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Ionic solids dissolve in water to form saturated solutions.

Solubility Equilibria. Ionic solids dissolve in water to form saturated solutions. The extent to which they dissolve depends on the forces of attraction between the Ions. As the charges increase, the forces get stronger, the solubility decreases.

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Ionic solids dissolve in water to form saturated solutions.

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  1. Solubility Equilibria • Ionic solids dissolve in water to form saturated solutions. • The extent to which they dissolve depends on the forces of attraction between the Ions. • As the charges increase, the forces get stronger, the solubility decreases. • As the ions get larger in size, the forces get weaker, the solubility increases. • Here is what happens when an ionic solid is dropped into water …...

  2. Ca2+ Cl1- Ca2+ Cl1- Cl1- Ca2+ Cl1- Cl1- Cl1- Ca2+ Cl1- Cl1- Notice for every Ca2+ which enters the solution, 2 Cl1- ions also dissolve

  3. This phenomena can be expressed using a chemical equation: CaCl2 (s) -----> Ca2+(aq) + 2 Cl1-(aq) Write the equation for the dissolution of a) boron sulfate. B2(SO4)3(s) ----> 2 B3+(aq) + 3 SO42-(aq) b) sodium chromate Na2CrO4(s) ----> 2 Na1+(aq) + CrO42-(aq) c) aluminum fluoride AlF3(s) ----> Al3+(aq) + 3 F1-(aq)

  4. Quantitative Aspects of Dissolution If 4.0 g of aluminium nitrate is dissolved in 450 mL of water what are the concentrations of aluminium and nitrate ions? First find number of moles of Al(NO3)3(s) n=m/MM = 4.0 g /212.996 g/mol = 0.019 mol Second write the equation for dissolution mol Al(NO3)3(s) ----> Al3+(aq) + 3 NO31-(aq) initial 0.019 final 0 0.019 0.019 x 3 = 0.056 mol

  5. Lastly convert mol to mol/L using C = n/V [Al(NO3)3] = 0.019/0.450 L = 4.2 x 10-2 molL-1 [Al3+] = 4.2 x 10-2 molL-1 [NO31-)] = 3 x 4.2 x 10-2 molL-1 = 1.3 x 10-1 molL-1 If 2.0 g of magnesium phosphate is dissolved in 4.50 L of water what are the concentrations of magnesium and phosphate ions?

  6. Mg3(PO4)2(s) ----> 3 Mg2+(aq) + 2 PO43-(aq) First Find n of Mg3(PO4)2(s) = m/MM = 2.0 g / 262.86 g/mol = 7.6x 10-3 mol Second find [Mg3(PO4)2] using C = n/V =7.6 x 10-3 mol/ 4.5 L = 1.7 x 10-3 molL-1 [Mg2+] = 3 x [Mg3(PO4)2] = 5.1 x 10-3 molL-1 [PO43-] = 2 x [Mg3(PO4)2] = 3.4 x 10-3 molL-1 1.7 x 10-3molL-1 5.1 x 10-3molL-1 3.4 x 10-3molL-1

  7. Soluble, Slightly Soluble and Insoluble In general, if more than 1 g of solute dissolves in 1.0 L of solution the substance is said to be soluble. There is no definite dividing line between slightly soluble and insoluble however these rules can be used to determine when solid precipitates result when 2 different ionic solutions are combined.

  8. Solubility Rules for Aqueous Solutions Soluble Compounds 1. All compounds of the alkali metals (Group 1A) are soluble. 2. All salts containing NH41+, NO31-, ClO41-, ClO31-, and C2H3O21- are soluble. 3. All chlorides, bromides, and iodides are soluble except when combined with Ag1+, Pb2+ and Hg22+ 4. All sulfates are soluble except those of Hg22+, Pb2+, Ca2+, Sr2+ and Ba2+.

  9. Solubility Rules for Aqueous Solutions Insoluble Compounds 5. All hydroxides and all metal oxides are insoluble except those of Group 1A and those of Ca2+, Sr2+, and Ba2+. 6. When metal oxides do dissolve, they react with water to form hydroxides. The oxide ion, O2-, does not exist in water for example Na2O(s) + H2O ------> 2NaOH(aq) 7. All compounds containing CO32-, PO43-, SO32-, S2- are insoluble except those of group 1A and NH41+. These rules will be provided for evaluations. They are found in the textbook. Page 471. See index under solubility rules.

