* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download 14.1 Dynamic Equilibrium, Keq , and the Mass Action Expression
Electrolysis of water wikipedia , lookup
Chemical potential wikipedia , lookup
Multi-state modeling of biomolecules wikipedia , lookup
Gas chromatography–mass spectrometry wikipedia , lookup
Supramolecular catalysis wikipedia , lookup
Hydrogen-bond catalysis wikipedia , lookup
Electrochemistry wikipedia , lookup
Asymmetric induction wikipedia , lookup
Marcus theory wikipedia , lookup
Nucleophilic acyl substitution wikipedia , lookup
Photoredox catalysis wikipedia , lookup
Acid dissociation constant wikipedia , lookup
Thermomechanical analysis wikipedia , lookup
Physical organic chemistry wikipedia , lookup
Thermodynamics wikipedia , lookup
Photosynthetic reaction centre wikipedia , lookup
Process chemistry wikipedia , lookup
Thermodynamic equilibrium wikipedia , lookup
Hydroformylation wikipedia , lookup
George S. Hammond wikipedia , lookup
Strychnine total synthesis wikipedia , lookup
Chemical reaction wikipedia , lookup
Lewis acid catalysis wikipedia , lookup
Rate equation wikipedia , lookup
Crystallization wikipedia , lookup
Stability constants of complexes wikipedia , lookup
Click chemistry wikipedia , lookup
Chemical thermodynamics wikipedia , lookup
Bioorthogonal chemistry wikipedia , lookup
Stoichiometry wikipedia , lookup
Transition state theory wikipedia , lookup
14.1 Dynamic Equilibrium, Keq , and the Mass Action Expression The Equilibrium Process Dr. Fred Omega Garces Chemistry 201 Miramar College 1 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Extent of a Reaction Chemical Reaction Most reactions do not occur with 100% conversion to products. At the molecular, when a reaction occurs to form products, some products will back react to form reactants. The extent of the reaction i.e., 20% or 80% can be determine by measuring concentration of each component in solution In general the extent of the reaction is a function of the following: Temperature, concentration and degree of organization, all of which are monitored by some constant value called the – The equilibrium constant, Keq 2 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 How Equilibrium is achieve Consider the reaction A D B Forward A g B ratef = kf[A] Reverse B f A rateb = kb[B] Overall A D B rate forward kf[A] = rate reverse kb[B] Example: N2O4 D NO2 Effect of temperature on the NO2 D N2O4 equilibrium. The tubes contain a mixture of NO2 and N2O4. As predicted by Le Chatelier’s principle, the equilibrium favors colorless N2O4 at lower temperatures. but shifts to the darker brown NO2 at higher temperature. 3 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Equilibrium: Mass Action Expression When equilibrium is establish, A D B “D” illustrates that rate forward = rate reverse or Kf[A] = Kb[B] – Rearranging this equation yields Kf/Kb = [B] / [A] which yields – The mass action expression : Keq = [B] / [A] For any generic chemical process at equilibrium aA + bB D pP + qQ A mass action expression can be written: p q P ] ∗ [Q ] [ K eq = a b [A] ∗ [B] This is also referred to as the Law of Mass Action 4 € Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Haber Process Consider the Haber Process: N2 (g) + 3H2 (g) D 2NH3 (g) (not 100 % process) As soon as NH3 is form, it back reacts and forms N2 and H2. Law of Mass Action Or Mass Action Expression 2 K eq = [NH3 ] 3 [N2 ] ∗ [H2 ] = Kc € 5 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Meaning of Keq : Sidebar Review of Fractions x y < 1 : If x < y x y > 1 : If x > y x y = 1 : If x = y the denominator dominates. € B) For x > y,€ the numerator dominates. C) For x = y, numerator and denominator are equal. A) For x < y, € 6 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Meaning of Keq Which is favored, Reactant or Product? p k eq € q x P ] • [Q ] Product] [ [ = = [A] • [B] [Reactant] A b y A) For Keq < 1, the ratio [Product]x < [Reactant]y B) For Keq > 1, the ratio [Product]x > [Reactant]y C) For Keq = 1, the ratio [Product]x = [Reactant]y A) For Keq < 1, at equilibrium Reactant is favored. B) For Keq > 1, at equilibrium Product is favored. C) For Keq = 1, at equilibrium Product and Reactant are equal. 7 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Understanding the Concept 1. What is the distinction between rates and the extent of the reaction? 2. What does the term dynamic equilibrium mean physically and at the atomic level? 3. What does the magnitude of Keq mean in terms of the chemical reaction? 4. If there is no change in concentration of reactants and product at equilibrium, why is it considered dynamic? 8 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 For Reactions Not at Equilibrium For reactions not yet at equilibrium, the Law of Mass action yield information in terms of the Reaction Quotient. Consider the following chemical process not at equilibrium. aA + bB D rR + pP A reaction quotient expression can be written: [R]r [P]p Q= [A]a [B]b Where the numerical value of Q will determine the direction the reaction will proceed. Q < Keq €Q > K eq 9 Reaction shifts to g Right Reaction shifts to f Left Dynamic Equilibrium , Keq and the Mass Action Expression January 13 LeChatelier’s Principle (Preview) Concentration Effect If a chemical system is at equilibrium and then a substance is added (either a reactant or product), the reaction will shift so as to re-establish equilibrium by subtracting part of the added substance. Conversely, removal of a substance will result in the reaction moving in the direction that forms more of the substance. Consider the Haber reaction that was discussed earlier. N2 + 3H2 D 2NH3 + E If some H2 is added to the reaction which was at equilibrium, the system self-adjust to remove the excess H2 by converting it to NH3 until equilibrium is re-establish; in the process some N2 is also consumed. 