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Solubility Products, K sp, and Ion Products Q sp

Solubility Products, K sp, and Ion Products Q sp. Ch 15, Part 4. The Other Half of the Prank. Take-Home Test Posted Due Wed Night. heterogeneous equilibrium. solubility products, K sp : Equilibrium Expression for K sp and Q sp. K sp and Q sp.

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Solubility Products, K sp, and Ion Products Q sp

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  1. Solubility Products, Ksp, and Ion Products Qsp Ch 15, Part 4

  2. The Other Half of the Prank Take-Home Test Posted Due Wed Night

  3. heterogeneous equilibrium • solubility products, Ksp: • Equilibrium Expression for Ksp and Qsp

  4. Ksp and Qsp • Ksp is the equilibrium expression used when the solution is saturated • When the solution is NOT saturated, Qsp is usually used to describe the equilibrium • Qsp is called the ION PRODUCT QUOTIENT • Equilibrium expression is the same • If solution is not saturated, precipitate will not form.

  5. Comparing Qsp and Ksp

  6. Molar Solubilities and Solubility Products • Solubility products, Ksp, of salts are indirect indication of their solubilities expressed in mol/L (called molar solubility). • However, the solubility products are more useful than molar solubility. • The molar solubilities are affected when there are common ions present in the solution. • We need to employ the solubility products to estimate the molar solubilities in these cases. • When a salt is dissolved in pure water, solubility products and molar solubilities are related. This is illustrated using calcium carbonate. If x is the concentration of Ca2+ (= [CO32-]) in the saturated solution, then • Ksp = x2 • Because Ksp = [Ca2+] [CO32-]

  7. Example 1 • The Ksp for AgCl is 1.8e-10 M2. • What is the molar solubility of AgCl in pure water? • Solution • Let x be the molar solubility, then • AgCl = Ag+ + Cl- xx • x = (1.8e-10 M2)1/2 = 1.3e-5 M

  8. Molar Solubility • The solubility product, Ksp is a better indicator than the usual solubility specification of g per 100 mL of solvent or moles per unit volume of solvent. • For the AgCl case, when the cation concentration is not the same as the anion concentration ([Ag+] =/= [Cl-]) solubility of AgCl can not be defined in terms of moles per L.

  9. Example 1 • In this case, the system can be divided into three zones. The condition [Ag+] [Cl-] = Ksp, is represented by a line which divides the plane into two zones. • When [Ag+] [Cl-] < Ksp, no precipitate will be formed. • When [Ag+] [Cl-] > Ksp, a precipitate will be formed. • When AgCl and NaCl dissolve in a solution, both salts give Cl- ions. The effect of [Cl-] on the solubility of AgCl is called the Common ion effect

  10. Example 2 • The Ksp for Ag2CrO4 is 9e-12 M3. • What is the molar solubility of Ag2CrO4 in pure water? • SolutionLet x be the molar solubility of Ag2CrO4, then • Ag2CrO4 = 2 Ag+ + CrO42- 2xx • (2 x)2(x) = Ksp • x = {(9e-12)/4}(1/3) = 1.3e-4 M • [Ag+] = 2.6e-4 M and the molar solubility is 1.3e-4 M. • A diagram similar to AgCl can be drawn but shape of the curve is different

  11. Example 3 • What is the pH in a saturated solution of Ca(OH)2?Ksp = 5.5e-6 M3 for Ca(OH)2. • Let [Ca2+] = x, then [OH-] = 2 x. The equilibrium and concentration are represented below: • Ca(OH)2 = Ca2+ + 2 OH- x 2 x • x = {5.5e-6/4}(1/3) [OH-] = 2 x.Note the stoichiometry of equilibrium. • Answer 12.35