  10. Sample Question 1- Will a precipitate form if saturated solutions of lead(II) nitrate and sodiumiodide are combined. Write a balanced chemical equation. This is a double displacement. First write down all positive and negative ions Pb2+NO31- + Na1+I1- nextbalance charges Pb(NO3)2 +NaI swap partners to produce products Pb(NO3)2 +NaI ===> Pb2+ I1-+ Na1+NO31- balance charges and number of particles Pb(NO3)2(aq)+ 2NaI(aq) -> PbI2(s) + 2NaNO3 (aq) rules 2 and 3 tell us PbI2 is a precipitate

  11. Sample Question 2 - Will a precipitate form if saturated solutions of calcium bromide and silver nitrate are combined. First write down all positive and negative ions Ca2+Br1- + Ag1+NO31- nextbalance charges CaBr2 +AgNO31- swap partners to produce products CaBr2 +AgNO3===>Ca2+NO31-+ Ag1+Br1- balance charges and number of particles CaBr2(aq)+ 2AgNO3(aq) ->2AgBr(s)+Ca(NO3)2(aq) rules 2 and 3 tell us AgBr is the precipitate

  12. Precipitates are slightly soluble and establish an Equilibrium between the solid and its ions in solution. In a previous example PbI2 was a precipitate. The equilibrium is: PbI2(s) <=====> Pb2+(aq) + 2I1-(aq) The equilibrium expression is: Ke = [Pb2+][I1-]2 remember in a heterogeneous system solids are not included This type of Ke has a special name it is called a solubility product or Ksp for short. The values of Ksp are temperature dependent and can be found in the textbook (802)

  13. Here is a demonstration of an equilibrium between solid lead(II) iodide and its ions

  14. Pb2+ PbI2 I1-

  15. Quantitative Aspects of Ksp Ksp values can be used to determine the amount of solute present in a saturated solution of slightly soluble material. Example - What is the concentration of each ion in a saturated solution of manganese(II) hydroxide at 25oC Ksp = 1.6 x 10-13?

  16. Mn(OH)2(s) Mn2+ (aq)+ 2 OH1- (aq) 2X X 3.4 x 10-5 M 6.8 x 10-5 M Let x = the amount of each ion which appears when the system is at equilibrium [Mn2+] [OH1-] 2 Ksp = = 1.6 x 10-13 [x] [2x] 2 = 1.6 x 10-13 4x3 = 1.6 x 10-13 X = 3.4 x 10-5 molL-1 so [Mn2+],[OH1-] are

  17. Find the mass of Mn(OH)2in 68.5 L of saturated solution at 250C. It was previously determined the [Mn(OH)2] = 3.4 x 10-5 molL-1 so the number of moles (n) = CV = 3.42 x 10-5 molL-1 x68.5 L = 2.334 x 10-3 mol The mass of Mn(OH)2 can be determined using m = n x MM so m = 2.334 x 10-3 mol x 88.95 g/mol m = 0.21 g of Mn(OH)2

  18. What is the mass of Pb2+ in 69 L of a saturated solution of lead(II) fluoride at 25oC? Ksp =3.6 x 10-8

  19. PbF2(s) Pb2+ (aq)+ 2 F1- (aq) 2X X 2.1 x 10-3 M 4.2 x 10-3 M Let x = the amount of each ion which appears when the system is at equilibrium [Pb2+] [F1-] 2 Ksp = = 3.6 x 10-8 [x] [2x] 2 = 3.6 x 10-8 = 3.6 x 10-8 4x3 X = 2.1 x 10-3 molL-1 so [Pb2+],[F1-] are

  20. [Pb2+] = 2.1 x 10-3 molL-1 so the number of moles (n) = CV = 2.1 x 10-3 molL-1 x69 L = 0.1435 mol The mass of Pb2+can be determined using m = n x MM so m = 0.1435 mol x 207.2 g/mol m = 3.0 x 10 g of Pb2+

  21. Calculating Ksp fromSolubility Data Find the Ksp of Mg3(PO4)2 if a saturated solution is analyzed and found to contain 1.2 x 10-5 molL-1 of PO43-.