10 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Pictorial View of LeChatelier Teeter•Totter At Equilibrium Stress applied Self Adjust Re-establish Equilibrium 11 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Keq and the Q The change in Q during a reaction and its relation to Keq: Consider the reaction: N2O4 D 2 NO2 The reaction coordinate diagram shows how the concentration of N2O4 and NO2 changes as the reaction approaches equilibrium. This is also reflected in Q. As the reaction proceeds to the right, N2O4 to NO2, the Q value increases, N2O4 becomes smaller and NO2 becomes larger. The reaction finally reaches equilibrium at teq, at which time the concentration of N2O4 and NO2 remains constant. Equilibrium is reached and the reaction quotient becomes equal to the equilibrium constant. Q = Keq. 12 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Heterogeneous Systems The equilibrium of substances with different phases. i.e., solid and aqueous coexisting Which solid is more concentrated? Solid and gas coexisting Which solid is more concentrated? Concentration of a solid is always a Constant; Even though the two containers contain different amounts of solid solute, as long as both solids are present at a given temperature, A) the concentration of the solution is the same, B) the partial pressure of the gas is the same. Concentration of solid remains the same (constant) ! The solid component is not taken into account in the Mass Action expression or the Reaction Quotient. 13 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Heterogeneous Systems CaCO3 Consider the following reaction: CaCO3(s) D CaO (s) + CO2 (g) k eq = but, [CaO] ∗ [CO2 ] [CaCO3 ] [CaO] and [CaCO3 ] Constant1 ] ∗ [CO2 ] [ k eq = [Constant2 ] k eq = [CO2 ] € 14 = Constant The Mass Action Expression for a heterogenerous system need not take into account the concentration of the solid. Even though the two containers have different amount of solids (CaCO3 and CaO), as long as both solids are present at the same temperature, the partial pressure of CO2 is the same at equilibrium. Dynamic Equilibrium , Keq and the Mass Action Expression January 13 In Class exercise The reaction is in a 10.0 L vessel. The initial concentration of [H2] is 12.0 molar and that of [N2] is 4.0 molar. i) Sketch the change of the concentration of N2 from 0 to 3.5 min. T = 200 K ii) At 2 minutes, Q is: a) increasing b) decreasing c) not changing d ) cannot be determine iii) At 5 minutes, Q is: a) increasing b) decreasing c) not changing d ) cannot be determine iv) Calculate the equilibrium constant at the 4.0-minute mark. 2.1•10-2 M-2 v) Calculate the equilibrium constant at the 8.0-minute mark. 2.1•10-2 M-2 15 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Keq as a Function of the Written Reaction What is the equilibrium constant if a reaction is written in the reverse direction? Consider, COCl2 and the reverse reaction CO (g) + Cl2 (g) (g) D CO (g) + Cl2 (g) D COCl2 (i) (ii) (g) What is the relationship of Keq between both reaction (i) & (ii)? K eq (i) = [CO] [Cl2 ] [COCl2 ] Inverse relationship - K eq (i) = K eq (ii) [COCl2 ] K eq (ii) = [CO] [Cl2 ] hence [CO] [Cl2 ] " [COCl2 ] % =$ [COCl2 ] # [CO] [Cl2 ] '& € 16 € K forward = −1 Dynamic Equilibrium , Keq and the Mass Action Expression -1 1 K reverse January 13 Effects of Keq and Variation of a Chem Equation Consider the variation of a chemical reaction: A D B Keq = [B] / [A] B D A Krev = [A] / [B] Krev = 1 / Keq K’ = [B]1/2 / [A]1/2 K’ = (Keq)1/2 K” = [A]2 / [B]2 K ” = (Keq)-2 1/2 A D 1/2 2B D 2A n A D Note: 17 n B B K’ = [B] n / [A] n K’ = (Keq) n Krev = 1 / Keq or Keq = 1 / Krev K’ = (Keq)1/2 or Keq = K’2 K ” = (Keq)-2 or Keq = (K ” )-1/2 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Expressing Q and Calculating K The relationship of the Mass Expression, Keq, can be extended to the Reaction Quotient, Q: 18 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Example 2: Keq relationship Consider the following equilibrium, at 480 °C: 2Cl2(g) + 2H2O (g) D 4 HCl(g) + O2 (g) Kp = 0.0752 a) What is the value of Kp for: 4 HCl(g) + O2 (g) D 2Cl2(g) + 2H2O (g) b) What is the value of Kp: Cl2(g) + H2O (g) D 2HCl(g) + 1/2 O2 (g) c) What is the value of Kc: 2 Cl2(g) + 2H2O (g) D 4HCl(g) + O2 (g) (To be discussed later) a) This equation is the reverse of the original b) This equation is half the original c) This equation will be solve in the next series of slides. 19 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Relationship Between Kc and Kp Consider the Reaction: COCl2 (g) D CO (g) + Cl2 (g) ( p CO) * ( p Cl2 ) KP = ( p COCl2 ) or [CO] [Cl2 ] Kc = [COCl2 ] It can be shown that: € K c = K p (RT) −Δn € K p = K c (RT) + Δn [RT is positive exponent (+Δn)] where Δn = ∑ (gas moles) − ∑ (gas moles) product reactant € 20 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Derivation Kc and Kp relationship [ P CO] ∗ [ P Cl2 ] kp = [ P COCl2 ] and kc CO] ∗ [Cl2 ] [ = [COCl2 ] but PV = nRT therefore € kp € € 21 n nRT = RT = [conc] ∗ RT V V CO] ∗ [ Cl ] {[CO] ⋅ RT} ∗ {[Cl ] ∗ RT} [ = = {[COCl ] ∗ RT} [ COCl ] p p p kp P= 2 2 2 2 CO] ∗ [Cl2 ] RT ∗ RT [CO] ∗ [Cl2 ] RT [ = ∗ = ∗ RT [COCl2 ] [COCl2 ] [CO] ∗ [Cl2 ] ∗ RT +1 = k ∗ RT +Δn kp = c [COCl2 ] 1 ∴ k p = k c ∗ RT +Δn and k c = k p ∗ RT −Δn where Δn = ∑ (gas moles) − ∑ (gas moles) product reactant Dynamic Equilibrium , Keq and the Mass Action Expression January 13 Summary: Equilibrium constant • The mass action expression is equal to the equilibrium constant. • The equilibrium constant value is generally written without units. • The equilibrium expression for a reaction is the reciprocal of that reaction when it is written in the reverse direction. • When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the factored reaction is the original expression raised to the nth power. Equation 1ref: R D P, Equation 2: nR D nP, Thus K2 = (k1ref)n The equilibrium constant express in terms of concentration, Kc, is related to the equilibrium constant expressed in terms of partial pressure, Kp, by the amount of gas moles changes during the course of the reaction. 