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AN INTRODUCTION TO SOLUBILITY PRODUCTS

AN INTRODUCTION TO SOLUBILITY PRODUCTS. 2008 SPECIFICATIONS. KNOCKHARDY PUBLISHING. KNOCKHARDY PUBLISHING. SOLUBILITY PRODUCTS. INTRODUCTION

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AN INTRODUCTION TO SOLUBILITY PRODUCTS

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  1. AN INTRODUCTION TO SOLUBILITY PRODUCTS 2008 SPECIFICATIONS KNOCKHARDY PUBLISHING

  2. KNOCKHARDY PUBLISHING SOLUBILITY PRODUCTS INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... www.knockhardy.org.uk/sci.htm Navigation is achieved by... either clicking on the grey arrows at the foot of each page or using the left and right arrow keys on the keyboard

  3. SOLUBILITY PRODUCTS • CONTENTS • Solubility of ionic compounds • Saturated solutions • Solubility products • Calculations • Common Ion Effect

  4. SOLUBILITY OF IONIC COMPOUNDS • ionic compounds tend to be insoluble in non-polar solvents • ionic compounds tend to be soluble in water • water is a polar solvent and stabilises the separated ions

  5. SOLUBILITY OF IONIC COMPOUNDS • ionic compounds tend to be insoluble in non-polar solvents • ionic compounds tend to be soluble in water • water is a polar solvent and stabilises the separated ions • some ionic compounds are very insoluble (AgCl, PbSO4, PbS) • even soluble ionic compounds have a limit as to how much dissolves

  6. SATURATED SOLUTIONS • solutions become saturated when solute no longer dissolves in a solvent • solubility varies with temperature • most solutes are more soluble at higher temperatures

  7. SATURATED SOLUTIONS • solutions become saturated when solute no longer dissolves in a solvent • solubility varies with temperature • most solutes are more soluble at higher temperatures Ionic crystal lattices can dissociate (break up) when placed in water. The ions separate as they are stabilised by polar water molecules.

  8. SATURATED SOLUTIONS • solutions become saturated when solute no longer dissolves in a solvent • solubility varies with temperature • most solutes are more soluble at higher temperatures Ionic crystal lattices can dissociate (break up) when placed in water. The ions separate as they are stabilised by polar water molecules. Eventually, no more solute dissolves and the solution becomes saturated. There is a limit to the concentration of ions in solution.

  9. SATURATED SOLUTIONS • solutions become saturated when solute no longer dissolves in a solvent • solubility varies with temperature • most solutes are more soluble at higher temperatures Ionic crystal lattices can dissociate (break up) when placed in water. The ions separate as they are stabilised by polar water molecules. Eventually, no more solute dissolves and the solution becomes saturated. There is a limit to the concentration of ions in solution.

  10. SOLUBILITY PRODUCT Even the most insoluble ionic compounds dissolve to a small extent. An equilibrium exists between the undissolved solid and its aqueous ions; (i)MX(s) M+(aq) + X¯(aq) (ii)BaSO4(s) Ba2+(aq) + SO42-(aq) (iii)PbCl2(s) Pb2+(aq) + 2Cl¯(aq)

  11. SOLUBILITY PRODUCT Even the most insoluble ionic compounds dissolve to a small extent. An equilibrium exists between the undissolved solid and its aqueous ions; (i)MX(s) M+(aq) + X¯(aq) (ii)BaSO4(s) Ba2+(aq) + SO42-(aq) (iii)PbCl2(s) Pb2+(aq) + 2Cl¯(aq) Applying the equilibrium law to (i) and assuming the concentration of MX(s) is constant in a saturated solution. [M+(aq)] [X¯(aq)] = a constant,Ksp [ ] is the concentration in mol dm-3 Kspis known as theSOLUBILITY PRODUCT

  12. SOLUBILITY PRODUCT AgCl(s) Ag+(aq) + Cl¯(aq) Ksp = [Ag+(aq)] [Cl¯(aq)] BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = [Ba2+(aq)] [SO42-(aq)] PbCl2(s) Pb2+(aq) + 2Cl¯(aq) Ksp = [Pb2+(aq)] [Cl¯(aq)]2 Notice that the concentration of Cl¯(aq) is raised to the power of 2 because there are two Cl¯(aq) ions in the equation

