia and 0. 200 M in ammonium chloride, how much does the p. H change? Assume no volume change due to addition of the solid Na. OH ~100% → Na + OH NH 4 + + OHR I C E NH 3 + - NH 3 NH 4+ + OH-

Buffering Action 2. 3. Using the concentrations of the salt and base and the Henderson. Hassselbach equation, the p. H can be calculated. Finally, calculate the change in p. H.

Buffering Action Original Solution 1. 00 L of solution containing 0. 100 M NH 3 and 0. 200 M NH 4 Cl Original p. H Acid or base added New p. H 0. 020 mol Na. OH 9. 08 +0. 13 0. 020 mol HCl -0. 14 8. 95 8. 81 • Notice that the p. H changes only slightly in each case.

Preparation of Buffer Solutions • Calculate the concentration of H+ and the p. H of the solution prepared by mixing 200 m. L of 0. 150 M acetic acid and 100 m. L of 0. 100 M sodium hydroxide solutions. • Determine the amounts of acetic acid and sodium hydroxide prior to the acid-base reaction. • Na. OH and CH 3 COOH react in a 1: 1 mole ratio. • After the two solutions are mixed, Calculate total volume. • The concentrations of the acid and base are: • Substitution of these values into the ionization constant expression (or the Henderson-Hasselbach equation) permits calculation of the p. H.

Preparation of Buffer Solutions • For biochemical situations, it is sometimes important to prepare a buffer solution of a given p. H. Starting with a solution that is 0. 100 M in aqueous ammonia prepare 1. 00 L of a buffer solution that has a p. H of 9. 15 using ammonium chloride as the source of the soluble ionic salt of the conjugate weak acid. NH 4 Cl ~100% NH 4 + + Cl- NH 3 H 2 O NH 4++ OH- • The Henderson-Hasselbalch equation is used to determine the ratio of the conjugate acid base pair • • p. OH can be determined from the p. H: p. Kb can be looked up in a table: [base] concentration is provided: Solve for [acid]: • Does this result make sense?

Titration Curves Strong Acid/Strong Base Titration Curves • These graphs are a plot of p. H vs. volume of acid or base added in a titration. • As an example, consider the titration of 100. 0 m. L of 0. 100 M perchloric acid with 0. 100 M potassium hydroxide. – In this case, we plot p. H of the mixture vs. m. L of KOH added. – Note that the reaction is a 1: 1 mole ratio.

Strong Acid/Strong Base Titration Curves (An M 1 V 1 M 2 V 2 problem) • Before any KOH is added the p. H of the HCl. O 4 solution is 1. 00. Remember perchloric acid is a strong acid that ionizes essentially 100%. m. L KOH 0 20 50 m. Mol KOH† 0 2 5 90 100 9 10 †m. Mol KOH = ml KOH∙Na. OH M *m. Mol HCl. O 4 = ml HCl. O 4∙ HCl. O 4 M 1: 1 mole ratio ‡m. Mol HCl. O 4 = ml HCl. O 4 - m. Mol KOH m. Mol HCl. O 4 m. L soln. Ұ 10* 100 8‡ 120 5‡ 150 1‡ 0‡ 190 200 Ұm. L p. H 1. 00 1. 18 1. 48 2. 28 7. 00 soln = m. L HCl. O 4 + m. L KOH = 100 m. L + total base added m. Mol H+ = m. Mol HCl. O 4 [H+] = m. Mol HCl. O 4 / m. L soln. p. H = -log[H+]

Strong Acid/Strong Base Titration Curves • We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

Weak Acid/Strong Base Titration Curves • As an example, consider the titration of 100. 0 m. L of 0. 100 M acetic acid, CH 3 COOH, (a weak acid) with 0. 100 M KOH (a strong base). – The acid and base react in a 1: 1 mole ratio. • Before the equivalence point is reached, both CH 3 COOH and KCH 3 COO are present in solution forming a buffer. – The KOH reacts with CH 3 COOH to form KCH 3 COO. • A weak acid plus the salt of a weak acid’s conjugate base form a buffer. • Hypothesize how the buffer production will effect the titration curve.

Weak Acid/Strong Base Titration Curves • The p. H changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. • [H+] of a solution of a weak acid (HA) is not equal to the concentration of the acid (HA) • [H+] depends on both its Ka and the analytical concentration of the acid (HA). • Only a fraction of a weak acid dissociates, so [H+] is less than [HA]; therefore, the p. H of a solution of a weak acid is higher than the p. H of a solution of a strong acid of the same concentration.