22 Dynamic Equilibrium , Keq and the Mass Action Expression January 13 14.2 Solving Equilibrium Problems The iΔe Process Dr. Fred Omega Garces Chemistry 201 Miramar College 1 Solving Equilibrium Problems January 13 Solving Equilibrium Problems There are many type of equilibrium problems which arise in the real world and in chemistry exams. These problems can be broken down to two basic types. 1. The first type is one in which the equilibrium concentrations are given and the equilibrium constant must be solved. Given: [Conc] at equilb Solve: Keq 2. The second type is one in which both the equilibrium constant and the initial concentration are given and the concentrations at equilibrium are solved for. Given: [Conc]o and Keq Solve: [Conc] at equilb 2 Solving Equilibrium Problems January 13 Type 1: Equilibrium The first type of problem is the type in which the equilibrium concentrations are given and the equilibrium constant, Keq, must be solved. Given: [Conc] at equilb Solve: Keq Steps involved in the equilibrium calculations. 1. Write the balanced equation for the reaction. 2. Write the equilibrium expression using the Mass Action Expression. 3. Plug the equilibrium concentration into the Mass Action and solve for the equilibrium constant, Keq . 3 Solving Equilibrium Problems January 13 1. Calculating Keq: Kc - Kp relationship (Type1) Carbon dioxide is placed in a 5.00-L high-pressure reaction vessel at 1400.°C. Given the following reaction, CO (g) + 1/2 O2 (g) D CO2 (g) , calculate the equilibrium constant Kc and Kp after equilibrium is achieved with 4.95 mol of CO2 , 0.0500 mol of CO, and 0.0250 mol of O2 found in the container. T = 1400°C V = 5.0 L CO(g) + i Δ mol e 1 2 O2 (g) ! CO2 (g) 0.050 mol 0.025mol 4.95mol 1.00 ⋅10 -2 M 5.00 ⋅10 -3 M 9.90 ⋅10 -1 M Vol = 5.0 L [e] € Supplemental: Calculate the partial pressure of each gas at equilibrium 4 Solving Equilibrium Problems January 13 2. Calculating Keq iCe - method (Type1 & 2) A mixture of 0.1000 mol of CO2 , 0.05000 mol of H2, and 0.1000 mol of H2O is placed in a 1.000 L vessel. The following equilibrium reaction is established: CO2 (g) + H2 (g) D CO(g)+ H2O (g) At equilibrium [CO2] = 0.0954 M. (a) Calculate the equilibrium concentrations of H2, CO, and H2O. (b) Calculate the Kc for the reaction (c) Calculate Kp for the reaction? CO2(g) V = 5.0 L i 0.1000M Δ −X [e] 0.0954 + H2 (g) 0.0500 −X ? ! CO (g) + +X ? H2O (g) 0.1000 +X ? € Supplemental: Calculate the partial pressure of each gas at equilibrium. 6 Solving Equilibrium Problems January 13 Type 2: Equilibrium The second type of problems provides information on both the equilibrium constant and the initial concentration. The concentrations at equilibrium must be solved for. Given: [Conc]o and Keq Solve: [Conc] at equilb Steps involve in the equilibrium calculations. 1. Write the balanced equation for the reaction. 2. Write the equilibrium expression using the Mass Action Expression. 3. List the initial conditions. 4. Set up the iΔe table and solve for the equilibrium concentration in terms of a variable (x). 5. Plug the equations expressing the equilibrium concentration into the mass action expression and solve for x. 6. Assign the value of x to the equilibrium concentration equation and determine the numerical value for the equilibrium concentration for each specie in the reaction. 8 Solving Equilibrium Problems January 13 3a. Calculation: Product formed @ Equilibrium (Type2) How much SO3 is formed when 0.50 mol of SO2 and 0.50 mol NO2 are in 1.00 L at 821 °C. Kc = 6.85 at this temperature. SO2(g) + NO2(g) i Δ 0.50 M −X 0.50M −X 0.00 +X 0.00 +X [e] 0.50 - X 0.50 - X +X +X ! SO3 (g) + NO(g) Supplemental: If this was a stoichiometry problem, what is the mole of SO3 when reaction is complete? 9 Solving Equilibrium Problems January 13 3a. Calculation: Product formed @ Equilibrium (Type2) How much SO3 is formed when 0.50 mol of SO2 and 0.50 mol NO2 are in 1.00 L at 821 °C. Kc = 6.85 at this temperature. 10 SO2(g) + NO2(g) i Δ 0.50 M −X 0.50M −X 0.00 +X 0.00 +X [e] 0.50 - X 0.50 - X +X +X ! SO3 Solving Equilibrium Problems (g) + NO(g) January 13 3b. Calculation: Product formed @ Equilibrium (Type2) How much SO3 is formed when 0.50 mol of SO2 and 1.00 mol NO2 are in 1.00 L at 821 °C. Kc = 6.85 at this temperature. (Non-perfect square) SO2(g) i Δ [e] 12 + NO2(g) ! SO3 (g) + NO(g) 0.50 M −X 1.00M −X 0.00 +X 0.00 +X 0.50 - X 1.00 - X +X +X Solving Equilibrium Problems January 13 4. Calculating Keq: Method of Approximation (Type2) NOCl decomposes to form the gases NO and Cl2. At 35°C the equilibrium constant is Kc = 1.6•10-5. If 1.0 mol of NOCl is placed in a 2.0 L flask what are the concentration of all specie at equilibrium? 2NOCl(g) V = 2.0 L 14 ! 