* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Soluble - HCC Learning Web
Survey
Document related concepts
History of electrochemistry wikipedia , lookup
Metastable inner-shell molecular state wikipedia , lookup
Ionic liquid wikipedia , lookup
Transition state theory wikipedia , lookup
Electrochemistry wikipedia , lookup
Membrane potential wikipedia , lookup
Acid–base reaction wikipedia , lookup
Acid dissociation constant wikipedia , lookup
Rutherford backscattering spectrometry wikipedia , lookup
Debye–Hückel equation wikipedia , lookup
Nanofluidic circuitry wikipedia , lookup
Determination of equilibrium constants wikipedia , lookup
Ionic compound wikipedia , lookup
Chemical equilibrium wikipedia , lookup
Transcript
Chapter 16 Solubility and Complex Ion Equilibria Section 16.1 Solubility Equilibria and the Solubility Product Solubility equilibrium is base on the assumption that solids dissolve in water to give the basic particles from which they are formed. Molecular solids dissolve to give individual aqueous molecules. Ionic solids dissociate to give their respective positive and negative ions: Section 16.1 Solubility Equilibria and the Solubility Product The ions in formed from the dissociation of ionic solids can carry an electrical current. Salt solutions, therefore, are good conductors of electricity. Molecular solids, however, do not dissociate in water to give ions, so no electrical current can be carried. Section 16.1 Solubility Equilibria and the Solubility Product Solubility The ratio of the maximum amount of solute to the volume of solvent in which this solute can dissolve. Generally expressed in two ways: grams of solute per 100 g of water moles of solute per Liter of solution Section 16.1 Solubility Equilibria and the Solubility Product Definition of Solubility A salt is considered soluble if it dissolves in water to give a solution with a concentration of at least 0.1 M at room temperature. A salt is considered insoluble if the concentration of an aqueous solution is less than 0.0001 M at room temperature. Salts with solubilities between 0.0001 M and 0.1 M are considered to be slightly soluble. Section 16.1 Solubility Equilibria and the Solubility Product Solubility Rules Soluble: Dissolve - Do NOT form a solid precipitate. 1.alkali metal ions and ammonium ion: Li+, Na+, K+, NH4+ 2.acetate ion: C2H3O213.nitrate ion: NO314.halide ions (X): Cl-, Br-, I- (Exceptions: AgX, HgX, and PbX2 are insoluble) 5.sulfate ion: SO42- (Exceptions: SrSO4, BaSO4, and PbSO4 are insoluble; AgSO4, CaSO4, and Hg2SO4 are slightly soluble) Section 16.1 Solubility Equilibria and the Solubility Product Solubility Rules Insoluble: Do NOT Dissolve - Do form a solid precipitate. 1.carbonate ion: CO322.chromate ion: CrO423.phosphate ion: PO434.sulfide ion: S2- (Exceptions: CaS, SrS, and BaS are soluble) 5.hydroxide ion: OH- (Exceptions: Sr(OH)2 and Ba(OH)2 are soluble; Ca(OH)2 is slightly soluble Section 16.1 Solubility Equilibria and the Solubility Product Low solubility salts Salts that have extremely low solubilities dissociate in water according to the principles of equilibrium. For example, the reaction for the dissociation of the s alt AgCl is: Saturated solution - Contains the maximum concentration of ions that can exist in equilibrium with the solid salt at a given temperature Section 16.1 Solubility Equilibria and the Solubility Product The reverse reaction for the dissolving of the salt would be the precipitation of the ions to form a solid: The system has reached equilibrium when the rate at which AgCl dissolves is equal to the rate at which AgCl precipitates. Section 16.1 Solubility Equilibria and the Solubility Product Solubility Equilibria Solubility product (Ksp) – equilibrium constant; has only one value for a given solid at a given temperature. Solubility – an equilibrium position. Section 16.1 Solubility Equilibria and the Solubility Product Difference between liquid equilibrium and liquid solid equilibrium NOTE: There is no denominator in the solubility product equilibrium constant. The key word to remember is PRODUCT which can remind you that you should have a multiplication (or product) of the concentrations of the ions. The reason that the solid reactant is not written is because its concentration effectively remains constant. Section 16.1 Solubility Equilibria and the Solubility Product Solubility Equilibria- Effect of Stoichiometry Coefficients from the balanced equations