Weak Acid/Strong Base Titration Curves (a RICE problem) 1. Determine the p. H of the acetic acid solution before the titration is begun. 2. Solve the algebraic equation for each addition of strong base, using a simplifying assumption that is appropriate for all weak acid and base ionizations. • At the equivalence point, the solution is 0. 500 M in KCH 3 COO, the salt of a strong base and a weak acid which hydrolyzes to give a basic solution. 3. The solution cannot have a p. H=7. 00 at equivalence point. – Both processes make the solution basic. Concentrations must now be calculated using the equation for Kb. Remember that Kw = Ka. Kb and that p. H + p. OH = 14

Weak Acid/Strong Base Titration Curves m. L KOH m. Mol OH† [CH 3 COOH] [CH 3 COO- ] • [H 3 O+] m. L soln. Ұ p. H 0 0 0. 100 1. 34 E-02 100 1. 87 20 2. 0 0. 0800 1. 67 E-02 8. 64 E-05 120 4. 06 50 5. 0 0. 0500 3. 33 E-02 2. 70 E-05 150 4. 57 90 9. 0 0. 0100 4. 74 E-02 3. 80 E-06 190 5. 42 100 10 5. 30 E-06 5. 00 E-02 1. 89 E-09 200 8. 72 110 10 0. 000 5. 00 E-02 2. 10 E-13 210 12. 68 After the equivalence point is reached, the p. H is determined by the excess KOH just as in the strong acid/strong base example.

Weak Acid/Strong Base Titration Curves • We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

• Weak Acid/Strong Base Titration Curves We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve. Identity of the weak acid or base being titrated strongly affects the shape of the titration curve. The shape of titration curves as a function of the p. Ka or p. Kb shows that as the acid or base being titrated becomes weaker (its p. Ka or p. Kb becomes larger), the p. H change around the equivalence point decreases significantly.

Weak Acid/Weak Base Titration Curves • Weak Acid/Weak Base Titration curves have very short vertical sections. • Visual indicators cannot be used. • The solution is buffered both before and after the equivalence point. • Comparison of the respective Ka and Kb values can be used to determine the p. H of the equivalence points of these titrations. a. Kbase = Kacid b. Kbase > Kacid c. Kbase < Kacid neutral solutions basic solutions acidic solutions

Indicators A good indicator must have the following properties: 1. Color change must be easily detected 2. Color change must be rapid 3. Indicator molecule must not react with the substance being titrated 4. The indicator should have a p. Kin that is within one p. H unit of the expected p. H at the equivalence point of the titration • Choosing an indicator for an acid-base titration 1. For titrations of strong acids and strong bases (and vice versa), any indicator with a p. Kin between 4 and 10 will do 2. For the titration of a weak acid, the p. H at the equivalence point is greater than 7, and an indicator such as phenolphthalein or thymol blue, with p. Kin > 7, should be used 3. For the titration of a weak base, where the p. H at the equivalence point is less than 7, an indicator such as methyl red or bromcresol blue, with p. Kin < 7, should be used

Indicators • The chemistry of indicators are described by the general equation • The ionization constant for the deprotonation of indicator H n is: [H+] [ n–] Kin = [H n] • The value of p. Kin determines the p. H at which the indicator changes color

Indicators • If the preceding expression is rearranged the range over which the indicator changes color can be discerned.

Indicators

The Solubility Product, Ksp • Silver chloride, Ag. Cl, is rather insoluble in water. • Careful experiments show that if solid Ag. Cl is placed in pure water and vigorously stirred, a small amount of the Ag. Cl dissolves in the water. • The equilibrium constant expression for this dissolution is called a solubility product constant. – Ksp = solubility product constant

The Solubility Product, Ksp • In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as:

The Solubility Product, Ksp • The same rules apply for compounds that have more than two kinds of ions. • One example of a compound that has more than two kinds of ions is calcium ammonium phosphate.

