AN INTRODUCTION TO SOLUBILITY PRODUCTS KNOCKHARDY PUBLISHING 2008

AN INTRODUCTION TO SOLUBILITY PRODUCTS KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS

KNOCKHARDY PUBLISHING SOLUBILITY PRODUCTS INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A 2 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 A 2 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

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

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

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 (Ag. Cl, Pb. SO 4, Pb. S) • even soluble ionic compounds have a limit as to how much dissolves

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

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.

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.

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) Ba. SO 4(s) Ba 2+(aq) + SO 42 -(aq) (iii) Pb. Cl 2(s) Pb 2+(aq) + 2 Cl¯(aq)

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) Ba. SO 4(s) Ba 2+(aq) + SO 42 -(aq) (iii) Pb. Cl 2(s) Pb 2+(aq) + 2 Cl¯(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 Ksp is known as the SOLUBILITY PRODUCT

SOLUBILITY PRODUCT Ag. Cl(s) Ag+(aq) + Cl¯(aq) Ksp = [Ag+(aq)] [Cl¯(aq)] Ba. SO 4(s) Ba 2+(aq) + SO 42 -(aq) Ksp = [Ba 2+(aq)] [SO 42 -(aq)] Pb. Cl 2(s) Pb 2+(aq) + 2 Cl¯(aq) Ksp = [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

SOLUBILITY PRODUCT Ag. Cl(s) Ag+(aq) + Cl¯(aq) Ksp = [Ag+(aq)] [Cl¯(aq)] Ba. SO 4(s) Ba 2+(aq) + SO 42 -(aq) Ksp = [Ba 2+(aq)] [SO 42 -(aq)] Pb. Cl 2(s) Pb 2+(aq) + 2 Cl¯(aq) Ksp = [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 Ksp for… Pb. S(s) Pb 2+(aq) + S 2 -(aq) Ksp = [Pb 2+(aq)] [S 2 -(aq)] Fe(OH)2(s) Fe 2+(aq) + 2 OH¯(aq) Ksp = [Fe 2+(aq)] [OH¯(aq)]2 Fe(OH)3(s) Fe 3+(aq) + 3 OH¯(aq) Ksp = [Fe 3+(aq)] [OH¯(aq)]3

SOLUBILITY PRODUCT Units The value of Ksp has units and it varies with temperature Ag. Cl Ksp = [Ag+(aq)] [Cl¯(aq)] units of… mol 2 dm-6 Ba. SO 4 Ksp = [Ba 2+(aq)] [SO 42 -(aq)] mol 2 dm-6 Pb. Cl 2 Ksp = [Pb 2+(aq)] [Cl¯(aq)]2 mol 3 dm-9

SOLUBILITY PRODUCT Units The value of Ksp has units and it varies with temperature Ag. Cl Ksp = [Ag+(aq)] [Cl¯(aq)] units of… Ba. SO 4 Ksp = [Ba 2+(aq)] [SO 42 -(aq)] mol 2 dm-6 Pb. Cl 2 Ksp = [Pb 2+(aq)] [Cl¯(aq)]2 mol 3 dm-9 Work out the units of Ksp for the following… Pb. S Ksp = [Pb 2+(aq)] [S 2 -(aq)] Fe(OH)2 Ksp = [Fe 2+(aq)] [OH¯(aq)]2 Fe(OH)3 Ksp = [Fe 3+(aq)] [OH¯(aq)]3 mol 2 dm-6

SOLUBILITY PRODUCT Units The value of Ksp has units and it varies with temperature Ag. Cl Ksp = [Ag+(aq)] [Cl¯(aq)] units of… mol 2 dm-6 Ba. SO 4 Ksp = [Ba 2+(aq)] [SO 42 -(aq)] mol 2 dm-6 Pb. Cl 2 Ksp = [Pb 2+(aq)] [Cl¯(aq)]2 mol 3 dm-9 Work out the units of Ksp for the following… Pb. S Ksp = [Pb 2+(aq)] [S 2 -(aq)] mol 2 dm-6 Fe(OH)2 Ksp = [Fe 2+(aq)] [OH¯(aq)]2 mol 3 dm-9 Fe(OH)3 Ksp = [Fe 3+(aq)] [OH¯(aq)]3 mol 4 dm-12

CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, Pb. S is 4 x 10 -28 mol 2 dm-6. Calculate the solubility of lead(II) sulphide.

CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, Pb. S is 4 x 10 -28 mol 2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is Pb. S(s) Pb 2+(aq) + S 2 -(aq)

CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, Pb. S is 4 x 10 -28 mol 2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is The expression for the solubility product is Pb. S(s) Pb 2+(aq) + S 2 -(aq) Ksp = [Pb 2+(aq)] [S 2 -(aq)]

CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, Pb. S is 4 x 10 -28 mol 2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is Pb. S(s) The expression for the solubility product is Pb 2+(aq) + S 2 -(aq) Ksp = [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 Ksp = [Pb 2+(aq)]2

CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, Pb. S is 4 x 10 -28 mol 2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is Pb. S(s) The expression for the solubility product is Pb 2+(aq) + S 2 -(aq) Ksp = [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 Ksp = Re-arranging; [Pb 2+(aq)] = Ksp = 4 x 10 -28 = [Pb 2+(aq)]2 2 x 10 -14 mol dm-3

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

CALCULATIONS Solubility products can be used to calculate the solubility of compounds. At 25°C the solubility product of lead(II) sulphide, Pb. S is 4 x 10 -28 mol 2 dm-6. Calculate the solubility of lead(II) sulphide. The equation for its solubility is Pb. S(s) The expression for the solubility product is Pb 2+(aq) + S 2 -(aq) Ksp = [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 Ksp = Re-arranging; [Pb 2+(aq)] = Ksp = 4 x 10 -28 = As you get one Pb 2+ from one Pb. S, the solubility of Pb. S Mr for Pb. S is 239; the solubility is 239 x 2 x 10 -14 g dm-3 [mass = moles x molar mass] [Pb 2+(aq)]2 2 x 10 -14 mol dm-3 = = 2 x 10 -14 mol dm-3 5. 78 x 10 -12 g dm-3

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.

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 molar mass = 5 x 10 -10 g dm-3 200

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 molar mass = = 5 x 10 -10 g dm-3 200 2. 5 x 10 -12 mol dm-3

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 molar mass = = The equation for its solubility is MY(s) 5 x 10 -10 g dm-3 200 2. 5 x 10 -12 mol dm-3 M+(aq) + Y-(aq)

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 molar mass = = The equation for its solubility is The expression for the solubility product is MY(s) Ksp = 5 x 10 -10 g dm-3 200 2. 5 x 10 -12 mol dm-3 M+(aq) + Y-(aq) [M+(aq)] [Y-(aq)]

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 molar mass = = The equation for its solubility is The expression for the solubility product is According to the equation; MY(s) Ksp = 5 x 10 -10 g dm-3 200 2. 5 x 10 -12 mol dm-3 M+(aq) + Y-(aq) [M+(aq)] [Y-(aq)] moles of M+(aq) = moles of Y-(aq) = moles of dissolved MY(s)

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 molar mass = = The equation for its solubility is The expression for the solubility product is According to the equation; MY(s) Ksp = 5 x 10 -10 g dm-3 200 2. 5 x 10 -12 mol dm-3 M+(aq) + Y-(aq) [M+(aq)] [Y-(aq)] moles of M+(aq) = moles of Y-(aq) = moles of dissolved MY(s) Substituting values Ksp = [2. 5 x 10 -12 ] The value of the solubility product Ksp = 6. 25 x 10 -24 mol 2 dm-6

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.

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) is a common ion as it is already in solution.

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) 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.

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 The solubility product at 25°C is Ag. Cl(s) Ag+(aq) + Cl¯(aq) Ksp = [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 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 The solubility product at 25°C is Ag. Cl(s) Ag+(aq) + Cl¯(aq) Ksp = [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 Ag. NO 3(aq) would produce the same effect as it would increase the concentration of Ag+(aq).

THE COMMON ION EFFECT If equal volumes of Ag. NO 3 (2 x 10 -5 mol dm-3) and Na. Cl (2 x 10 -5 mol dm-3) solutions are mixed, will Ag. Cl be precipitated? Ksp = 1. 2 x 10 -10 mol 2 dm-6

THE COMMON ION EFFECT If equal volumes of Ag. NO 3 (2 x 10 -5 mol dm-3) and Na. Cl (2 x 10 -5 mol dm-3) solutions are mixed, will Ag. Cl be precipitated? Ksp = 1. 2 x 10 -10 mol 2 dm-6 • in Ag. NO 3 the concentration of Ag+ is 2 x 10 -5 mol dm-3 in Na. Cl the concentration of Cl¯ is 2 x 10 -5 mol dm-3

THE COMMON ION EFFECT If equal volumes of Ag. NO 3 (2 x 10 -5 mol dm-3) and Na. Cl (2 x 10 -5 mol dm-3) solutions are mixed, will Ag. Cl be precipitated? Ksp = 1. 2 x 10 -10 mol 2 dm-6 • in Ag. NO 3 the concentration of Ag+ is 2 x 10 -5 mol dm-3 in Na. Cl 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

THE COMMON ION EFFECT If equal volumes of Ag. NO 3 (2 x 10 -5 mol dm-3) and Na. Cl (2 x 10 -5 mol dm-3) solutions are mixed, will Ag. Cl be precipitated? Ksp = 1. 2 x 10 -10 mol 2 dm-6 • in Ag. NO 3 the concentration of Ag+ is 2 x 10 -5 mol dm-3 in Na. Cl 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

THE COMMON ION EFFECT If equal volumes of Ag. NO 3 (2 x 10 -5 mol dm-3) and Na. Cl (2 x 10 -5 mol dm-3) solutions are mixed, will Ag. Cl be precipitated? Ksp = 1. 2 x 10 -10 mol 2 dm-6 • in Ag. NO 3 the concentration of Ag+ is 2 x 10 -5 mol dm-3 in Na. Cl 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 Ksp for Ag. Cl. . . NO PRECIPITATION OCCURS