2NO + Cl2 (g) i Δ 0.50 M − 2X 0 + 2X 0 +X [e] 0.5 - 2X + 2X +X Solving Equilibrium Problems (g) K c = 1.6⋅ 10 -5 January 13 Assumption Check When making assumptions, if a reaction has a relatively small keq and a relatively large initial reactant concentration, then the concentration change (x) can often be neglected without introducing significant error. This does not mean x = 0, because then this would mean there is no reaction. It means that if a reaction proceeds very little (small k) and if you start with a high reactant concentration, very little will be used up, so the following holds. [react]o -x ≈ [react]eq ≈ [react]o When making the assumption that x is negligible, you must check that the error introduced is not significant. If the assumption results in a change (error) in concentration of less than 5%, the error is not significant and the assumption is justified. To test the assumption, use the following formula: (Δ conc / initial concentration) • 100 < 5% In the previous problem, the assumption is check by the following calculations: [2x / (0.5) ] • 100 = [ 2(1•10-2) / 0.5 ]•100 = 4% < 5% The assumption is valid in this example. 16 Solving Equilibrium Problems January 13 Assumption Check [React]initial / keq In general, simplifying assumptions works when the initial concentration of the reactant is high but not when it is low. To summarize, we assume that x (Δ [React]) can be neglected if Keq is small relative to [React]initial. The benchmark in justifying the assumption isYour job is to determine what is the threshold for [React]initial / keq so that the assumption is justified. If [React]initial / ka > 400, the assumption is justified; neglecting x introduces an error of < 5% If [React]initial / ka < 400, the assumption is not justified; neglecting x introduces an error of > 5% 17 Solving Equilibrium Problems January 13 5. Calculating Keq: Quadratic or iteration? At 250 ° C, the reaction PCl5 (g) D PCl3 (g) + Cl2 (g) has an equilibrium constant Kc = 1.80. If 0.100 mol PCl5 is added to a 5.00-L vessel, what are the concentrations of all specie at equilibrium ? What % of the reactant remains? PCl5 (g) ! PCl3 (g) + Cl2(g) i Δ 0.02 M −X 0 +X 0 +X [e] 0.02 - X +X +X 18 Solving Equilibrium Problems January 13 6. Calculation: Product formed @ Equilibrium (Type2) Exactly 4 mol of SO3 is sealed in a 5.0 L container @1500K. Kc is 9.34•103 for the rxn, what are the conc. of all specie at equilb? 20 Solving Equilibrium Problems January 13 7. ... Another equilibrium problem (difficult) An engineer is studying the oxidation of SO2 to produce the precursor, SO3, in the manufacture of sulfuric acid. The Kp = 1.7 • 108 at 600. K for the reaction. 2SO2 (g) + O2(g) D 2SO3 (g) i) At equilibrium pSO3 = 300. atm and pO2 = 100. atm. Calculate pSO2 ii) The engineer places a mixture of 0.0040 mol of SO2(g) and 0.0028 mol of O2(g) in a 1.0-L container and raises the temperature to 1000K. The reaction reaches equilibrium and 0.0020 mol of SO3 (g) is present. What would the engineer calculate the Kc and PSO2 for the reaction to be 1000K. Does the engineer expects the reaction to be exothermic or endothermic? 22 Solving Equilibrium Problems January 13 Procedure for Solving Equilibrium Problems 1. 2. 3. 4. Write the balanced equation for the reaction. Write the equilibrium expression using the Mass Action Expression. List the initial conditions. Determine the type of equilibrium problem you have. Type 1 - Given the equilibrium concentration, solve for Keq . Type 2 - Given Keq and initial concentration, solve for conc. @ equilb. Type1 Plug the equilibrium concentration into the Mass Action and solve for the equilibrium constant, Keq . Type2 a) Set up the iΔe table and solve for the equilibrium concentration in terms of a variable (x). b) Plug the equations expressing the equilibrium concentration into the mass action expression and solve for x c) Assign the value of x to the equilibrium concentration equation and determine the numerical value for the equilibrium concentration for each specie in the reaction. 23 Solving Equilibrium Problems January 13 14.3 LeChatelier Predicting Direction of a Chemical Reaction Dr. Fred Omega Garces Chemistry 201 Two equal-weight boys at equilibrium on a teeter totter. The boy at the right is given a five-pound weight thereby disturbing the equilibrium. The boy on the left scoots farther back on the teeter totter to restore the equilibrium. 1 LeChatlier and Chemical Direction Miramar College January 13 Quotient: Review For reactions not yet at equilibrium, the Law of Mass action yield information in terms of the Reaction Quotient. Consider the following chemical process not at equilibrium. aA + bB D rR + pP A reaction quotient expression can be written: [R]r [P]p Q= a b [A] [B] Where the numerical value of Q will determine the direction the reaction will proceed. € 2 Q < Keq Reaction shifts to g Right Q > Keq Reaction shifts to f Left LeChatlier and Chemical Direction January 13 Reaction Quotient: Direction of Reaction Consider, T =532 °C, Kc = 0.19 what direction will the reaction proceed if the initial concentration is change to- N2 (g) i) 0.30 M + 3H2 D (g) 0.20 M 2NH3 (g) 0.10 M 2 0.10M] [ Q= [0.30M] ∗ [0.20M] 3 = € 3 LeChatlier and Chemical Direction January 13 LeChatelier Principle: A Review Teeter•Totter At Equilibrium Stress applied Self Adjust Re-establish Equilibrium 5 LeChatlier and Chemical Direction January 13 Equilibrium: Stress / Relief on Reactant Altering Chemical Concentrations Stress on Reactant, Rxn shift right 6 LeChatlier and Chemical Direction Relief on Reactant, Rxn shift left January 13 Equilibrium: Stress / Relief on Product Altering Chemical Concentrations Stress on Product, Rxn shift left 7 LeChatlier and Chemical Direction Relief on Product, Rxn shift right January 13 Exothermic Heats of Solution Exothermic Process. Energy is a product Heating a solution in which the ΔHsoln is exothermic Energy is a product, ΔHsoln ( - ), results in a shift of the reaction to the left, or more solute precipitating out of solution. 