  13. SOLUBILITY PRODUCT AgCl(s) Ag+(aq) + Cl¯(aq) Ksp = [Ag+(aq)] [Cl¯(aq)] BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = [Ba2+(aq)] [SO42-(aq)] PbCl2(s) Pb2+(aq) + 2Cl¯(aq) Ksp = [Pb2+(aq)] [Cl¯(aq)]2 Notice that the concentration of Cl¯(aq) is raised to the power of 2 Complete the equilibrium equation and write an expression for Ksp for… PbS(s) Pb2+(aq) + S2-(aq) Ksp = [Pb2+(aq)] [S2-(aq)] Fe(OH)2(s) Fe2+(aq) + 2OH¯(aq) Ksp = [Fe2+(aq)] [OH¯(aq)]2 Fe(OH)3(s) Fe3+(aq) + 3OH¯(aq) Ksp = [Fe3+(aq)] [OH¯(aq)]3

  14. SOLUBILITY PRODUCT AgCl(s) Ag+(aq) + Cl¯(aq) Ksp = [Ag+(aq)] [Cl¯(aq)] BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = [Ba2+(aq)] [SO42-(aq)] PbCl2(s) Pb2+(aq) + 2Cl¯(aq) Ksp = [Pb2+(aq)] [Cl¯(aq)]2 Notice that the concentration of Cl¯(aq) is raised to the power of 2 Complete the equilibrium equation and write an expression for Ksp for… PbS(s) Pb2+(aq) + S2-(aq) Ksp = [Pb2+(aq)] [S2-(aq)] Fe(OH)2(s) Fe2+(aq) + 2OH¯(aq) Ksp = [Fe2+(aq)] [OH¯(aq)]2 Fe(OH)3(s) Fe3+(aq) + 3OH¯(aq) Ksp = [Fe3+(aq)] [OH¯(aq)]3

  15. SOLUBILITY PRODUCT AgCl(s) Ag+(aq) + Cl¯(aq) Ksp = [Ag+(aq)] [Cl¯(aq)] BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = [Ba2+(aq)] [SO42-(aq)] PbCl2(s) Pb2+(aq) + 2Cl¯(aq) Ksp = [Pb2+(aq)] [Cl¯(aq)]2 Notice that the concentration of Cl¯(aq) is raised to the power of 2 Complete the equilibrium equation and write an expression for Ksp for… PbS(s) Pb2+(aq) + S2-(aq) Ksp = [Pb2+(aq)] [S2-(aq)] Fe(OH)2(s) Fe2+(aq) + 2OH¯(aq) Ksp = [Fe2+(aq)] [OH¯(aq)]2 Fe(OH)3(s) Fe3+(aq) + 3OH¯(aq) Ksp = [Fe3+(aq)] [OH¯(aq)]3

  16. SOLUBILITY PRODUCT AgCl(s) Ag+(aq) + Cl¯(aq) Ksp = [Ag+(aq)] [Cl¯(aq)] BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = [Ba2+(aq)] [SO42-(aq)] PbCl2(s) Pb2+(aq) + 2Cl¯(aq) Ksp = [Pb2+(aq)] [Cl¯(aq)]2 Notice that the concentration of Cl¯(aq) is raised to the power of 2 Complete the equilibrium equation and write an expression for Ksp for… PbS(s) Pb2+(aq) + S2-(aq) Ksp = [Pb2+(aq)] [S2-(aq)] Fe(OH)2(s) Fe2+(aq) + 2OH¯(aq) Ksp = [Fe2+(aq)] [OH¯(aq)]2 Fe(OH)3(s) Fe3+(aq) + 3OH¯(aq) Ksp = [Fe3+(aq)] [OH¯(aq)]3

  17. SOLUBILITY PRODUCT Units The value of Ksp has units and it varies with temperature AgCl Ksp = [Ag+(aq)] [Cl¯(aq)] units of… mol2 dm-6 BaSO4 Ksp = [Ba2+(aq)] [SO42-(aq)] mol2 dm-6 PbCl2 Ksp = [Pb2+(aq)] [Cl¯(aq)]2 mol3 dm-9