  22. Mg3(PO4)2(s) 3Mg2+ (aq)+ 2 PO43- (aq) 2X 3X 1.8 x 10-5 M 1.2 x 10-5 M Let x = the amount of each ion which appears when the system is at equilibrium 2x = 1.2 x 10-5 X = 6.0 x 10-6 [Mg2+] = 3x = 1.8 x 10-5 Ksp = [Mg2+] 3[PO43-]2 Ksp = [1.8 x 10-5]3[1.2 x 10-5]2 = 8.4 x 10-25

  23. 455 mL of a saturated solution of Ag2CrO4 is analyzed and found to contain 1.48 mg of silver. Using this data find the Ksp of Ag2CrO4.

  24. Ag2CrO4(s) 2Ag1+ (aq)+ CrO42- (aq) 2X X 3.01 x 10-5 M 1.5 x 10-5 M First find C of Ag1+ using C = n/V where n = m/MM, n = 1.48 x 10-3 g/107.9 g/mol C = 1.37 x 10-5 / 0.455 L = 3.01 x 10-5 = 2x x = 1.5 x 10-5 Ksp = [Ag1+]2[CrO42-] Ksp = [3.01 x 10-5]2[1.5 x 10-5 ] = 1.37 x 10-14

  25. 15 mL of a saturated solution of Ca(OH)2 is used to neutralize 0.45 g of a monoprotic standard acid with a molar mass of 204 g. Using this data find the Ksp of Ca(OH)2.

  26. When the solution turns pink the number # of mol of acid = # of mol of base equivalence pt. Titrant - Ca(OH)2 solution of known concentration 0.10 of potassium hydrogen phthalate distilled water few drops phenolphthalein

  27. Equilibrium Problems 1. What volume of a saturated solution of Cd(OH)2, Ksp is 7.2 x 10-15, is required to neutralize 1.3 g of potassium hydrogen phthalate (M is 204.23 g/mol). (2.6 x 102 L). 2. What is the Ksp of yttrium carbonate if 14.5 L of a saturated solution is analyzed and found to contain 1.3 x 10-3 g of this solute? (1.0 x 10-31) 3. If the Ksp of silver phosphate is 8.9 x 10-17 what mass of silver would be found in 29.9 L of saturated solution? (0.41 g) 4. What is the solubility of strontium arsenate. Ksp is 4.3 x 10-19?(8.3 x 10-5 mol/L) 5. What mass of lithium would you find in 34.1 L of a saturated solution of lithium phosphate if the Ksp is 2.4 x 10-11? (0.69 g)

  28. Common Ion Effect Ag2CO3(s) 2Ag1+ (aq)+ CO32- (aq) Consider this equilibrium If either the Ag1+ ion or the CO32- ion is added to this equilibrium it will upset the equilibrium and a shift will occur in this direction This shift will cause a decrease in the amount of Ag2CO3(s) capable of dissolving

  29. It is impossible to add just one of these ions so electrolytes containing one of these ions have a second ion which doesn’t affect the equilibrium. This other ion is known as a spectator ion. Examples- AgNO3, K2CO3, AgC2H3O2, Na2CO3 are some of the salts which could be used to shift this equilibrium. Notice they are all soluble materials.

  30. Ag2CO3(s) The presence of sodium carbonate pushes the equilibrium in the backwards direction which reduces the amount of silver carbonate which can dissolve Ag2CO3(s) 2Ag1+ (aq)+ CO32- (aq) CO32- Na1+ Na1+ CO32- Na1+ Na1+ 0 0.15 molL-1 initial Amount capable of dissolving 2x x 0.15 + x 2x @ equilibrium X is too small to affect the 0.15 so Determine mass of Ag2CO3(s) which could dissolve in 3.5 L of a 0.15 molL-1 solution of sodium carbonate. The common ion is CO32-

  31. 0.15 molL-1 initial Amount capable of dissolving 2x x Ag2CO3(s) 2Ag1+ (aq)+ CO32- (aq) 0.15 + x 2x @ equilibrium Ksp = [Ag1+]2[CO32-] 8.1 x 10-12 = [2x]2[0.15] 4x2 = 5.4 x 10-11 x = 3.67 x 10-6 molL-1 n = CV = 3.67 x 10-6 molL-1 x 3.5 L m=n x MM = 1.29 x 10-5 mol x 275.7 g/mol m=3.5 x 10-3 g of Ag2CO3(s)