8 LeChatlier and Chemical Direction January 13 Endothermic Heats of Solution Endothermic Process. Energy is a reactant Heating a solution in which the ΔHsoln is endothermic, Energy is a reactant, ΔHsoln ( + ), results in a Shift of the reaction to the right, or more solute dissolving in solution. 9 LeChatlier and Chemical Direction January 13 Solubilities of Solids Vs. Temperature Solubility of several ionic solid as a function of temperature. Most salts have positive heats of solution. When the salt solution is heated, more solute dissolves. Some salts have negative enthalpy of solution, (exothermic process) i.e., Ce2(SO4)3. When these salt solutions are heated, the solute becomes less soluble. Other example: Mg(OH)2 and Starch 10 LeChatlier and Chemical Direction January 13 2ii) Temperature & Solubility: Gases Temperature - (Gas) Consider the extent in which O2 or CO2 dissolves in water. What are the conditions which will increase the solubility of gas in water. As the temperature increase, both solute and solvent will be moving faster, the gas solute however will now have enough energy to leave the liquid interface [Solute] D gas [Solute] Solution Is this an exothermic or endothermic process? 11 LeChatlier and Chemical Direction January 13 Gas solute; Exothermic ΔHsoln As the temperature increase, both solute and solvent will be move faster. The gas solute however will now have enough energy to leave the liquid interface because IMF can be overcome 12 LeChatlier and Chemical Direction January 13 Disaster: (1700 dead) from Gas Solubility Lake Nyos in Cameroon, the site of a natural disaster. In 1986 a huge bubble of CO2 escaped from the lake and asphyxiated more than 1700 people. http://hubpages.com/hub/The-Exodus-Decoded 13 LeChatlier and Chemical Direction In the African nation of Cameroon in 1986 a huge bubble of CO2 gas escaped from Lake Nyos and moved down a river valley at 20 m/s (about 45 mph). Because CO2 is denser than air, it hugged the ground and displaced the air in its path. More than 1700 people suffocated. The CO2 came from springs of carbonated groundwater at the bottom of the lake. Because the lake is so deep, the CO2 mixed little with the upper layers of water, and the bottom layer became supersaturated with CO2. When this delicate situation was changed, perhaps because of an earth-quake or landslide, the CO2 came out of the lake water just like it does when a can of soda is opened. January 13 The effect of a Change in Temperature in terms of Reaction Quotient and LeChatelier In an endothermic process, energy is a reactant and an increase in temperature results in a shift of the reaction to the right. When the temperature is decreased the reaction shifts to the left. In an exothermic process, energy is a product and an increase in temperature results in a shift of the reaction to the left. When the temperature is decreased the reaction shifts to the right. The question is raised, why does the reaction adjust itself, when according to the Mass Action Expression, the concentrations of chemicals are not altered with temperature changes? What causes the Mass Action Expression not to equal Keq(new) (≠ keq(old)). Consider an Endothermic reaction: E + R D P: Keq = [Prod] / [React] . If the temperature is raised, the reaction shifts to the right, (Q < Keq ) A shift to the right means that [Prod] = h (will raised) and [React] = i (will lowers). This will only occur if the [Prod] / [React] (or Q) is now less than Keq @ new the temperature. In other words when the temperature is change (increase T), the equilibrium constant changes, the current [P] / [R] ratio is still equal to the old Keq which is now Q (the reaction quotient) and the reaction shifts to re-establish equilibrium that is to attain the new Keq. 14 LeChatlier and Chemical Direction January 13 Temperature Effect (increase) & Reaction Quotient Exothermic Rxn: Increase in Temperature Endothermic Rxn: Increase in Temperature K eq(old Temp) = Q Direction Reaction Keq Direction Reaction K eq(old Temp) = Q @ When the temperature is When the temperature is New raised for an exothermic raised for an endothermic Temperature reaction, the Keq constant reaction, the Keq constant changes. Because the changes. Because the current concentrations yields current concentrations yields a reaction quotient greater a reaction quotient less than than Keq (new) the reaction Keq (new) the reaction must must shift to the left.) shift to the right. 15 LeChatlier and Chemical Direction January 13 Temperature Effect (decrease) & Reaction Quotient Exothermic Rxn: Decrease in Temperature K eq(old Temp) = Q Direction Reaction Endothermic Rxn: Decrease in Temperature Keq @ When the temperature is New lowered for an exothermic Temperature reaction, the Keq constant changes. Because the current concentrations yields a reaction quotient lower than Keq (new) the reaction must shift to the right. 16 LeChatlier and Chemical Direction Direction Reaction K eq(old Temp) = Q When the temperature is lowered for an endothermic reaction, the Keq constant changes. Because the current concentrations yields a reaction quotient greater than Keq (new) the reaction must shift to the left. January 13 Temperature Effect on [CoCl4]2- D [Co(H2O)] Consider the reaction: [CoCl4]2- + 6H2O D blue ΔH < 0 [Co(H2O)] + 4Clpink In this experiment when the solution was placed in cold water, the solution turned vivid pink. The pink color indicates a shift to the right for the reaction shown above. This will only occur if the new Keq at the instant the temperature is altered, is now higher than the old Keq, (which is now called Q). Therefore the reactant concentration decreases, the product concentration increases as the reaction adjust itself so that the Mass action equals Keq: Q g Keq K 17 eq(old Temp) = Q Direction Reaction LeChatlier and Chemical Direction K eq (new Temp) January 13 3i) Pressure on Solubility: Solids / Liquid Pressure - (Solid and Liquid) The solubility of solids and liquids are hardly affected by pressure. Solids and liquids are already very close to each other. An increase in pressure will not affect solubility. 