Ksp Equilibrium of insoluble salts SSS3 4th Class

Insoluble salts Salts that dissociate very, very, very little in water When writing net ionic equations these salts are NON-Electrolytes, so the formula is not written as ions.

Ksp Ksp ------Equilibrium constant for “insoluble salts” (Nothing is totally insoluble) Solve problems using ICE Three types of problems Solve for Equilibrium product constant (Ksp) Solve for molar solubility (x in ICE) Precipitation (think Le Chatlier’s) if K> Q, shifts to products so no precipitation. If K< Q, shifts to favor reactants so precipitation will occur

Solving Problems Write the equations Write the equilibrium expression Solve using ICE Ksp = [Ag+] [Cl- ] AgCl(s)  Ag+ (aq) + Cl- (aq) I 0 0 C +x +x E x x Ksp = x 2

Solubility Equilibria Solubility product (Ksp) – equilibrium constant; has only one value for a given solid at a given temperature. Solubility – an equilibrium position. Bi2S3(s) 2Bi3+(aq) + 3S2–(aq) I 0 0 C 2x 3x E 2x 3x Ksp = (2x)2 (3x)3 Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Problem 1 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. No No. In order to relate Ksp values to solubility directly, the salts must contain the same number of ions. For example, for a binary salt, Ksp = s2 (s = solubility); for a ternary salt, Ksp = 4s3. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Problem 2 Calculate the solubility of silver chloride in water. Ksp = 1.6 × 10–10 1.3×10-5 M Calculate the solubility of silver phosphate in water. Ksp = 1.8 × 10–18 1.6×10-5 M a) 1.3×10-5 M b) 1.6×10-5 M Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved 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. The solubilities are the same. Since HCl is a strong acid, it is completely dissociated in water. There are no common ions between AgCl and HNO3. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved 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. The silver phosphate is more soluble in an acidic solution. This is because the phosphate ion is a relatively good base and will react with the proton from the acid (essentially to completion). The phosphate ion does not react nearly as well with water. This is an example of the effect of LeChâtelier's principle on the position of the solubility equilibrium. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved Problem 3 Calculate the solubility of AgCl in: Ksp = 1.6 × 10–10 100.0 mL of 4.00 x 10-3 M calcium chloride. 2.0×10-8 M 100.0 mL of 4.00 x 10-3 M calcium nitrate. 1.3×10-5 M a) 2.0×10-8 M Note: [Cl-] in CaCl2 is twice the [CaCl2] given. 100.0 mL is not used in the calculation. b) 1.3×10-5 M Copyright © Cengage Learning. All rights reserved

Precipitation (Mixing Two Solutions of Ions) Ksp < Q; precipitation occurs and will continue until the concentrations are reduced to the point that they satisfy Ksp. Ksp > Q; no precipitation occurs. Copyright © Cengage Learning. All rights reserved

Multiple choice 1 MnS(s) + 2 H+ <-> Mn2+ + H2S(g) At 25 °C the solubility product constant, Ksp, for MnS in 5 x 10¯15 and the acid dissociation constants K1 and K2 for H2S are 1 x 10¯7 and 1 x 10¯13, respectively. What is the equilibrium constant for the reaction represented by the equation above at 25 °C? (A) 1 x 10¯13 / 5 x 10¯15 (B) 5 x 10¯15 / 1 x 10¯7 (C) 1 x 10¯7 / 5 x 10¯20 (D) 5 x 10¯15 / 1 x 10¯20 (E) 1 x 10¯20 / 5 x 10¯15

Multiple choice 2 What is the molar solubility in water of Ag2CrO4? (The Ksp for Ag2CrO4 is 8 x 10¯12.) A) 8 x 10¯12 M B) 2 x 10¯12 M C) (4 x 10¯12 M)1/2 D) (4 x 10¯12 M)1/3 E) (2 x 10¯12 M)1/3  

Free Response At 25ºC the solubility product constant, Ksp, for strontium sulfate, SrSO4, is 7.610-7. The solubility product constant for strontium fluoride, SrF2, is 7.9´10-10. (a) What is the molar solubility of SrSO4 in pure water at 25ºC? (b) What is the molar solubility of SrF2 in pure water at 25ºC? (c) An aqueous solution of Sr(NO3)2 is added slowly to 1.0 litre of a well-stirred solution containing 0.020 mole F- and 0.10 mole SO42- at 25ºC. (You may assume that the added Sr(NO3)2 solution does not materially affect the total volume of the system.) 1. Which salt precipitates first? 2. What is the concentration of strontium ion, Sr2+, in the solution when the first precipitate begins to form? (d) As more Sr(NO3)2 is added to the mixture in (c) a second precipitate begins to form. At that stage, what percent of the anion of the first precipitate remains in solution?