  32. How many times more soluble is Ca(OH)2 in distilled water than in a 0.28 molL-1 solution of potassium hydroxide.(Ksp of Ca(OH)2 is6.5 x 10-6) Ca(OH)2(s) The presence of KOH pushes the equilibrium in the backwards direction which reduces the amount of Ca(OH)2 which can dissolve Ca(OH)2(s) Ca2+ (aq)+ 2 OH1- (aq) OH1- K1+ OH1 K1+ OH1- OH1 K1+ K1+ First determine the solubility in the KOH 0 0.28 molL-1 initial Amount capable of dissolving x 2x x 0.28 + 2x @ equilibrium Since the amount of OH1- which enters the solution from dissolving the Ca(OH)2 is very small it is insignificant

  33. 0.28 molL-1 initial Amount capable of dissolving x 2x Ca(OH)2(s) Ca2+ (aq)+ 2 OH1- (aq) x 0.28 + 2x @ equilibrium The amount of Ca(OH)2 which dissolves is represented by x so solving it answers half of the question. Ksp = x(0.28)2 so using tables for Ksp x = 6.5 x 10-6 / (0.28)2 = 8.29 x 10-5 molL-1

  34. Now determine the solubility in pure water Ca(OH)2(s) Ca2+ (aq)+ 2 OH1- (aq) initial 0 molL-1 0 molL-1 Amount capable of dissolving x 2x x 2x @ equilibrium Ksp = x(2x)2 so using tables for Ksp 6.5 x 10-6 = 4x3 x = 1.18 x 10-2 molL-1 To determine how much more soluble it is in pure water take 1.18 x 10-2/ 8.29 x 10-5 = 140x’s

  35. How many times more soluble is HgI2 in distilled water than in a 0.21 molL-1 solution of mercuric nitrate (Hg(NO3)2).1.1 x 10-28Common ion is HgI2(s) The presence of Hg2+ pushes the equilibrium in the backwards direction which reduces the amount of mercuric iodide which can dissolve HgI2(s) Hg2+ (aq)+ 2 I1- (aq) NO31- NO31- Hg2+ Hg2+ NO31- NO31- Hg2+ First determine the solubility in the Hg(NO3)2 0.21 molL-1 0 initial Amount capable of dissolving x 2x 0.21 + x 2x @ equilibrium Since the amount of Hg2+ which enters the solution from dissolving the HgI2 is very small it is insignificant

  36. 0.21 molL-1 0 initial Amount capable of dissolving x 2x HgI2(s) Hg2+ (aq)+ 2 I1- (aq) 0.21 + x 2x @ equilibrium Ksp = 0.21(2x)2 so using tables for Ksp 1.1 x 10-28 = (0.21)4x2 x = 1.14 x 10-14 molL-1

  37. Now determine the solubility in pure water HgI2(s) Hg2+ (aq)+ 2 I1- (aq) initial 0 molL-1 0 molL-1 Amount capable of dissolving x 2x x 2x @ equilibrium Ksp = x(2x)2 so using tables for Ksp 1.1 x 10-28 = 4x3 x = 3.02 x 10-10 molL-1 To determine how much more soluble it is in pure water take 3.02 x 10-10 /1.14 x 10-14 = 2.6 x 104x’s

  38. When Do Precipitates Form?. When different solutions are combined sometimes an insoluble precipitate will appear. This happens if the quantities of specific ions become large enough to form solid ionic crystals. Mathematically this can be determined using the ion product from the Ksp expression for the suspected precipitate. To demonstrate this let’s try a problem

  39. Will a precipitate form if 24 mL of a 0.024 molL-1 solution of Pb(NO3)2 is combined with 84 mL of a 0.0024 molL-1 solution of CaI2? First write the balanced chemical equation assuming a double displacement reaction. Make sure to use the appropriate charges. Ca2+ + 2 I1-+ Pb2++ 2 NO31- How do these ions interact? What are some of the possible precipitates? Ca(NO3)2andPbI2 Check out the solubility rules to determine which of these is only slightly soluble

  40. Solubility Rules for Aqueous Solutions Soluble Compounds 1. All compounds of the alkali metals (Group 1A) are soluble. 2. All salts containing NH41+, NO31-, ClO41-, ClO31-, and C2H3O21-are soluble. 3. All chlorides, bromides, and iodides are soluble except when combined with Ag1+, Pb2+ and Hg22+ 4. All sulfates are soluble except those of Hg22+, Pb2+, Ca2+, Sr2+ and Ba2+.