18 LeChatlier and Chemical Direction January 13 3 ii) Pressure on Solubility: Gas Pressure - (Gas) Solubility of gas is greatly affected by pressure Henry’s Law: PA = KHXA Gas solutes are very sensitive to pressure. Because gas particles are separated by void space, an increase in pressure will increase the solubility of the gas. Divers must be careful when diving to great depths because the potential of dissolved N2 gas in the blood will lead to the bends. 19 LeChatlier and Chemical Direction January 13 Pressure Affect: Teeter Totter Analogy [Solute] Pressure Sensitive D gas [Solute] Solution In utilizing LeChatelier Principle to determine the direction of solubility for a gaseous solute with variation in pressure, the first thing that must be establish is which side is more sensitive to pressure. In our case the gas is more sensitive than the solution. 20 LeChatlier and Chemical Direction January 13 The effect of a Change Volume of reaction vessel in terms of Reaction Quotient and LeChatelier For a chemical reaction, LeChatelier Principle can be verified in terms of the Reaction Quotient. Consider the reaction: N2 + 3H2 D 2NH3 Let [H2] = 0.1207 M, [N2] = 0.0402 M and [NH3] = 0.00272 M for a 1-L vessel at equilibrium. In which direction will the reaction shift if the reaction vessel is decreased by half such that V = Vo/2. The concentrations will now double for all specie at the instant the volume is decreased. Molarity = mol / L, [H2] = 0.2414 M, [N2] = .0804 M and [NH3] = 0.00544 M . Plugging in to the mass action expression and solving for Q, In the reaction quotient equation Q < Keq which means that in order to regain equilibrium, the product must increase and the reactant decrease, a shift to the right in the overall reaction. 21 LeChatlier and Chemical Direction January 13 Summary of Pressure, Temperature Affect on Solubility 23 ΔH (s, l or g) Temp Direction Solubility (+) Endothermic h g Prod h increase (+) Endothermic i f React i decrease (-) Exothermic h f React i decrease (-) Exothermic i g Prod h increase Pressure Direction Solubility Gas solute h g Prod h increase Gas solute i f React i decrease LeChatlier and Chemical Direction January 13 Example Summary Consider the following system at equilibrium: 6H2O (g) + 6CO2 (g) D 2 C6H12O6 (s) + 6O2 (g) Complete the following table. Indicate changes in moles and concentrations by entering I, D, N, or ? in the table. (I = increase, D = decrease, N = no change, ? = insufficient information to determine) 24 LeChatlier and Chemical Direction January 13 14.4 (Chp16) Solubility and Solubility Product What drives substances to dissolve and others to remain as a precipitate? Dr. Fred Omega Garces Chemistry 201 Miramar College 1 Solubility and Solubility Product January 13 When a Substance Dissolve When a reaction occurs in which an insoluble product is produced, the Keq is called a solubility-product. BaSO4 (s) D Ba+2 + SO42- Keq = Ksp = [Ba+2] [SO42-] (aq) Do not confuse solubility (s) with solubility product Ksp. Picture of BaSO4 dissolving Solubility depends on Temp., conc., and pH Solubility Product depends on Temp. Only !!! 2 Solubility and Solubility Product January 13 Solubility Solubility: What is the meaning of solubility ? i. The ability of substance to dissolve in a solvent. ii. Quantity that dissolves to form a saturated solution. s g per 100 mL or g /L g (molar solubility, grams per liter saturated solution.) g • mol g g mol g Molarity The greater the solubility, the smaller the amount of precipitation. 3 Solubility and Solubility Product January 13 Factors Affecting Solubility (s) 1 Nature of Solute (Concentration). - Like dissolves like (IMF) 2 Temperature Factor i) Solids/Liquids- Solubility increases with Temperature Increase K.E. increases motion and collision between solute / solvent. ii) gas - Solubility decreases with Temperature Increase K.E. result in gas escaping to atmosphere. 3 Pressure Factor i) Solids/Liquids - Very little effect Solids and Liquids are already lose together, extra pressure will not increase solubility. ii) Gas- Solubility increases with Pressure. Increase pressure squeezes gas solute into solvent. 4 Solubility and Solubility Product January 13 In a Chemical Reaction For a reaction to take place Reactants should be soluble in solventThis ensures that reactants combine (come in contact) with each other efficiently. When reaction proceedsProducts which are formed have different properties than that of the reactants. One such property is the solubility. When a precipitate forms upon mixing two solution it is a result of a new specie that is present in the solution. 