Determination of Solubility Product Constants • One liter of saturated silver chloride solution contains 0. 00192 g of dissolved Ag. Cl at 25 o. C. Calculate the molar solubility of, and Ksp for, Ag. Cl. • The molar solubility can be easily calculated from the data: • The equation for the dissociation of silver chloride, the appropriate molar concentrations, and the solubility product expression are: • Substitution of the molar concentrations into the solubility product expression gives:

Determination of Solubility Product Constants • 1. One liter of saturated calcium fluoride solution contains 0. 0167 gram of Ca. F 2 at 25 o. C. Calculate the molar solubility of, and Ksp for, Ca. F 2. Calculate the molar solubility of Ca. F 2. • From the molar solubility, we can find the ion concentrations in saturated Ca. F 2. Then use those values to calculate the Ksp. – Note: You are most likely to leave out the factor of 2 for the concentration of the fluoride ion!

Uses of Solubility Product Constants • • The solubility product constant can be used to calculate the solubility of a compound at 25 o. C. Calculate the molar solubility of barium sulfate, Ba. SO 4, in pure water and the concentration of barium and sulfate ions in saturated barium sulfate at 25 o. C. For barium sulfate, Ksp= 1. 1 x 10 -10.

Uses of Solubility Product Constants • The solubility product constant for magnesium hydroxide, Mg(OH)2, is 1. 5 x 10 -11. Calculate the molar solubility of magnesium hydroxide and the p. H of a saturated magnesium hydroxide solution at 25 o. C. • Be careful, do not forget the stoichiometric coefficient of 2! • Substitute the algebraic expressions into the solubility product expression. • Solve for the p. OH and p. H.

The Common Ion Effect and Solubility • Solubility product expression – Equilibrium concentrations of cation and anion are inversely related – As the concentration of the anion increases, the maximum concentration of the cation needed for precipitation to occur decreases, and vice versa – Ksp is constant • Common ion effect – The solubility of an ionic compound depends on the concentrations of other salts that contain the same ions. – This dependency is an example of the common ion effect; adding a common cation or anion shifts a solubility equilibrium in the direction predicted by Le. Châtelier’s principle. – The solubility of any sparingly soluble salt is almost always decreased by the presence of a soluble salt that contains a common ion.

The Common Ion Effect in Solubility Calculations • Calculate the molar solubility of barium sulfate, Ba. SO 4, in 0. 010 M sodium sulfate, Na 2 SO 4, solution at 25 o. C. Compare this to the solubility of Ba. SO 4 in pure water. (Example 20 -3). (What is the common ion? How was a common ion problem solved in Chapter 19? ) 1. Write equations to represent the equilibria. 2. Substitute the algebraic representations of the concentrations into the Ksp expression and solve for x. • The molar solubility of Ba. SO 4 in 0. 010 M Na 2 SO 4 solution is 1. 1 x 10 -8 M. • The molar solubility of Ba. SO 4 in pure water is 1. 0 x 10 -5 M. – Ba. SO 4 is 900 times more soluble in pure water than in 0. 010 M sodium sulfate! – Adding sodium sulfate to a solution is a fantastic method to remove Ba 2+ ions from solution! • If your drinking water were suspected to have lead ions in it, suggest a method to prove or disprove this suspicion.

Qualitative Analysis Using Selective Precipitation

The Ion Product • The ion product (Q) of a salt is the product of the concentrations of the ions in solution raised to the same powers as in the solubility product expression. • The ion product describes concentrations that are not necessarily equilibrium concentrations, whereas Ksp describes equilibrium concentrations. • The process of calculating the value of the ion product and comparing it with the magnitude of the solubility product is a way to determine if a solution is unsaturated, or supersaturated and whether a precipitate will form when solutions of two soluble salts are mixed.

The Ion Product • Three possible conditions for an aqueous solution of an ionic solid 1. Q < Ksp: the solution is unsaturated, and more of the ionic solid will dissolve 2. Q = Ksp: the solution is saturated and at equilibrium 3. Q > Ksp: the solution is supersaturated, and ionic solid will precipitate

The Reaction Quotient in Precipitation Reactions • • The reaction quotient, Q, and the Ksp of a compound are used to calculate the concentration of ions in a solution and whether or not a precipitate will form. We mix 100 m. L of 0. 010 M potassium sulfate, K 2 SO 4, and 100 m. L of 0. 10 M lead (II) nitrate, Pb(NO 3)2 solutions. Will a precipitate form? 1. Write out the solubility expressions. 2. Calculate the Qsp for Pb. SO 4. – Assume that the solution volumes are additive. – Concentrations of the important ions are: 3. Finally, calculate Qsp for Pb. SO 4 and compare it to the Ksp.