  12. Example 4 • At 760◦C, Kc = 33.3 for the reaction • PCl5(g) ⇀↽ PCl3(g) + Cl2(g) • If a mixture that consists of 0.54 mol PCl3 and 0.85 mol Cl2 is placed in a 9 L reaction vessel and heated to 760◦C, what is the equilibrium composition of PCl5? • Initially, • [PCl3] = 0.54 mol/9 L = 0.06 mol/L • [Cl2] = 0.85 mol/9 L = 0.0944444 mol/L

  13. Example 4 Since 33.4543 is larger than the initial con- centrations, it can be discarded. Thus [PCl5] = 0.000169385 mol/L 33.3 x = x2 − 0.154444 x + 0.00566667 0 = x2 − 33.4544 + 0.00566667

  14. AP Question #1 • Propanoic acid, HC3H5O2, ionizes in water according to the equation below. • HC3H5O2(aq) ↔ C2H5O2-(aq) + H+(aq) • Ka = 1.34 x 10-5 • Write the equilibrium expression

  15. AP Question #1 • Propanoic acid, HC3H5O2, ionizes in water according to the equation below. • HC3H5O2(aq) ↔ C2H5O2-(aq) + H+(aq) • Ka = 1.34 x 10-5 • Write the equilibrium expression

  16. AP Example #1b

  17. AP Example #1 • Calculate the pH of a 0.265 M solution of propanoic acid • A 0.496 g sample of sodium propanate, NaC3H5O2, is added to 50.0 mL sample of a 0.265 M solution of propanoic acid. Assuming no change in volume of the solution occurs, calculate each of the following: • The concentration of the propanate ion, C3H5O2- (aq), in the solution. • The concentration of the H+ (aq) in the solution.

  18. AP Example #1C

  19. AP Example #1c

  20. AP Example #2 • The methanote ion, HCO2- (aq) reacts with water to form methanoic acid and hydroxide ion, as shown in the following equation. • HCO2-(aq) + H2O (l)HCO2H (aq) + OH- (aq) • Given that [OH-] is 4.18 x 10-6 M in a 0.309 M solution of sodium methanoate, calculate each of the following • The value of Kb for the methanoate ion, HCO2- (aq) • The value of Ka for methanoic acid, HCO2H

  21. AP Example #3 (2003) • 1. C6H5NH2(aq) + H2O(l) C6H5NH3+(aq) + OH–(aq) Aniline, a weak base, reacts with water according to the reaction represented above. (a) Write the equilibrium constant expression, Kb, for the reaction represented above. (b) A sample of aniline is dissolved in water to produce 25.0 mL of 0.10 M solution. The pH of the solution is 8.82. Calculate the equilibrium constant, Kb, for this reaction. (c) The solution prepared in part (b) is titrated with 0.10 M HCl. Calculate the pH of the solution when 5.0 mL of the acid has been titrated. (d) Calculate the pH at the equivalence point of the titration in part (c). (e) The pKa values for several indicators are given below. Which of the indicators listed is most suitable for this titration? Justify your answer.

  22. AP Example #3 (2003) • (e) The pKa values for several indicators are given below. Which of the indicators listed is most suitable for this titration? Justify your answer.

  23. AP Example #3 Answers (a) Kb = (b) pOH = 14 – pH = 14 – 8.82 = 5.18 -log[OH–] = 5.18; [OH–] = 6.61´10–6M [OH–] = [C6H5NH3+] Kb = = 4.4*10–10

  24. AP Example #3 (Answers) (c) 25 mL ´ = 2.5 mmol C6H5NH2 5 mL ´ = 0.5 mmol H+ added 2.0 mmol base remains in 30.0 mL solution 4.4*10–10 = X = 1.80´10–9 = [OH–] [H+] = = 5.6*10–6; pH = 5.26 (d) when neutralized, there are 2.5 mmol of C6H5NH3+ in 50.0 mL of solution, giving a [C6H5NH3+] = 0.050 M his cation will partially ionize according to the following equilibrium: C6H5NH3+(aq) C6H5NH2(aq) + H+(aq) at equilibrium, [C6H5NH2] = [H+] = X [C6H5NH3+] = (0.050–X) = Ka = 2.3´10-5 X = 1.06´10–3 = [H+] pH = –log[H+] = 2.98

  25. AP Example #3 (Answers) • (e) erythrosine; the indicator will change color when the pH is near its pKa, since the equivalence point is near pH 3, the indicator must have a pKa near this value.