5 Solubility and Solubility Product January 13 Precipitation Reaction Consider the reaction i AgNO 3 (aq) + HCl (aq) D AgCl(s) + HNO3(aq) (aq) D AgCl Net ionic equation: Ag+(aq) + Cl- If written in this form, then mass action is: (s) Write the reverse, (standard convention) AgCl Then (s) D Ag+(aq) + Cl- (aq) K eq = Keq = Ksp = [Ag+] [Cl-] 1 [Ag + ][Cl− ] Ksp - solubility product constant An indicator of the solubility of substance of interest. € Yields info. on the amount of ions allowed in solution. (Must be determine experimentally) NaOCH2CH3 (s) + H2O (l) D Na+ (aq) + OH- (aq) + CH3CH2OH (aq) 6 Solubility and Solubility Product Ksp = [Na+] [OH-] [CH3CH2OH] January 13 Solubility Equilibrium: Example What is the solubility product for the following reaction: D 3Ag+ (aq) + PO43- (aq) s = 4.4•10-5M Ag3PO4 i Excess Δ - 4.4•10-5M 3(4.4•10-5M ) Solid 1.32•10-4M e 7 0 0 + 4.4•10-5M 4.4•10-5 M Solubility and Solubility Product January 13 Solubility from Solubility Chart The diagram below shows the solubility of some salts as a function of temperature. This graph can be used to determine the Ksp for any one of the salts shown. Consider KNO3, the solubility of KNO3 at 20°C according to the graph is approximately 30 g per 100ml H2O. The strategy is to find the molar concentration and then use this information to determine the Ksp for the salt at the specified temperature. The solubility of KNO3 at 20°C is 30 g per 100ml H2O. What is the Ksp for KNO3 at this temperature? Assume density of solution = 1.0 g / cc MWKNO3 = 101.11 g/mol 9 Solubility and Solubility Product January 13 Solubility Vs. Ksp Consider ; La (IO3)3 Ksp = 6.2 • 10-12 11 < Ba(IO3)2 Ksp = 1.5 • 10-9 Solubility and Solubility Product < AgIO3 Ksp = 3.1 •10-8 January 13 Precipitation and Common ion Effect How can one predict if a precipitate forms when mixing solutions? Solubility rules Soluble Substances containing Exceptions Insoluble substances containing Exceptions nitrates, (NO3-) chlorates (ClO3-) perchlorates (ClO4-) acetates (CH3COO-) None carbonates (CO32-) phosphates (PO43-) Slightly soluble halogens (X-) X- = Cl-, Br-, I- Ag, Hg, Pb hydroxides (OH-) alkali, NH4+, Ca, Sr, Ba sulfates (SO42-) Ba, Hg, Pb alkali & NH4+ None Solubility rule provides information on the specie that will form a precipitate. In general, a molar solubility of 0.01M or greater for a substance is considered soluble. http://www.ausetute.com.au/solrules.html 13 Solubility and Solubility Product January 13 Precipitation Experiment Consider the reaction: 3 Ca2+ + 2PO43- (aq) D Ca3(PO4)2 (s) According to the solubility rules, this should form a precipitate (ppt) . The question is what concentration is required before a precipitate forms ? To solve, It is always convenient to write the equation as a dissolution of solid Ca3(PO4)2 (s) D 3 Ca2+ + 2PO43- Q = [Ca+2]3 • [PO43-]2 (aq) next compare Q to Ksp In solubility product problems, Q is the ion-product instead of reaction quotient 14 Solubility and Solubility Product January 13 Ion Product; Q Q - indicator of direction of reaction for reaction not at equilb. Direction Ksp Reaction Q (i) Unsaturated. Q is small, system consist mostly of ion. No ppt. forms (s) g (aq) 15 Q Sat. Point Solubility and Solubility Product Direction Reaction Q (h) Supersaturated. Q is large, system will adjust to reduce high ion concentration. Solid forms (s) f (aq) January 13 In Class: Precipitation Determination Question, (19.88 Sil): Will any solid Ba(IO3)2 form when 6.5 mg of BaCl2 is dissolved in 0.500L of 0.033M NaIO3? ([Ba+2] = 6.2e-5 M) Ksp Ba(IO3)2 = 1.5 •10-9 , BaCl2 MW = 208.23 g/mol Ba(IO3)2 = Ba(IO3)2 16 (s) D Ba2+(aq) + 2 IO3- Solubility and Solubility Product (aq) January 13 Common Ion Effect For a system containing a precipitate in equilibrium with its ions, solubility of solid can be reduced if an ionic compound is dissolved in the solution. Ag+ I- Ag+ I- AgI(s) AgI(s) Ag+ I- Add Ivia NaI AgI(s) AgI(s) D Ag+(aq) + I-(aq) add common ion i.e., NaI to solution Direction of Reaction Reduce solubility i.e., less AgI will dissolve in soln’ which means more ppt. forms in solution 18 Solubility and Solubility Product January 13 ...according to LeChatelier AgI(s) Ag+ I- AgI(s) D Ag+ + IQ analogy: AgI(s) NaI(s) Ag+ AgI(s) Q = [Ag+] [I-] Ag+ I- Ksp < Q I- Ksp More AgI forms and Solubility is lowered 19 Increase I- Q Direction of Rxn (ppt occur) Solubility and Solubility Product January 13 Factor Affecting Solubility: Solubility & Common Ion Calculate the solubility of copper(I) iodide, CuI (Ksp = 1.1•10-12) a) water b) 0.05 M sodium iodide a) 20 i Δ e CuI (s) D Lots -s Solid Cu+(aq) + I- (aq) 0 0 +s +s +s +s b) i Δ e Solubility and Solubility Product CuI (s) D Lots -s Solid Cu+(aq) + 0 +s +s I- (aq) 0.05 +s 0.05 + s January 13 Factor Affecting Solubility: Solubility & Common Ion Calculate the solubility of copper(I) iodide, CuI (Ksp = 1.1•10-12) a) water b) 0.05 M sodium iodide a) i Δ e CuI (s) D Lots -s Solid Cu+(aq) + I- (aq) 0 0 +s +s +s +s b) i Δ e CuI (s) D Cu+(aq) + I- (aq) Lots -s Solid 0 +s +s 0.05 +s 0.05 + s a) Ksp = 1.1⋅10 -12 =[Cu+ ] [I - ] = s2 1.05⋅10 -6 = s Without common ion, solubility is, s = 1.05 •10-6 M 21 Solubility and Solubility Product January 13 Calculation: Solubility, Iteration Method Reger Ex 12.19: What is the solubility of calcium fluoride in 0.025 M sodium fluoride solution? Ksp = 6.2•10-8 M i Δ e 23 CaF2 (s) D Ca+2(aq) + 2F- (aq) Lots 0 0.025M -s +s +2s Solid +s 0.025M +2s Solubility and Solubility Product January 13 Calculation: Solubility of Two salts in same solution At 50°C, the solubility products, Ksp, of PbSO4 and SrSO4 are 1.6•10-8 M and 2.8•10-7M, respectively. What are the values of [SO42-], [Pb2+], and [Sr2+] in a solution at equilibrium with both substances ? PbSO4 (s) D Pb2+(aq) + SO42 -(aq) ; Ksp = 1.6•10-8 = [Pb2+][SO42-] SrSO4 (s) D Sr+ (aq) + SO4 2-(aq) ; Ksp = 2.8•10-7 = [Sr2+][SO42-] Let x = [Pb2+] , y = [Sr2+] , x + y = [SO42-] 25 Solubility and Solubility Product January 13 Calculation: Same type problem different data At 25°C, the solubility product of AgI and PbI2 is 8.3•10-17 M and 7.9•10-9M, respectively. What are the values of [I-], [Ag+], and [Pb2+] in a solution at equilibrium with both