  18. SOLUBILITY PRODUCT Units The value of Ksp has units and it varies with temperature AgCl Ksp = [Ag+(aq)] [Cl¯(aq)] units of… mol2 dm-6 BaSO4 Ksp = [Ba2+(aq)] [SO42-(aq)] mol2 dm-6 PbCl2 Ksp = [Pb2+(aq)] [Cl¯(aq)]2 mol3 dm-9 Work out the units of Ksp for the following… PbS Ksp = [Pb2+(aq)] [S2-(aq)] Fe(OH)2 Ksp = [Fe2+(aq)] [OH¯(aq)]2 Fe(OH)3 Ksp = [Fe3+(aq)] [OH¯(aq)]3

  19. SOLUBILITY PRODUCT Units The value of Ksp has units and it varies with temperature AgCl Ksp = [Ag+(aq)] [Cl¯(aq)] units of… mol2 dm-6 BaSO4 Ksp = [Ba2+(aq)] [SO42-(aq)] mol2 dm-6 PbCl2 Ksp = [Pb2+(aq)] [Cl¯(aq)]2 mol3 dm-9 Work out the units of Ksp for the following… PbS Ksp = [Pb2+(aq)] [S2-(aq)] mol2 dm-6 Fe(OH)2 Ksp = [Fe2+(aq)] [OH¯(aq)]2 mol3 dm-9 Fe(OH)3 Ksp = [Fe3+(aq)] [OH¯(aq)]3 mol4 dm-12

  20. CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide.

  21. CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq)

  22. CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) The expression for the solubility product is Ksp = [Pb2+(aq)] [S2-(aq)]

  23. CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) The expression for the solubility product is Ksp = [Pb2+(aq)] [S2-(aq)] According to the equation, you get one Pb2+(aq) for every one S2-(aq); the concentrations will be equal [Pb2+(aq)] = [S2-(aq)] Substituting and rewriting the expression for Ksp Ksp = [Pb2+(aq)]2

  24. CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) The expression for the solubility product is Ksp = [Pb2+(aq)] [S2-(aq)] According to the equation, you get one Pb2+(aq) for every one S2-(aq); the concentrations will be equal [Pb2+(aq)] = [S2-(aq)] Substituting and rewriting the expression for Ksp Ksp = [Pb2+(aq)]2 Re-arranging; [Pb2+(aq)] = Ksp = 4 x 10-28 = 2 x 10-14 mol dm-3

  25. CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) The expression for the solubility product is Ksp = [Pb2+(aq)] [S2-(aq)] According to the equation, you get one Pb2+(aq) for every one S2-(aq); the concentrations will be equal [Pb2+(aq)] = [S2-(aq)] Substituting and rewriting the expression for Ksp Ksp = [Pb2+(aq)]2 Re-arranging; [Pb2+(aq)] = Ksp = 4 x 10-28 = 2 x 10-14 mol dm-3 As you get one Pb2+ from one PbS, the solubility of PbS = 2 x 10-14 mol dm-3

  26. CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) The expression for the solubility product is Ksp = [Pb2+(aq)] [S2-(aq)] According to the equation, you get one Pb2+(aq) for every one S2-(aq); the concentrations will be equal [Pb2+(aq)] = [S2-(aq)] Substituting and rewriting the expression for Ksp Ksp = [Pb2+(aq)]2 Re-arranging; [Pb2+(aq)] = Ksp = 4 x 10-28 = 2 x 10-14 mol dm-3 As you get one Pb2+ from one PbS, the solubility of PbS = 2 x 10-14 mol dm-3 Mr for PbS is 239; the solubility is 239 x 2 x 10-14 g dm-3 = 5.78 x 10-12 g dm-3 [mass = moles x molar mass]

  27. CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, PbS is 4 x 10-28 mol2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb2+(aq) + S2-(aq) The expression for the solubility product is Ksp = [Pb2+(aq)] [S2-(aq)] According to the equation, you get one Pb2+(aq) for every one S2-(aq); the concentrations will be equal [Pb2+(aq)] = [S2-(aq)] Substituting and rewriting the expression for KspKsp = [Pb2+(aq)]2 Re-arranging; [Pb2+(aq)] = Ksp = 4 x 10-28 = 2 x 10-14 mol dm-3 As you get one Pb2+ from one PbS, the solubility of PbS = 2 x 10-14 mol dm-3 Mr for PbS is 239; the solubility is 239 x 2 x 10-14 g dm-3 = 5.78 x 10-12 g dm-3 [mass = moles x molar mass]