become exponents Bi2S3(s) 2Bi3+(aq) + 3S2–(aq) 2 2 K sp = Bi S 3+ Copyright © Cengage Learning. All rights reserved 3 12 Section 16.1 Solubility Equilibria and the Solubility Product CONCEPT CHECK! In comparing several salts at a given temperature, does a higher Ksp value always mean a higher solubility? Explain. If yes, explain and verify. If no, provide a counter-example. Unlike Ka and Kb for acids and bases, the relative values of Ksp cannot be used to predict the relative solubilities of salts if the salts being compared produce a different number of ions. Unlike Ka and Kb for acids and bases, the relative values of Ksp cannot be used to predict the relative solubilities of salts if the salts being compared produce a different number of ions. 13 Section 16.1 Solubility Equilibria and the Solubility Product Solubility of salt Calculate the solubility of CaF2 in g/L (Ksp = 4.0 x 10-8) First, write the BALANCED REACTION: Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION: In the above equation, however, we have two unknowns, [Ca2+] and [F-]2. So, we have to write one in terms of the other using mole ratios. According to the balanced equation, for every one mole of Ca2+ formed, 2 moles of F- are formed. To simplify things a little, let's assign the the variable X for the solubility of the Ca2+: Section 16.1 Solubility Equilibria and the Solubility Product If we SUBSTITUTE these values into the equilibrium expression, we now only have one variable to worry about, X: We can now SOLVE for X: Section 16.1 Solubility Equilibria and the Solubility Product EXERCISE! Calculate the solubility of silver chloride in water. Ksp = 1.6 × 10–10 [Ag+]=[Cl-] so you solubility = √Ksp 1.3×10-5 M Calculate the solubility of silver phosphate in water. Ksp = 1.8 × 10–18 =[Ag+]*[PO4+3]3 [Ag+]=3x [PO4+3]=x 1.6×10-5 M Copyright © Cengage Learning. All rights reserved 16 Section 16.1 Solubility Equilibria and the Solubility Product We assigned X as the solubility of the Ca2+ which is equal to the solubility of the salt, CaF2. However, our units right now are in molarity (mol/L), so we have to convert to grams: Section 16.1 Solubility Equilibria and the Solubility Product Calculate Ksp from Solubility The solubility of AgCl in pure water is 1.3 x 10-5 M. Calculate the value of Ksp. First, write the BALANCED REACTION: Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION: It is given in the problem that the solubility of AgCl is 1.3 x 10-5. Since the mole ratio of AgCl to both Ag+ and Cl- is 1:1, the solubility of each of the ions is equal to the solubility of AgCl. SUBSTITUTE the solubility given into the equilibrium expression to get Ksp: 1.3 x 10-5 M X 1.3 x 10-5 M Section 16.1 Solubility Equilibria and the Solubility Product EXERCISE! Calculate the solubility of AgCl in: Ksp = 1.6 × 10–10 a) 100.0 mL of 4.00 x 10-3 M calcium chloride. 2.0×10-8 M b) 100.0 mL of 4.00 x 10-3 M calcium nitrate. 1.3×10-5 M Copyright © Cengage Learning. All rights reserved 19 Section 16.1 Solubility Equilibria and the Solubility Product Ion Product (Qsp) The product of the concentrations of the ions at any moment in time (not necessarily at equilibrium). Imagine we have a saturated solution of AgCl. The equilibrium reaction for the dissociation of this salt is: Section 16.2 Precipitation and Qualitative Analysis Precipitation (Mixing Two Solutions of Ions) Q > Ksp; precipitation occurs and will continue until the concentrations are reduced to the point that they satisfy Ksp. Q < Ksp; no precipitation occurs. Copyright © Cengage Learning. All rights reserved 21 Section 16.1 Solubility Equilibria and the Solubility Product Common Ion Effect The decrease in the solubility of a salt that occurs when the salt is dissolved in a solution that already contains another source of one of its ions. For example, if AgCl is added to a NaCl solution (which contains the common ion, Cl-) the solubility of the AgCl decreases. Section 16.2 Precipitation and Qualitative Analysis Effect of adding a common ion [Ag+] is equal to [Cl-] at equilibrium because the mole ratio of Ag+ to Cl- is 1:1. What would happen to the solution if a tiny bit of AgNO3 (a soluble salt) were added? Since AgNO3 is soluble, it dissociates completely to give Ag+ and NO3- ions. There would now be two sources of the Ag+ ion, from the AgCl and from the AgNO3: Section 16.2 Precipitation and Qualitative Analysis Adding AgNO3 increases the Ag+ concentration and the solution