  41. PbI2 is a possible precipitate but only if the [Pb2+][I1-]2 > Ksp of PbI2 Where did this come from? Remember an insoluble substance establishes an equilibrium between a solid and the dissolved ions. For this precipitate the equilibrium is PbI2(s) Pb2+(aq)+ 2I1-(aq) Ksp = [Pb2+][I1-]2, This expression shows the maximum amount of Pb2+and I1- in solution which can coexist.

  42. Now determine the [Pb2+] and [I1-] Initially 24 mL of a 0.024 molL-1 solution of Pb(NO3)2 is combined with 84 mL of a 0.0024 molL-1 solution of CaI2, so [Pb2+] = 0.024 molL-1, [I1-] = 2 mol I1-x 0.0024 molL-1 since CaI2 ------> Ca2+ + 2I1- When the solutions are combined there is a dilution occurring so to find the new concentration use C1V1 = C2V2. For [Pb2+] C1 = 0.024 molL-1,V1 = 24 mL, V2=(24 + 84)mL C2 = (0.024 molL-1 )(24 mL) / 108 mL so [Pb2+] = 0.00533 molL-1.

  43. For [I1-] C1 = 0.0048 molL-1,V1 = 84 mL, V2=(24 + 84)mL C2 = (0.0048 molL-1 ) (84 mL) / 108 mL so [I1-] = 3.73 x 10-3molL-1. A precipitate forms if [Pb2+][I1-]2> Ksp ( 0.00533 molL-1 )(3.73 x 10-3molL-1 )2 = 7.4 x 10-8 since 7.4 x 10-8 > 8.5 x 10-9 Yes, a precipitate forms Now try this problem

  44. Will a precipitate form if 62 mL of a 0.036 molL-1 solution of Ca(NO3)2 is combined with 83 mL of a 0.0074 molL-1 solution of Na2CO3? First write the balanced chemical equation assuming a double displacement reaction. Make sure to use the appropriate charges. Ca2+ + CO32-+ Na1++ 2 NO31- How do these ions interact? What are some of the possible precipitates? CaCO3andNaNO3 Check out the solubility rules to determine which of these is only slightly soluble

  45. Some of the Solubility Rules 1. All compounds of the alkali metals (Group 1A) are soluble. 2. All salts containing NH41+, NO31-, ClO41-, ClO31-, and C2H3O21-are soluble. 3. All chlorides, bromides, and iodides are soluble except when combined with Ag1+, Pb2+ and Hg22+ 6. All compounds containing CO32-, PO43-, SO32-, S2- are insolubleexcept those of group 1A and NH41+.

  46. CaCO3 is a possible precipitate but only if the [Ca2+][CO32-] > Ksp Where did this come from? Remember an insoluble substance establishes an equilibrium between a solid and the dissolved ions. For this precipitate the equilibrium is CaCO3(s) Ca2+(aq)+ CO32-(aq) Ksp = [Ca2+][CO32-], This expression shows the maximum amount of Ca2+and CO32- in solution which can coexist.

  47. Now determine the [Ca2+] and [CO32-] Initially 62 mL of a 0.036 molL-1 solution of Ca(NO3)2 is combined with 83 mL of a 0.0074 molL-1 solution of Na2CO3, so [Ca2+] = 0.036 molL-1, [CO32-] = 0.0074 molL-1 When the solutions are combined there is a dilution occurring so to find the new concentration use C1V1 = C2V2. For [Ca2+] C1 = 0.036 molL-1,V1 = 62 mL, V2=(62 + 83)mL C2 = (0.036 molL-1 )(62 mL) / 145 mL so [Ca2+] = 0.0154 molL-1.

  48. For [CO32-] C1 = 0.0074 molL-1,V1 = 83 mL, V2=(62 + 83)mL C2 = (0.0074 molL-1 ) (83 mL) / 145 mL so [CO32-] = 4.24 x 10-3molL-1. A precipitate forms if [Ca2+][CO32-]> Ksp (1.54 x 10-2molL-1 )(4.24 x 10-3molL-1 )= 6.5 x 10-5 since 6.5 x 10-5 > 4.5 x 10-9 Yes, a precipitate forms

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