The Reaction Quotient in Precipitation Reactions • Suppose we wish to remove mercury from an aqueous solution that contains a soluble mercury compound such as Hg(NO 3)2. We can do this by precipitating mercury (II) ions as the insoluble compound Hg. S. What concentration of sulfide ions, from a soluble compound such as Na 2 S, is required to reduce the Hg 2+ concentration to 1. 0 x 10 -8 M? For Hg. S, Ksp=3. 0 x 10 -53. • What volume of the solution (1. 0 x 10 -8 M Hg 2+ ) contains 1. 0 g of mercury? • Equlibria that simultaneously involve two or more different equilibrium constant expressions are simultaneous equilibria.

Simultaneous Equilibria Involving Slightly Soluble Compounds • If 0. 10 mole of ammonia and 0. 010 mole of magnesium nitrate, Mg(NO 3)2, are added to enough water to make one liter of solution, will magnesium hydroxide precipitate from the solution? For Mg(OH)2, Ksp = 1. 5 x 10 -11. Kb for NH 3 = 1. 8 x 10 -5. Calculate Qsp for Mg(OH)2 and compare it to Ksp. – Mg(NO 3)2 is a soluble ionic compound so [Mg 2+] = 0. 010 M. – Aqueous ammonia is a weak base that we can calculate [OH-]. – 1. 2. Once the concentrations of both the magnesium and hydroxide ions are determined, the Qsp can be calculated and compared to the Ksp. 3. Equlibria that simultaneously involve two or more different equilibrium constant expressions are simultaneous equilibria.

Simultaneous Equilibria Involving Slightly Soluble Compounds • How many moles of solid ammonium chloride, NH 4 Cl, must be used to prevent precipitation of Mg(OH)2 in one liter of solution that is 0. 10 M in aqueous ammonia and 0. 010 M in magnesium nitrate, Mg(NO 3)2 ? (Note the similarity between this problem and Example 20 -12. ) • Calculate the maximum [OH-] that can exist in a solution that is 0. 010 M in Mg 2+. • Using the maximum [OH-] that can exist in solution, determine the number of moles of NH 4 Cl required to buffer 0. 10 M aqueous ammonia so that the [OH-] does not exceed 3. 9 x 10 -5 M. • Check these values by calculating Qsp for Mg(OH)2. • Use the ion product for water to calculate the [H+] and the p. H of the solution.

In General How to Solve and Predict Reversible Reactions 1. Are the compounds involved strong or weak electrolytes a. If strong then there is a 100% dissociation and no way back use a single headed arrow and when adding the simultaneous equations together only and add the products. i. e. Na. OH → Na+ + OHb. if a weak electrolyte use a double headed arrow and when adding simultaneous equations together add both reactants and products i. e. CH 3 COOH ↔ CH 3 COO- + H+ 2. Determine if reactions are neutralization, dissociation, or equilibrium reactions I. Neutralization a. Class I – strong electrolyte/strong electrolyte: produces salt and water. Use ICE table with units of amount (mmol or moles), cancel terms, determine amounts of products. Stop. b. Class II – weak electrolyte/weak electrolyte: Compare Kas or Kbs to determine which predominates products or reactants. Use the equlibrium equation and ICE table with units of concentration (mmol/m. L or mol/L) to determine equilibrium concentrations. Stop. c. Class III – strong electrolyte/weak electrolyte: produces salt of the conjugate and water and further actions, Dissociation or Equilibrium reactions must be determined continue to 2 II II. Dissociate any products and begin again at repeat step 1 then continue to 2 III. Products of the dissociation though by name appear to be the same compounds they are not identical in that they have a different source and are treated separately.

III. Equilibrium – write new stoichiometrically balanced chemical equilibrium equations. 3. Write the equilibrium equation and solve for x and determine [H+] or [OH-] 4. Find p. H or p. OH 5. Determine the p. OH from p. H or p. H from p. OH 6. Determine the [H+] or [OH-] not found in step 3 Ksp Write a stoichiometrically balanced equation for the limited dissociation of the solid If provided with the Ksp and looking for molar solubility of products Write the stoichiometric ratio, i. e. 1: 2 Multiply the ratio through by x, i. e. x: 2 x Write the equilibrium equation, i. e. Ksp = [M 2+][Y-]2 Substitute in the stoichiometrically determined x values for the [ ]s, Ksp = (x)(2 x)2 Solve for x If provided a mass determined to be in solution and looking for Ksp convert mass to moles convert moles to molarity Write the equilibrium equation, i. e. Ksp = [M 2+][Y-]2 Solve for Ksp