is no longer at equilibrium. The ion product (Qsp) at that moment is bigger than the solubility product (Ksp). The reaction will eventually return to equilibrium but when it does, the [Ag+] is no longer equal to the [Cl-]. Instead, the [Ag+] will be larger than the [Cl-]. Section 16.2 Precipitation and Qualitative Analysis Let's go back to the saturated AgCl solution. What would happen this time if a tiny bit of NaCl (a soluble salt) were added? Since NaCl is soluble, it dissociates completely to give Na+ and Cl- ions. There would now be two sources of the Clion, from AgCl and from NaCl: Adding NaCl increases the Cl- concentration and the solution is no longer at equilibrium. The ion product (Qsp) at that moment is bigger than the solubility product (Ksp). The reaction will eventually return to equilibrium but when it does, the [Ag+] is no longer equal to the [Cl-]. Instead, the [Cl-] will be larger than the [Ag+]. Section 16.1 Solubility Equilibria and the Solubility Product CONCEPT CHECK! How does the solubility of silver chloride in water compare to that of silver chloride in an acidic solution (made by adding nitric acid to the solution)? Explain. The solubilities are the same. Copyright © Cengage Learning. All rights reserved 26 Section 16.1 Solubility Equilibria and the Solubility Product CONCEPT CHECK! How does the solubility of silver phosphate in water compare to that of silver phosphate in an acidic solution (made by adding nitric acid to the solution)? Explain. The silver phosphate is more soluble in an acidic solution. Copyright © Cengage Learning. All rights reserved 27 Section 16.1 Solubility Equilibria and the Solubility Product CONCEPT CHECK! How does the Ksp of silver phosphate in water compare to that of silver phosphate in an acidic solution (made by adding nitric acid to the solution)? Explain. The Ksp values are the same. Copyright © Cengage Learning. All rights reserved 28 2 Section 16.1 9 Solubility Equilibria and the Solubility Product Effect of pH on Solubility Sometimes it is necessary to account for other reactions aqueous ions might undergo. – For example, if the anion is the conjugate base of a weak acid, it will react with H3O+. – You should expect the solubility to be affected by pH. By adding and complexing out ions you can affect the pH of solution which could affect ppt reactions. 3 Section 16.1 0 Solubility Equilibria and the Solubility Product – Consider the following equilibrium. CaC2O 4 (s ) H2O 2 2 Ca (aq) C 2O 4 (aq) – Because the oxalate ion is conjugate base, it will react with H3O+ (added acid to lower pH). 2 C 2O 4 (aq ) H 3O (aq) H2O HC 2O 4 (aq) H 2O(l) – According to Le Chatelier’s principle, as C2O42- ion is removed by the reaction with H3O+, more calcium oxalate dissolves (increase solubility). – Therefore, you expect calcium oxalate to be more soluble in acidic solution (lower pH) than in pure water. The acidity will react with the oxalate and shift the equil toward the right and allow more calcium oxalate to dissolve. Section 16.2 Precipitation and Qualitative Analysis Selective Precipitation (Mixtures of Metal Ions) Use a reagent whose anion forms a precipitate with only one or a few of the metal ions in the mixture. Example: Solution contains Ba2+ and Ag+ ions. Adding NaCl will form a precipitate with Ag+ (AgCl), while still leaving Ba2+ in solution. Copyright © Cengage Learning. All rights reserved 31 Section 16.2 Precipitation and Qualitative Analysis Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S At a low pH, [S2–] is relatively low and only the very insoluble HgS and CuS precipitate. When OH– is added to lower [H+], the value of [S2–] increases, and MnS and NiS precipitate. Copyright © Cengage Learning. All rights reserved 32 Section 16.2 Precipitation and Qualitative Analysis Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S Copyright © Cengage Learning. All rights reserved 33 Section 16.2 Precipitation and Qualitative Analysis Separating the Common Cations by Selective Precipitation Copyright © Cengage Learning. All rights reserved 34 Section 16.3 Equilibria Involving Complex Ions Complex Ion Equilibria Charged species consisting of a metal ion surrounded by ligands. Ligand: Lewis base Formation (stability) constant. Equilibrium constant for each step of the formation of a complex ion by the addition of an individual ligand to a metal ion or complex ion in aqueous solution. Copyright © Cengage Learning. All rights reserved 35 3 Section 16.2 6 Precipitation and Qualitative Analysis Complex-Ion Equilibria – For example, the silver ion, Ag+, can react with ammonia to form the Ag(NH3)2+ ion. Ag 2(: NH 3 ) ( H 3 N : Ag : NH 3 ) 3 Section 16.2 7 Precipitation and Qualitative Analysis – A complex is defined as a compound containing complex ions. – A ligand is a Lewis base (makes electron pair available) that bonds to a metal ion to form a complex ion. Lewis Acid is the cation. 