  26. AP Example #4 (2004) 1. Answer the following questions relating to the solubilities of two silver compounds, Ag2CrO4 and Ag3PO4. Silver chromate dissociates in water according to the equation shown below. Ag2CrO4(s)  2 Ag+(aq) + CrO42–(aq) Ksp = 2.6 × 10–12 at 25°C • Write the equilibrium-constant expression for the dissolving of Ag2CrO4(s). • Calculate the concentration, in mol L-1, of Ag+(aq) in a saturated solution of Ag2CrO4 at 25°C. • Calculate the maximum mass, in grams, of Ag2CrO4 that can dissolve in 100. mL of water at 25°C.

  27. AP Example #4 (2004) 1. Answer the following questions relating to the solubilities of two silver compounds, Ag2CrO4 and Ag3PO4. Silver chromate dissociates in water according to the equation shown below. Ag2CrO4(s)  2 Ag+(aq) + CrO42–(aq) Ksp = 2.6 × 10–12 at 25°C • A 0.100 mol sample of solid AgNO3 is added to a 1.00 L saturated solution of Ag2CrO4 . Assuming no volume change, does [CrO42–] increase, decrease, or remain the same? Justify your answer. In a saturated solution of Ag3PO4 at 25°C, the concentration of Ag+(aq) is 5.3 × 10–5M. The equilibrium­constant expression for the dissolving of Ag3PO4(s) in water is shown below. Ksp = [Ag+]3 [PO43-] • Write the balanced equation for the dissolving of Ag3PO4 in water. Calculate the value of Ksp for Ag3PO4 at 25°C. • A 1.00 L sample of saturated Ag3PO4 solution is allowed to evaporate at 25°C to a final volume of 500. mL. What is [Ag+] in the solution? Justify your answer.

  28. AP Example #4 (Answers) a) Write the equilibrium constant expression for the dissolving of Ag2CrO4. Ksp = [Ag2+]2[CrO42-] b) Calculate the concentration in mol L-1, of Ag+ in a saturated solution of Ag2CrO4 at 25˚C.

  29. AP Example #4 (Answers) • c) Calculate the maximum mass in grams of Ag2CrO4 that can dissolve in 100. mL of water at 25˚C. d) A 0.100 mol sample of solid AgNO3 is added to 1.00 L saturated solution of Ag2CrO4. Assuming no volume change, does [CrO42-] increase, decrease, or remain the same? Justify your answer. Common ion effect. Adding AgNO3 is adding Ag+ to the product side. This will shift the equilibrium to the left (LeChatliers Principle). Thus the concentration of the chromate will go down. You could justify this with math but since they don’t ask it then don’t.

  30. AP Example #5 (2001) • 1. Answer the following questions relating to the solubility of the chlorides of silver and lead. (a) At 10C, 8.9  10-5 g of AgCl(s) will dissolve in 100. mL of water. (i) Write the equation for the dissociation of AgCl(s) in water. (ii) Calculate the solubility, in mol L–1, of AgCl(s) in water at 10C. (iii) Calculate the value of the solubility-product constant, Ksp for AgCl(s) at 10C.