  28. CALCULATIONS The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25°C.

  29. CALCULATIONS The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25°C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200

  30. CALCULATIONS The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25°C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 = 2.5 x 10-12 mol dm-3

  31. CALCULATIONS The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25°C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 = 2.5 x 10-12 mol dm-3 The equation for its solubility is MY(s) M+(aq) + Y-(aq)

  32. CALCULATIONS The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25°C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 = 2.5 x 10-12 mol dm-3 The equation for its solubility is MY(s) M+(aq) + Y-(aq) The expression for the solubility product is Ksp = [M+(aq)] [Y-(aq)]

  33. CALCULATIONS The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25°C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 = 2.5 x 10-12 mol dm-3 The equation for its solubility is MY(s) M+(aq) + Y-(aq) The expression for the solubility product is Ksp = [M+(aq)] [Y-(aq)] According to the equation; moles of M+(aq) = moles of Y-(aq) = moles of dissolved MY(s)

  34. CALCULATIONS The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25°C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 = 2.5 x 10-12 mol dm-3 The equation for its solubility is MY(s) M+(aq) + Y-(aq) The expression for the solubility product is Ksp = [M+(aq)] [Y-(aq)] According to the equation; moles of M+(aq) = moles of Y-(aq) = moles of dissolved MY(s) Substituting values Ksp = [2.5 x 10-12 ] [2.5 x 10-12 ] The value of the solubility product Ksp = 6.25 x 10-24 mol2 dm-6

  35. CALCULATIONS The solubility of ionic compound MY at 25°C is 5 x 10-10 g dm-3 . The relative mass of MY is 200. Calculate the solubility product of the salt MY at 25°C. Solubility of MY in mol dm-3 = solubility in g = 5 x 10-10 g dm-3 molar mass 200 = 2.5 x 10-12 mol dm-3 The equation for its solubility is MY(s) M+(aq) + Y-(aq) The expression for the solubility product is Ksp = [M+(aq)] [Y-(aq)] According to the equation; moles of M+(aq) = moles of Y-(aq) = moles of dissolved MY(s) Substituting values Ksp = [2.5 x 10-12 ] [2.5 x 10-12 ] The value of the solubility product Ksp = 6.25 x 10-24 mol2 dm-6

  36. THE COMMON ION EFFECT Adding a common ion, (one which is present in the solution), will result in the precipitation of a sparingly soluble ionic compound. eg Adding a solution of sodium chloride to a saturated solution of silver chloride will result in the precipitation of silver chloride.

  37. THE COMMON ION EFFECT Adding a common ion, (one which is present in the solution), will result in the precipitation of a sparingly soluble ionic compound. eg Adding a solution of sodium chloride to a saturated solution of silver chloride will result in the precipitation of silver chloride. Adding the ionic compound MA to a solution of MX increases the concentration of M+(aq). M+(aq) is a common ion as it is already in solution.

  38. THE COMMON ION EFFECT Adding a common ion, (one which is present in the solution), will result in the precipitation of a sparingly soluble ionic compound. eg Adding a solution of sodium chloride to a saturated solution of silver chloride will result in the precipitation of silver chloride. Adding the ionic compound MA to a solution of MX increases the concentration of M+(aq). M+(aq) is a common ion as it is already in solution. The extra M+ ions means that the solubility product is exceeded. To reduce the value of [M+(aq)][X-(aq)] below the Ksp, some ions are removed from solution by precipitating.

  39. THE COMMON ION EFFECT Adding a common ion, (one which is present in the solution), will result in the precipitation of a sparingly soluble ionic compound. eg Adding a solution of sodium chloride to a saturated solution of silver chloride will result in the precipitation of silver chloride. Silver chloride dissociates in water as follows AgCl(s) Ag+(aq) + Cl¯(aq) The solubility product at 25°C is Ksp = [Ag+(aq)] [Cl¯(aq)] = 1.2 x 10-10 mol2 dm-6 If the value of the solubility product is exceeded, precipitation will occur.

