1. AN INTRODUCTION TO SOLUBILITYPRODUCTS KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS

2. 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 orusing the left and right arrow keys on the keyboard KNOCKHARDY PUBLISHING SOLUBILITY PRODUCTS

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

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, PbSO 4, 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)BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) (iii) PbCl 2 (s) Pb 2+ (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)BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) (iii) PbCl 2 (s) Pb 2+ (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, K sp [ ] is the concentration in mol dm -3 K sp is known as the SOLUBILITY PRODUCT

12. SOLUBILITY PRODUCT AgCl(s) Ag + (aq) + Cl¯(aq) K sp = [Ag + (aq)] [Cl¯(aq)] BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) K sp = [Ba 2+ (aq)] [SO 4 2- (aq)] PbCl 2 (s) Pb 2+ (aq) + 2Cl¯(aq) K sp = [Pb 2+ (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) K sp = [Ag + (aq)] [Cl¯(aq)] BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) K sp = [Ba 2+ (aq)] [SO 4 2- (aq)] PbCl 2 (s) Pb 2+ (aq) + 2Cl¯(aq) K sp = [Pb 2+ (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 K sp for… PbS(s) Pb 2+ (aq) + S 2- (aq) K sp = [Pb 2+ (aq)] [S 2- (aq)] Fe(OH) 2 (s) Fe 2+ (aq) + 2OH¯(aq) K sp = [Fe 2+ (aq)] [OH¯(aq)] 2 Fe(OH) 3 (s) Fe 3+ (aq) + 3OH¯(aq) K sp = [Fe 3+ (aq)] [OH¯(aq)] 3

14. SOLUBILITY PRODUCT AgCl(s) Ag + (aq) + Cl¯(aq) K sp = [Ag + (aq)] [Cl¯(aq)] BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) K sp = [Ba 2+ (aq)] [SO 4 2- (aq)] PbCl 2 (s) Pb 2+ (aq) + 2Cl¯(aq) K sp = [Pb 2+ (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 K sp for… PbS(s) Pb 2+ (aq) + S 2- (aq) K sp = [Pb 2+ (aq)] [S 2- (aq)] Fe(OH) 2 (s) Fe 2+ (aq) + 2OH¯(aq) K sp = [Fe 2+ (aq)] [OH¯(aq)] 2 Fe(OH) 3 (s) Fe 3+ (aq) + 3OH¯(aq) K sp = [Fe 3+ (aq)] [OH¯(aq)] 3

15. SOLUBILITY PRODUCT AgCl(s) Ag + (aq) + Cl¯(aq) K sp = [Ag + (aq)] [Cl¯(aq)] BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) K sp = [Ba 2+ (aq)] [SO 4 2- (aq)] PbCl 2 (s) Pb 2+ (aq) + 2Cl¯(aq) K sp = [Pb 2+ (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 K sp for… PbS(s) Pb 2+ (aq) + S 2- (aq) K sp = [Pb 2+ (aq)] [S 2- (aq)] Fe(OH) 2 (s) Fe 2+ (aq) + 2OH¯(aq) K sp = [Fe 2+ (aq)] [OH¯(aq)] 2 Fe(OH) 3 (s) Fe 3+ (aq) + 3OH¯(aq) K sp = [Fe 3+ (aq)] [OH¯(aq)] 3

16. SOLUBILITY PRODUCT AgCl(s) Ag + (aq) + Cl¯(aq) K sp = [Ag + (aq)] [Cl¯(aq)] BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) K sp = [Ba 2+ (aq)] [SO 4 2- (aq)] PbCl 2 (s) Pb 2+ (aq) + 2Cl¯(aq) K sp = [Pb 2+ (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 K sp for… PbS(s) Pb 2+ (aq) + S 2- (aq) K sp = [Pb 2+ (aq)] [S 2- (aq)] Fe(OH) 2 (s) Fe 2+ (aq) + 2OH¯(aq) K sp = [Fe 2+ (aq)] [OH¯(aq)] 2 Fe(OH) 3 (s) Fe 3+ (aq) + 3OH¯(aq) K sp = [Fe 3+ (aq)] [OH¯(aq)] 3