3 Section 16.2 8 Precipitation and Qualitative Analysis Complex-Ion Formation The aqueous silver ion forms a complex ion with ammonia in steps. Ag (aq ) NH 3 (aq) Ag( NH 3 ) (aq ) Ag( NH 3 ) (aq ) NH 3 (aq) Ag( NH 3 )2 (aq ) – When you add these equations, you get the overall equation for the formation of Ag(NH3)2+. Ag (aq ) 2NH 3 (aq) Ag( NH 3 )2 (aq ) 3 Section 16.2 9 Precipitation and Qualitative Analysis Ag (aq ) 2NH 3 (aq) Ag( NH 3 )2 (aq ) The formation constant, Kf, is the equilibrium constant for the formation of a complex ion from the aqueous metal ion and the ligands. – The formation constant for Ag(NH3)2+ is: [ Ag( NH 3 )2 ] Kf 2 [ Ag ][NH 3 ] – The value of Kf for Ag(NH3)2+ is 1.7 x 107. Section 16.2 Precipitation and Qualitative Analysis Complex-Ion Formation – The large value means that the complex ion is quite stable. – When a large amount of NH3 is added to a solution of Ag+, you expect most of the Ag+ ion to react to form the complex ion (large Kf - equil lies far to right). – Handle calculations same way as any other K M 4 Section 16.2 1 a tPrecipitation and Qualitative Analysis e r Complex-Ion Formation i The dissociation constant, Kd , is the reciprocal, a or inverse, value of K . f l + is The equation for the dissociation of Ag(NH ) 3 2 w a Ag ( NH ) ( aq ) Ag (aq ) 2NH 3 (aq) 3 2 s d The equilibrium constant equation is e 2 v 1 [ Ag ][NH 3 ] e Kd l K f [ Ag( NH 3 )2 ] o p 4 Section 16.2 2 Precipitation and Qualitative Analysis Equilibrium Calculations with Kf What is the concentration of Ag+(aq) ion in 1.00 liters of solution that is 0.010 M AgNO3 and 1.00 M NH3? The Kf for Ag(NH3)2+ is 1.7 x 107. Section 16.3 Equilibria Involving Complex Ions Complex Ion Equilibria Be2+(aq) + F–(aq) BeF+(aq) K1 = 7.9 × 104 BeF+(aq) + F–(aq) BeF2(aq) K2 = 5.8 × 103 BeF2(aq) + F–(aq) BeF3– (aq) K3 = 6.1 × 102 BeF3– (aq) + F–(aq) Copyright © Cengage Learning. All rights reserved BeF42– (aq) K4 = 2.7 × 101 43 Section 16.3 Equilibria Involving Complex Ions Complex Ions and Solubility Two strategies for dissolving a water–insoluble ionic solid. If the anion of the solid is a good base, the solubility is greatly increased by acidifying the solution. In cases where the anion is not sufficiently basic, the ionic solid often can be dissolved in a solution containing a ligand that forms stable complex ions with its cation. Copyright © Cengage Learning. All rights reserved 44 Section 16.3 Equilibria Involving Complex Ions solubility and complex ions Section 16.3 Equilibria Involving Complex Ions Complex Ions Are Formed From Lewis Acid Metal Interactions Section 16.3 Equilibria Involving Complex Ions CONCEPT CHECK! Calculate the solubility of silver chloride in 10.0 M ammonia given the following information: Ksp (AgCl) = 1.6 × 10–10 Ag+ + NH3 AgNH3+ + NH3 AgNH3+ Ag(NH3)2+ K = 2.1 × 103 K = 8.2 × 103 0.48 M Calculate the concentration of NH3 in the final equilibrium mixture. 9.0 M Copyright © Cengage Learning. All rights reserved 47 Section 16.3 Equilibria Involving Complex Ions Combining the two equilibria Ag+ + NH3 <=> AgNH3+ AgNH3+ + NH3 <=> Ag(NH3)2+ K = 2.1 × 103 K = 8.2 × 103 Ag+ + 2NH3 <=> [Ag(NH3)2] Kf= 2.1 × 103 x 8.2 × 103 = 1.6 x 107 Section 16.3 Equilibria Involving Complex Ions AgCl(s) + 2 NH3(aq) <=> [Ag (NH3)2]+ + Clwhere K = [Ag (NH3)2] [Cl-] / [NH3]2 is a combination of 3 equilibriums: A) AgCl(s) <=> Ag+ + Clwhose Ksp = [Ag+] [Cl-] = Ksp = 1.8 X 10-10 B) Ag+ + 2NH3 <=> [Ag(NH3)2] whose Kf = [Ag(NH3)2] / [Ag+] [NH3]2 = 1.6 X 10^7 Section 16.3 Equilibria Involving Complex Ions if multiplying the ratio (A) time (B) gives your ratio: ([Ag+] [Cl-]) * ([Ag(NH3)2] / [Ag+] [NH3]2 ) =[Ag (NH3)2] [Cl-]/[NH3]2 then multiplying the Ksp times the Kf gives your K (1.8 X 10-10) (1.6 X 107) = 2.88 X 10-3 (with a constant that large, we need to use the quadratic) Section 16.3 Equilibria Involving Complex Ions Set Up a Rice Table AgCl(s) 2 NH3(aq) Ag (NH3)2+ Cl- X X 10 -2X K = [Ag (NH3)2] [Cl-] / [NH3]2 2.88 X 10-3 = [X] [X] / [10 - 2X]2