  40. THE COMMON ION EFFECT Adding a common ion, (one which is present in the solution), will result in the precipitation of a sparingly soluble ionic compound. eg Adding a solution of sodium chloride to a saturated solution of silver chloride will result in the precipitation of silver chloride. Silver chloride dissociates in water as follows AgCl(s) Ag+(aq) + Cl¯(aq) The solubility product at 25°C is Ksp = [Ag+(aq)] [Cl¯(aq)] = 1.2 x 10-10 mol2 dm-6 If the value of the solubility product is exceeded, precipitation will occur. The value can be exceeded by adding EITHER of the two soluble ions. If sodium chloride solution is added, the concentration of Cl¯(aq) will increase and precipitation will occur. Likewise, addition of silver nitrate solution AgNO3(aq) would produce the same effect as it would increase the concentration of Ag+(aq).

  41. THE COMMON ION EFFECT If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3) solutions are mixed, will AgCl be precipitated? Ksp = 1.2 x10-10 mol2 dm-6

  42. THE COMMON ION EFFECT If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3) solutions are mixed, will AgCl be precipitated? Ksp = 1.2 x10-10 mol2 dm-6 • in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3 in NaCl the concentration of Cl¯ is 2 x 10-5 mol dm-3

  43. THE COMMON ION EFFECT If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3) solutions are mixed, will AgCl be precipitated? Ksp = 1.2 x10-10 mol2 dm-6 • in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3 in NaCl the concentration of Cl¯ is 2 x 10-5 mol dm-3 • when equal volumes are mixed, the concentrations are halved [Ag+] = 1 x 10-5 mol dm-3[Cl¯] = 1 x 10-5 mol dm-3

  44. THE COMMON ION EFFECT If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3) solutions are mixed, will AgCl be precipitated? Ksp = 1.2 x10-10 mol2 dm-6 • in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3 in NaCl the concentration of Cl¯ is 2 x 10-5 mol dm-3 • when equal volumes are mixed, the concentrations are halved [Ag+] = 1 x 10-5 mol dm-3 [Cl¯] = 1 x 10-5 mol dm-3 • [Ag+] [Cl¯] = [1 x 10-5 mol dm-3] x [1 x 10-5 mol dm-3] = 1 x 10-10 mol2 dm-6

  45. THE COMMON ION EFFECT If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3) solutions are mixed, will AgCl be precipitated? Ksp = 1.2 x10-10 mol2 dm-6 • in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3 in NaCl the concentration of Cl¯ is 2 x 10-5 mol dm-3 • when equal volumes are mixed, the concentrations are halved [Ag+] = 1 x 10-5 mol dm-3 [Cl¯] = 1 x 10-5 mol dm-3 • [Ag+] [Cl¯] = [1 x 10-5 mol dm-3] x [1 x 10-5 mol dm-3] = 1 x 10-10 mol2 dm-6 • because this is lower than the Ksp for AgCl... NO PRECIPITATION OCCURS

  46. THE COMMON ION EFFECT If equal volumes of AgNO3 (2 x 10-5 mol dm-3) and NaCl (2 x 10-5 mol dm-3) solutions are mixed, will AgCl be precipitated? Ksp = 1.2 x10-10 mol2 dm-6 • in AgNO3 the concentration of Ag+ is 2 x 10-5 mol dm-3 in NaCl the concentration of Cl¯ is 2 x 10-5 mol dm-3 • when equal volumes are mixed, the concentrations are halved [Ag+] = 1 x 10-5 mol dm-3[Cl¯] = 1 x 10-5 mol dm-3 • [Ag+] [Cl¯] = [1 x 10-5 mol dm-3] x [1 x 10-5 mol dm-3] = 1 x 10-10 mol2 dm-6 • because this is lower than the Ksp for AgCl... NO PRECIPITATION OCCURS

  47. AN INTRODUCTION TO SOLUBILITY PRODUCTS THE END © 2009 JONATHAN HOPTON & KNOCKHARDY PUBLISHING

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