substances ? (Answer behind) AgI i Lots Δ -s e Lots (s) D Ag + (aq) + 0 +s +s I-(aq) PbI2 i Lots Δ -t e Lots 2t +s +s+2t K sp = [Ag+ ] ∗ [I -] [I ] = - 7.9 ⋅ 10-9 8.3 ⋅ 10-17 € = [Pb+2][I- ]2 [Ag+ ] [I- ] = [I -]2 = [Ag+ ] [s] [s+2t] = 9.518 •107 € [Pb+2][I- ]2 = 7.9 ⋅ 10-9 =[t] [s+2t]2 Assume s << t 7.9 ⋅ 10-9 = [t] [2t]2 = 4[t]3 28 Pb D 0 +t +t +2 (aq) + 2I- (aq) s +2t s+2t K sp = [Pb+2 ] ∗ [I -]2 K sp [t] [s+2t]2 (s) ; t =1.25 •10−3 = [Pb+2] K sp [Pb+2 ] [Ag+ ] [I- ] = 8.3 ⋅ 10-17 = [s] [s+t] Assume s << t 8.3 ⋅ 10-17 = [s] [t] = [s] [2.50 •10−3] ; [s] = 3.32 •10−14 M [Ag+ ]=3.32 •10−14 M, [Pb+2] =1.25 •10−3 M, [I- ] =2.5 •10−3 M Solubility and Solubility Product January 13 Calculation: Same type problem different data At 25°C, the solubility product of AgI and Hg2I2 is 8.5•10-17 M and 5.3•10-29M, respectively. What are the values of [I-], [Ag+], and [Hg22+] in a solution at equilibrium with both substances ? 29 Solubility and Solubility Product January 13 Common Ion Effect B&L 17.44 (7th Ed.): The Ksp for cerium iodate, Ce(IO3)3, is 3.2 •10-10 a) Calculate the molar solubility of Ce(IO3)3 in pure water. b) What concentration of NaIO3 in solution would be necessary to reduce the Ce3+ concentration in a saturated solution of Ce(IO3)3 by a factor of 10 below that calculation in part (a). i Δ e 30 Ce(IO3)3 Lots -s Lots (s) D Ce+3(aq) + 0 +s +s 3 IO3- (aq) 0 +3s +3s Solubility and Solubility Product January 13 Qualitative Analysis Ions are precipitated selectively by adding a precipitation ion until the Ksp of one compound is exceeded without exceeding the Ksp of the others. An extension of this approach is to control the equilibrium of the slightly soluble compound by simultaneously controlling an equilibrium system that contains the precipitating ion. Qualitative analysis of ion mixtures involves adding precipitating ions to separate the unknown ions into ion groups. The groups are then analyzed further through precipitation and complex ion formation. 32 Solubility and Solubility Product January 13 Selective Precipitation Qualitative Analysis Q < Ksp, solid dissolve until Q=Ksp Q = Ksp, equilib Q > Ksp, ppt until Q=Ksp Grp1: Insoluble Chlorides Ag+, Pb2+, Hg22+ Grp2: Acid-Insoluble sulfides Cu2+, Bi3+, Cd2+, Pb2+, Hg2+, As3+, Sb3+, Sn4+ Grp3: Base-Insoluble sulfides Cu2+, Al3+, Fe2+, Fe3+, Co2+, Ni2+, Cr3+, Zn2+, Mn2+ Grp4: Insoluble Phosphates Ba2+, Ca2+, Mg2+ Grp5: Alkali Metals and NH4+ Na+, K+, NH4+ 33 Solubility and Solubility Product January 13 Lab Practical, Qualitative Analysis Qualitative Analysis of Household Chemicals 34 Solubility and Solubility Product January 13 Complex Ion Process Metals forming Complexes: Metals by virtue of electron pairs will act as Lewis acids and bind a substrate to form a Complex Consider the following: Ag(NH3)2+ (aq) Complex ion AgCl(s) D Ag+(aq) + Cl- (aq) Ag+(aq) + 2NH3(aq) D Ag(NH3)2+ (aq) AgCl(s) + 2NH3(aq) D Ag(NH3)2+ (aq) + Cl- (aq) Normally insoluble AgCl can be made soluble By the addition of NH3. The presence of NH3 drives the top reaction to the right and increase the solubility of AgCl An assembly of metal ion and the Lewis base (NH3) is called a complex ion. The formation of this complex is describe by Kf = Formation Constant, Kf Ag+(aq) + 2NH3(aq) D Ag(NH3)2+ (aq) [Ag(NH3 )2+ ] [Ag+ ] ∗[NH3 ]2 = 1.7 • 10 7 Solubility of metal salts affected presence of Lewis Base: e.g., NH3, CN-, OH37 € Solubility and Solubility Product January 13 Formation Constants Table 38 Complex Kf [Ag(CN)2]– Complex Kf 5.6e18 [Cr(OH)4]– [Ag(EDTA)]3– 2.1e7 [Ag(en)2]+ Complex Kf 8e29 [HgI4]2– 6.8e29 [CuCl3]2– 5e5 [Hg(ox)2]2– 9.5e6 5.0e7 [Cu(CN)2]– 1.0e16 [Ni(CN)4]2– 2e31 [Ag(NH3)2]+ 1.6e7 [Cu(CN)4]3– 2.0e30 [Ni(EDTA)]2– 3.6e18 [Ag(SCN)4]3– 1.2e10 [Cu(EDTA)]2– 5e18 [Ni(en)3]2+ 2.1e18 [Ag(S2O3)2]3– 1.7e13 [Cu(en)2]2+ 1e20 [Ni(NH3)6]2+ 5.5e8 [Al(EDTA)]– 1.3e16 [Cu(CN)4]2– 1e25 [Ni(ox)3]4– 3e8 [Al(OH)4]– 1.1e33 [Cu(NH3)4]2+ 1.1e13 [PbCl3]– 2.4e1 [Al(Ox)3]3– 2e16 [Cu(ox)2]2– 3e8 [Pb(EDTA)]2– 2e18 [Cd(CN)4]2– 6.0e18 [Fe(CN)6]4– 1e37 [PbI4]2– 3.0e4 [Cd(en)3]2+ 1.2e12 [Fe(EDTA)]2– 2.1e14 [Pb(OH)3]– 3.8e14 [Cd(NH3)4]2+ 1.3e7 [Fe(en)3]2+ 5.0e9 [Pb(ox)2]2– 3.5e6 [Co(EDTA)]2– 2.0e16 [Fe(ox)3]4– 1.7e5 [Pb(S2O3)3]4– 2.2e6 [Co(en)3]2+ 8.7e13 [Fe(CN)6]3– 1e42 [PtCl4]2– 1e16 [Co(NH3)6]2+ 1.3e5 [Fe(EDTA)]– 1.7e24 [Pt(NH3)6]2+ 2e35 [Co(ox)3]4– 5e9 [Fe(ox)3]3– 2e20 [Zn(CN)4]2– 1e18 [Co(SCN)4]2– 1.0e3 [Fe(SCN)]2+ 8.9e2 [Zn(EDTA)]2– 3e16 [Co(EDTA)]– 1e36 [HgCl4]2– 1.2e15 [Zn(en)3]2+ 1.3e14 [Co(en)3]3+ 4.9e48 [Hg(CN)4]2– 3e41 [Zn(NH3)4]2+ 4.1e8 [Co(NH3)6]3+ 4.5e33 [Hg(EDTA)]2– 6.3e21 [Zn(OH)4]2– 4.6e17 [Co(ox)3]3– 1e20 [Hg(en)2]2+ 2e23 [Zn(ox)3]4– 1.4e8 [Cr(EDTA)]– 1e23 Solubility and Solubility Product January 13 Calculations: Formation of Complex Ion By using the values of Ksp for AgI and Kf for Ag(CN)2-, calculate the equilibrium constant for the following reaction: AgI(s) + 2CN-(aq) D Ag(CN)2- (aq) + I-(aq) Note that this equation is the sum of 39 AgI (s) D Ag+(aq) + I -(aq) ; Ksp = 8.3•10-17 = [Ag+][I -] Ag+(aq) + 2CN-(aq) D Ag(CN)2-(aq) ; Kf = 1.0•1021 = [Ag(CN)2-]/[Ag+][CN -]2 Solubility and Solubility Product January 13 Calculations: Formation of Complex Ion By using the values of Ksp for AgI and Kf for Ag(CN)2-, calculate the equilibrium constant for the following reaction: AgI(s) + 2CN-(aq) D Ag(CN)2- (aq) + I-(aq) Note that this equation is the sum of AgI (s) D Ag+(aq) + I -(aq) ; Ksp = 8.3•10-17 = [Ag+][I -] Ag+(aq) + 2CN-(aq) D Ag(CN)2-(aq) ; Kf = 1.0•1021 = [Ag(CN)2-]/[Ag+][CN -]2 AgI(s) + 2CN-(aq) D Ag(CN)2- (aq) + I-(aq) K = Ksp• Kf = ( [Ag+][I -] ) • ( [Ag(CN)2]/[Ag+][CN -]2 ) K = Ksp• Kf = 8.3•10-17 • 1.0•1021 K = 8.3•104 40 Solubility and Solubility Product January 13