The Solubility product is related to SPARINGLY SOLUBLE SALTS ONLY!!!!! Examples are silver chloride, lead(II) chloride, calcium sulphate (in other words, it deals with salts commonly called insoluble salts) Consider a sparingly soluble salt MX Imagine the reaction that occurs when the solid is placed in water aMX (s) + H2O [pic] aMn+ (aq) + bXm- (aq) The equilibrium expression would be [Mn+]a [Xm-]b [MX]a [H2O] However the concentrations of both the solid AND the water are constants, so a new equilibrium expression called Ksp is derived
Ksp = [Mn+]a [Xm-]b Note: The product of the concentration of ions of a sparingly soluble salt in water OTHER than at equilibrium is called the IONIC PRODUCT Example 1. AgCl [pic] Ag+ + Cl- Ksp = [Ag+][Cl-] Example 2 Cu(OH)2 [pic] Cu2+ + 2OH- Ksp = [Cu2+][OH-]2 Checkpoint A Write the Ksp expressions for the sparingly soluble salts below a. AgF b. HgCl2c. Cu(OH)2 Common ion effect The solubility of one salt is reduced by the presence of another salt having a common ion. For example consider the reaction AgCl (s) [pic] Ag+ (aq) + Cl- (aq)
The ions present in solution are Ag+ and Cl- from the salt silver chloride. Imagine a soluble salt such as ammonium chloride is added to the solution, with reference to Le Chatelier’s Principle, what would happen?? The addition of more chloride ions would increase the [Cl-] in the equilibrium shown above and the system would then try to decrease the [Cl-] by shifting the equilibrium to the LEFT. By shifting to the left, this causes an increase in the amount of solid silver chloride formed, i. e. the solubility of the silver chloride decreases in the presence of ammonium chloride.
What would be observed is an increased amount of precipitation when the salt ammonium chloride (or any salt containing a common ion) is added to a solution containing silver chloride. Checkpoint B Imagine the sparingly soluble salt PbSO4 is added to water a) First write the equilibrium reaction that would occur when the salt is added to water. b) Tick the salt(s) below which via the common ion effect would cause an increased amount of precipitation if it is added to the water already containing PbSO4 i) CaCO3 ii) Pb(NO3)2 iii) MgSO4 iv) Na2SO4 v) NaCl
How to calculate the solubility product of a sparingly soluble salt If the molar solubility of CaI2 at 25 °C is 7. 9 x 10-7 mol dm-3, what would be its Ksp value? Step 1 Write the equilibrium expression CaI2 [pic] Ca2+ + 2I- Step 2 Write the Ksp expression [Ca2+][I-]2 Step 3 Determine the # of moles of each ion present ratio of CaI2 : Ca2+ : I- = 1 : 1 : 2 therefore # of moles of Ca2+ = 7. 9 x 10-7 and # of moles of I- = 7. 9 x 10-7 x 2 = 1. 58 x 10-5 Step 4 Use the # of moles of each ion to determine the Ksp value Ksp = 7. x 10-7 x (1. 58 x 10-5)2 = 1. 97 x 10-16 To calculate the molar solubility from the Ksp value Ksp PbI2 is 1. 39 x 10-8. What is its molar solubility? Step 1 Write the equilibrium expression PbI2 [pic] Pb2+ + 2I- Step 2 Use the letter “a” to represent the # of mol of PbI2 then a moles of PbI2 gives a moles of Pb2+ and 2a moles of I- Step 3 Write the Ksp expression using the algebraic terms Ksp = [Pb2+][I-]2 = a * (2a)2 = 4a3 ( 1. 39 x 10-8 = 4a3 Step 4Solve for a 3v(1. 39 x 10-8) = 0. 0015 4
Step 5 Give molar solubility molar solubility of PbI2 is 0. 0015 mol dm-3 How to determine the decrease in solubility of a sparingly soluble salt via common ion effect If the molar solubility of CaI2 is 7. 9 x 10-7 mol dm-3 in water and then 0. 1M NaI is added to the solution, what would be the new solubility of the CaI2? Step 1 Write the Ksp value & expression of CaI2 (taken from previous example) Ksp = [Ca2+][I-]2 = 1. 97 x 10-16 Step 2 Substitute the concentration of the common ion with the value given
NaI is a soluble salt which dissociates FULLY therefore the concentration of I- = 0. 1 (ignore the I- ions coming from the sparingly soluble salt. ) Step 3 Determine the concentration of Ca2+ Ca2+ = Ksp = 1. 97 x 10-16 = 1. 97 x 10-14 [I-]2 0. 12 Step 4 Determine molar solubility ratio of Ca2+ : CaI2 = 1 : 1 therefore # of moles of CaI2 = molar solubility = 1. 97 x 10-14 mol dm-3 therefore the solubility of CaI2 decreases from 7. 9 x 10-7 to approx 2 x 10-14
Checkpoint C [pic] Checkpoint D [pic] Real life illustrations of solubility product Kidney stones are crystals of calcium oxalate. When the concentrations of calcium ions and oxalate ions are too high, the Ksp of calcium oxalate is exceeded and this causes the precipitation of calcium oxalate crystals. Note kidney stones are generally of the order 3 millimetres or more to cause obstruction of the urethra and eventually pain. By minimising the uptake of oxalate ions and to a lesser extent calcium ions, this reduces the risk of kidney stone formation.
Qualitative analysis also exploits the principle of solubility product of sparingly soluble salts. For a precipitate to be formed, the ionic product must exceed the solubility product. Then precipitation occurs until the ionic product no longer exceeds the Ksp value. Reagents used for qualitative analysis must be of a certain minimum concentration, if not, the ionic product would not be enough to equal the Ksp value and no precipitate would be seen and a wrong conclusion of the ion present would occur. ———————– That is a 40 MILLION fold decrease in solubility!!!