(a) SrSO4(s)  Sr2+(aq) + SO42-(aq) At equilibrium: [Sr2+] = X M = [SO42-] X2 = Ksp = 7.6x10-7 X = 8.7x10-4 mol/L, solubility of SrSO4 (b) SrF2(s)  Sr2+(aq) + 2 F-(aq) At equilibrium: [Sr2+] = X M = [F-] = 2X M KSP = [Sr2+][F-]2 = (X)(2X)2 = 7.9´10-10 X = 5.8x10-4 mol/L, solubility of SrF2 (c) Solve for [Sr2+] required for precipitation of each salt. Ksp = [Sr2+][F-]2 = 7.9x10-10 Ksp = [Sr2+][SO42-] = 7.6x10-7 Since 2.0x10-6 M < 7.6x10-6 M, SrF2 must precipitate first. When SrF2 precipitates, [Sr2+] = 2.0x10-6 M (d) The second precipitate to form is SrSO4, which appears when [Sr2+]= 7.6x10-6 M (based on calculations in Part c.) When [Sr2+] = 7.6x10-6 M, [F-] is determined as follows: = (7.6x10-6)(z)2 = 7.9x10-10 ; z = 1.0x10-2 M

Free Response 2 1994 A MgF2(s)  Mg2+(aq) + 2 F-(aq) In a saturated solution of MgF2 at 18ºC, the concentration of Mg2+ is 1.21x10-3 molar. The equilibrium is represented by the equation above. (a) Write the expression for the solubility-product constant, Ksp, and calculate its value at 18ºC. (b) Calculate the equilibrium concentration of Mg2+ in 1.000 liter of saturated MgF2 solution at 18ºC to which 0.100 mole of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible. (c) Predict whether a precipitate of MgF2 will form when 100.0 milliliters of a 3.00x10-3-molar Mg(NO3)2 solution is mixed with 200.0 milliliters of a 2.00xl0-3-molar NaF solution at 18ºC. Calculations to support your prediction must be shown. (d) At 27ºC the concentration of Mg2+ in a saturated solution of MgF2 is 1.17x10-3 molar. Is the dissolving of MgF2 in water an endothermic or an exothermic process? Give an explanation to support your conclusion.

(b) X = concentration loss by Mg2+ ion Answer: (a) Ksp = [Mg2+][F-]2 = (1.21x10-3)(2.42x10-3)2 = 7.09x10-9 (b) X = concentration loss by Mg2+ ion 2X = concentration loss by F- ion [Mg2+] = (1.21x10-3 - X) M [F-] = (0.100 + 2.42x10-3 - 2X) M since X is a small number then (0.100 + 2.42x10-3 - 2X)  0.100 Ksp = 7.09x10-9 = (1.21x10-3 - X)(0.100)2 X = 1.2092914x10-3 [Mg2+]= 1.21x10-3-1.20929´x0-3 = 7.09x10-7M (c) [Mg2+] = 3.00x10-3M ´ 100.0 mL/300.0 mL = 1.00´10-3M [F-] = 2.00x10-3 M ´ 200.0 mL/300.0 mL = 1.33´10-3 M trial Ksp = (1.00x10-3)(1.33x10-3)2 = 1.78´10-9 trial Ksp < = 7.09x10-9,  no ppt. (d) @ 18ºC, 1.21x10-3 M MgF2 dissolves @ 27ºC, 1.17x10-3 M MgF2 dissolves MgF2  Mg2+ + 2 F- + heat dissolving is exothermic; if heat is increased it forces the equilibrium to shift left (according to LeChatelier’s Principle) and less MgF2 will dissolve.

Equilibrium in review Kc, Kp, Ka, Kb, Ksp Temperature change is the only variable to change the value of K Use K > Q or K< Q to justify a shift or Le Chatlier’s Principle The units for K are not significant All problems can be solved using ICE

Review There are TWO reaction types in chemistry Reactions that go to completion (stoichiometry-ICE) Net ionic equations Electrolytes dissociation in water Redox Reactions that establish equilibrium (Equilibrium - ICE) Reactions with gases that establish equilibrium due to pressure and temperature Phase change Weak Acids, Weak Bases, Insoluble salts

Preparing for the test Practice Practice Practice Work old test Work problems without calculator Study Guides Books Rework problems- Don’t just look at it! You have to get your speed up Don’t give up! You have worked too hard to let this opportunity pass by Answer every problem!