Predicting Precipitate Formation The Ksp value is the product of the upper limit of the product of the concentration of the soluble ions. When the product of the concentration of the ions exceed the value of Ksp they cannot exist in equilibrium anymore. They will form a precipitate in order to reduce the concentrations of the ions in solution back to the equilibrium value. To determine whether the Ksp is exceeded, just substitute the ion concentrations into an expression similar to the Ksp expression and determine an experimental Trial Ion Product. (T.I.P.) If the T.I.P. exceeds the Ksp then a precipitate forms. If the T.I.P. is larger than the values of Ksp then the ions are above the saturation line and the ions are in a super-saturated situation. The ions will precipitate until their individual concentrations multiplied together equal the value of Ksp. If the T.I.P. does not exceed the value of Ksp then the ions are in an unsaturated situation, which means there aren't enough ions dissolved to form a precipitate. If the T.I.P. equals the value of Ksp then the solution is saturated. The solution can hold no more dissolved ions without a precipitate forming. Problem Does a ppt. of AgCl form when 1 mL of 0.1 mol/L AgNO3 is added to a beaker containing 1 L of tap water with a Cl- ion concentration of 1.0 x 10-5 mol/L? [Cl-] = 1.0 x 10-5 mol/L [Ag+] calculation. Start with the given info, 1 mL of 0.1 mol/L Set up a ratio that states, if there are 0.1 moles of Ag+ in 1000 mL how many moles will be in 1 mL? 0.1 mol  =  x 1000mL   1 mL x = 0.0001 moles of Ag+ are present. (This is how many moles there are, this is not the concentration) The [Ag+] will be this many moles of Ag+ dissolved in the volume of solution in this question. Therefore [Ag+] =      0.0001 mole  = 0.0001 mole = 1.0 x 10-4 mol/L              1000 mL + 1 mL       1.001 L T.I.P. = [Ag+][Cl-]           = (1.0 x 10-4)(1.0 x 10-5)           = 1.0 x 10-9 The Ksp for AgCl is 1.8 x 10-10 therefore T.I.P > Ksp so a ppt. will form. Follow-up Problem Does a ppt of AgI form when 10 mL of 0.1 mol/L AgNO3 gets added to 90 mL of a solution containing 1.0 x 10-10 mol/L of KI? What is the highest concentration of I- ions that can exist in equilibrium with 0.01 mol/L Ag+ ions? How many grams of KI are present in 1 litre of a solution with the I- ion concentration found above? AgI(s) <====> Ag+(aq) + I-(aq) Ksp = [Ag+][I-] 8.3 x 10-17 = (1.0 x 10-2)[I-] There [I-] = 8.3 x 10-17 = 8.3 x 10-15 mol/L                    1.0 x 10-2 If [I-] exceeds 8.3 x 10-15 mol/L then a ppt. will form. K+ + I- ------> KI If we have 8.3 x 10-15 moles of I- we also have 8.3 x 10-15 moles of K+ and hence we also have 8.3 x 10-15 moles of KI. g = moles x molecular mass    = 8.3 x 10-15 mole x 166.04 g/mole    = 1.38 x 10-12 grams Follow up Problems 1. What is the highest concentration of Cl- ions that can exist in equilibrium with 0.001 mol/L Tl+ ions at 25oC? 2. Calculate the [CrO42-] required to begin precipitation of these metal ions from solutions containing 1 g of the metal ions per L of solution. a) Ag+ b) Ca2+ c) Pb2+
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