  31. AP Example #5 (2001) (b) At 25C, the value of Ksp for PbCl2(s) is 1.6  10-5 and the value of Ksp for AgCl(s) is 1.8  10-10. (i) If 60.0 mL of 0.0400 M NaCl(aq) is added to 60.0 mL of 0.0300 M Pb(NO3)2(aq), will a precipitate form? Assume that volumes are additive. Show calculations to support your answer. (ii) Calculate the equilibrium value of [Pb2+(aq)] in 1.00 L of saturated PbCl2 solution to which 0.250 mole of NaCl(s) has been added. Assume that no volume change occurs. (iii) If 0.100 M NaCl(aq) is added slowly to a beaker containing both 0.120 M AgNO3(aq) and 0.150 M Pb(NO3)2(aq) at 25C, which will precipitate first, AgCl(s) or PbCl2(s)? Show calculations to support your answer.

  32. AP Example #5 (2001) (a) At 10° C, 8.9 x10-5 g of AgCl(s) will dissolve in 100 mL of water. (i) Write the equation for the dissociation of AgCl(s) in water. AgCl(s) <===> Ag+ + Cl- (ii) Calculate the solubility, in mol/L, of AgCl(s) in water at 10° C. (8.9 x10-5 g AgCl)(mol AgCl/169 g AgCl) = 5.27 x10-7 mol 5.27 x10-7 mol/0.1 L = 5.27 x10-6 M (iii) Calculate the value of the solubility product constant, Ksp, for AgCl(s) at 10° C. Ksp = [Ag +][Cl -] = [5.27 x10-6]2 = 2.78 x10-11

  33. AP Example #5 (2001) • (b) At 25° C, the value of Ksp for PbCl2(s) is 1.6 x10-5 and the value for AgCl(s) is 1.8 x10-10. • (i) If 60.0 mL of 0.0400 M NaCl(aq) is added to 60.0 mL of 0.0300 M Pb(NO3)2(aq), will a precipitate form? Assume that volumes are additive. Show calculations to support your answer. • (0.06 L)(0.04 M) = 2.4 x10-3 mol Cl-/0.120 L - = 0.02 mol M Cl- • (0.06 L)(0.03 M) = 1.8 x10-3 mol Pb2+/0.120 L = 0.015 M Pb2+ • [Pb2+][Cl-]2 = Q = [0.015][0.02]2 = 6.0 x10-6 • No ppt 1.6 x10-5 > 6.0 x10-6

  34. AP Example #5 (2001) (ii) Calculate the equilibrium value of [Pb2(aq)] in 1.00 L of saturated PbCl2 solution to which 0.2050 mole of NaCl(s) has been added. Assume that no volume change occurs. 1.6 x10-5 = [Pb2+][0.25]2 = 2.56 x10-4 (iii) If 0.100 M is added slowly to a beaker containing both 0.120 M AgNO3(aq) and 0.150 M Pb(NO3)2 at 25° C, which will precipitate first, AgCl(s) or PbCl2(s)? Show calculations to support your answer. QAgCl = [0.12][0.1]2 = 1.2 x10-3 QPbCl2 = [0.15][0.1]2 = 1.5 x10-3

  35. AP Example #6 • 5. Answer the questions below that relate to the five aqueous solutions at 25C shown above. • (a) Which solution has the highest boiling point? Explain. • (b) Which solution has the highest pH? Explain. • (c) Identify a pair of the solutions that would produce a precipitate when mixed together. Write the formula of the precipitate. • (d) Which solution could be used to oxidize the Cl–(aq) ion? Identify the product of the oxidation. • (e) Which solution would be the least effective conductor of electricity? Explain.

  36. AP Example #6 (a) Which solution has the highest boiling point? Explain. Pb(NO3)2: greatest number of moles of particles (b) Which solution has the highest pH? Explain C2H3O2 - + H2O <==> HC2H3O2 + OH - (c) Identify a pair of solutions that would produce a precipitate when mixed together. Write the formula of the precipitate. Pb2+ + Cl - ===> PbCl2 (d) Which solution could be used to oxidize the Cl -(aq) ion? Identify the product of the solution. Cl - + MnO4 ===> Cl2 + Mn 2+ (e) Which solution would be the least effective conductor of electricity? Explain. C2H5OH: stays in solution as a molecular species

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