17. SOLUBILITY PRODUCT Units UnitsThe value of K sp has units and it varies with temperature AgCl K sp = [Ag + (aq)] [Cl¯(aq)] units of…mol 2 dm -6 BaSO 4 K sp = [Ba 2+ (aq)] [SO 4 2- (aq)]mol 2 dm -6 PbCl 2 K sp = [Pb 2+ (aq)] [Cl¯(aq)] 2 mol 3 dm -9

18. SOLUBILITY PRODUCT Units UnitsThe value of K sp has units and it varies with temperature AgCl K sp = [Ag + (aq)] [Cl¯(aq)] units of…mol 2 dm -6 BaSO 4 K sp = [Ba 2+ (aq)] [SO 4 2- (aq)]mol 2 dm -6 PbCl 2 K sp = [Pb 2+ (aq)] [Cl¯(aq)] 2 mol 3 dm -9 Work out the units of K sp for the following… PbS K sp = [Pb 2+ (aq)] [S 2- (aq)] Fe(OH) 2 K sp = [Fe 2+ (aq)] [OH¯(aq)] 2 Fe(OH) 3 K sp = [Fe 3+ (aq)] [OH¯(aq)] 3

19. SOLUBILITY PRODUCT Units UnitsThe value of K sp has units and it varies with temperature AgCl K sp = [Ag + (aq)] [Cl¯(aq)] units of…mol 2 dm -6 BaSO 4 K sp = [Ba 2+ (aq)] [SO 4 2- (aq)]mol 2 dm -6 PbCl 2 K sp = [Pb 2+ (aq)] [Cl¯(aq)] 2 mol 3 dm -9 Work out the units of K sp for the following… PbS K sp = [Pb 2+ (aq)] [S 2- (aq)]mol 2 dm -6 Fe(OH) 2 K sp = [Fe 2+ (aq)] [OH¯(aq)] 2 mol 3 dm -9 Fe(OH) 3 K sp = [Fe 3+ (aq)] [OH¯(aq)] 3 mol 4 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 mol 2 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 mol 2 dm -6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb 2+ (aq) + S 2- (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 mol 2 dm -6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb 2+ (aq) + S 2- (aq) The expression for the solubility product isK sp = [Pb 2+ (aq)] [S 2- (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 mol 2 dm -6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb 2+ (aq) + S 2- (aq) The expression for the solubility product isK sp = [Pb 2+ (aq)] [S 2- (aq)] According to the equation, you get one Pb 2+ (aq) for every one S 2- (aq); the concentrations will be equal [Pb 2+ (aq)] = [S 2- (aq)] Substituting and rewriting the expression for K sp K sp = [Pb 2+ (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 mol 2 dm -6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb 2+ (aq) + S 2- (aq) The expression for the solubility product isK sp = [Pb 2+ (aq)] [S 2- (aq)] According to the equation, you get one Pb 2+ (aq) for every one S 2- (aq); the concentrations will be equal [Pb 2+ (aq)] = [S 2- (aq)] Substituting and rewriting the expression for K sp K sp = [Pb 2+ (aq)] 2 Re-arranging;[Pb 2+ (aq)] = K sp = 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 mol 2 dm -6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb 2+ (aq) + S 2- (aq) The expression for the solubility product isK sp = [Pb 2+ (aq)] [S 2- (aq)] According to the equation, you get one Pb 2+ (aq) for every one S 2- (aq); the concentrations will be equal [Pb 2+ (aq)] = [S 2- (aq)] Substituting and rewriting the expression for K sp K sp = [Pb 2+ (aq)] 2 Re-arranging;[Pb 2+ (aq)] = K sp = 4 x 10 -28 = 2 x 10 -14 mol dm -3 As you get one Pb 2+ 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 mol 2 dm -6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb 2+ (aq) + S 2- (aq) The expression for the solubility product isK sp = [Pb 2+ (aq)] [S 2- (aq)] According to the equation, you get one Pb 2+ (aq) for every one S 2- (aq); the concentrations will be equal [Pb 2+ (aq)] = [S 2- (aq)] Substituting and rewriting the expression for K sp K sp = [Pb 2+ (aq)] 2 Re-arranging;[Pb 2+ (aq)] = K sp = 4 x 10 -28 = 2 x 10 -14 mol dm -3 As you get one Pb 2+ from one PbS, the solubility of PbS = 2 x 10 -14 mol dm -3 M r 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 mol 2 dm -6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is PbS(s) Pb 2+ (aq) + S 2- (aq) The expression for the solubility product isK sp = [Pb 2+ (aq)] [S 2- (aq)] According to the equation, you get one Pb 2+ (aq) for every one S 2- (aq); the concentrations will be equal [Pb 2+ (aq)] = [S 2- (aq)] Substituting and rewriting the expression for K sp K sp = [Pb 2+ (aq)] 2 Re-arranging;[Pb 2+ (aq)] = K sp = 4 x 10 -28 = 2 x 10 -14 mol dm -3 As you get one Pb 2+ from one PbS, the solubility of PbS = 2 x 10 -14 mol dm -3 M r 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 isK sp = [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 isK sp = [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 isK sp = [M + (aq)] [Y - (aq)] According to the equation;moles of M + (aq) = moles of Y - (aq) = moles of dissolved MY(s) Substituting values K sp = [2.5 x 10 -12 ] [2.5 x 10 -12 ] The value of the solubility product K sp = 6.25 x 10 -24 mol 2 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 isK sp = [M + (aq)] [Y - (aq)] According to the equation;moles of M + (aq) = moles of Y - (aq) = moles of dissolved MY(s) Substituting values K sp = [2.5 x 10 -12 ] [2.5 x 10 -12 ] The value of the solubility product K sp = 6.25 x 10 -24 mol 2 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. egAdding 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. egAdding 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. egAdding 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 K sp, 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. egAdding 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 K sp = [Ag + (aq)] [Cl¯(aq)] = 1.2 x 10 -10 mol 2 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. egAdding 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 K sp = [Ag + (aq)] [Cl¯(aq)] = 1.2 x 10 -10 mol 2 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 AgNO 3 (aq) would produce the same effect as it would increase the concentration of Ag + (aq).

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

42. THE COMMON ION EFFECT If equal volumes of AgNO 3 (2 x 10 -5 mol dm -3 ) and NaCl (2 x 10 -5 mol dm -3 ) solutions are mixed, will AgCl be precipitated? K sp = 1.2 x10 -10 mol 2 dm -6 in AgNO 3 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 AgNO 3 (2 x 10 -5 mol dm -3 ) and NaCl (2 x 10 -5 mol dm -3 ) solutions are mixed, will AgCl be precipitated? K sp = 1.2 x10 -10 mol 2 dm -6 in AgNO 3 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 AgNO 3 (2 x 10 -5 mol dm -3 ) and NaCl (2 x 10 -5 mol dm -3 ) solutions are mixed, will AgCl be precipitated? K sp = 1.2 x10 -10 mol 2 dm -6 in AgNO 3 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 mol 2 dm -6

45. THE COMMON ION EFFECT If equal volumes of AgNO 3 (2 x 10 -5 mol dm -3 ) and NaCl (2 x 10 -5 mol dm -3 ) solutions are mixed, will AgCl be precipitated? K sp = 1.2 x10 -10 mol 2 dm -6 in AgNO 3 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 mol 2 dm -6 because this is lower than the K sp for AgCl... NO PRECIPITATION OCCURS

46. THE COMMON ION EFFECT If equal volumes of AgNO 3 (2 x 10 -5 mol dm -3 ) and NaCl (2 x 10 -5 mol dm -3 ) solutions are mixed, will AgCl be precipitated? K sp = 1.2 x10 -10 mol 2 dm -6 in AgNO 3 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 mol 2 dm -6 because this is lower than the K sp for AgCl... NO PRECIPITATION OCCURS

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