are the common ions from Ag 2 CrO 4. Also [Ag + ] must be twice [Ag 2 CrO 4 ] from the balanced dissolution equation:

Transcription

1 Problem 16.9 A saturated solution of Ag 2 CrO 4 prepared by dissolving Ag 2 CrO 4 solid (whereby there is excess solid remaining) has [CrO 4 ] = 6.5 x 10-5 M. Calculate K sp for Ag 2 CrO 4. Ag + and CrO 4 are the common ions from Ag 2 CrO 4. Also [Ag + ] must be twice [Ag 2 CrO 4 ] from the balanced dissolution equation: Ag 2 CrO 4 (s) 2 Ag + + CrO 4 [Ag + ] = 2 x [CrO 4 ] = (2) (6.6 x 10-5 ) = 1.3 x 10-4 M K sp = [Ag + ] 2 [CrO 4 ] = (1.3 x 10-4 ) 2 (6.5 x 10-5 ) = 1.1 x Problem A) Which has the greater molar solubility: AgCl with K sp = 1.8 x 10-10, or Ag 2 CrO 4 with K sp = 1.1 x 10-12? B) Which has the greater solubility in grams per litre? A) We dissolve an amount of each solid in water to create saturated solutions of each. For AgCl, we know that x moles of AgCl will dissolve to give [Ag + ] = [Cl - ] = x. The molar solubility (x) of AgCl can be found by: K sp = [Ag + ][Cl - ] = x 2 x 2 = 1.8 x x = 1.8 x x = 1.3 x 10-5 mol/l For the Ag 2 CrO 4, we know that x moles of Ag 2 CrO 4 will dissolve to give [Ag + ] = 2x and [CrO 4 ] = x. The molar solubility (x) of Ag 2 CrO 4 can be found by: 1

2 Problem continued K sp = [Ag + ] 2 [CrO 4 ] = (2x) 2 (x) = (4x 2 ) (x) = 4x 3 x 3 = 1.1 x / 4 x = x x = 6.5 x 10-5 mol/l The molar solubility of Ag 2 CrO 4 is greater. B) The gram solubility (g/l) of any solid is the molar mass (g/mol) times the molar solubility (mol/l), so: gram solubility AgCl = g/mol x 1.3 x 10-5 mol/l gram solubility AgCl = 1.9 x 10-3 g/l gram solubility Ag 2 CrO 4 = g/mol x 6.5 x 10-5 mol/l gram solubility Ag 2 CrO 4 = 2.2 x 10-2 g/l The gram solubility of Ag 2 CrO 4 is greater Factors that Affect Solubility The Common-Ion Effect From an equilibrium point of view, we saw the dissolution of a general solid is: M m X x (s) m M n+ (aq) + x X y- (aq) K sp = [M n+ ] m [X y- ] x So, from Le Chatelier s Principle, we know if the solution already has a non-zero concentration of one of the ions in it when we add the solid, the stress on the dissolution reaction is this added ion concentration. To relieve the stress the equilibrium will shift towards the reactants (the solid), meaning less solid will dissolve than ordinarily would happen. The common ion effect can be illustrated by this Figure. The solubility of MgF 2 decreases on addition of a common ion F - 2

3 Problem Calculate the molar solubility of MgF 2 in 0.10 mol/l MgCl 2 at 25 C given that K sp for MgF 2 is 7.4 x In 0.10 mol/l MgCl 2, we can assume the molar solubility of MgF 2 is x. At equilibrium then, there will be 2x moles of F -, but there must be x moles of Mg 2+ because there are 0.10 moles of Mg 2+ already present from MgCl 2. Set up an equilibrium table: MgF 2 (s) Mg F - Equil [ ] x 2x Now K sp = 7.4 x = [Mg 2+ ] [F - ] 2 = ( x) (2x) 2 - because K sp is so small, then x is very small and x can be approx. to x = (0.10) (2x) 2 x 2 = 7.4 x / 0.4 x = x x = 1.4 x 10-5 mol/l The molar solubility of MgF 2 has decreases to 1.4 x 10-5 because of the presence of the common ion Mg 2+. The ph of a solution If a solid dissolved to give a basic anion in solution, addition of acid will increase the solubility of the solid. This can be explained by Le Chatelier s Principle. Consider calcium carbonate, CaCO 3 CaCO 3 (s) Ca 2+ (aq) + CO 3 (aq) K sp = 5.0 x 10-9 Carbonate, CO 3, is a basic anion that will react with a proton to give HCO 3 - CO 3 (aq) + H 3 O + (aq) HCO 3- (aq) + H 2 O (l) This reaction will shift to the right (products) as the ph becomes more acidic. This shift in the equilibrium is accompanied by a decrease in carbonate. Since this decrease in carbonate is happening in the same container as the dissolution reaction, the dissolution reaction must shift to the right as well to compensate for the loss of carbonate. This means more solid will dissolve as we add more acid. 3

4 Problem Which of the following compounds are more soluble in acidic solution than in pure water? a) AgCN Since the anion CN - is a basic anion, this solid is more soluble in acidic solution. b) PbI 2 I - is a neutral anion, and so the solid is not more soluble in acid. c) Al(OH) 3 Since the anion OH - is a basic anion, this solid is more soluble in acidic solution. d) ZnS Since the anion S is a basic anion, this solid is more soluble in acidic solution. Formation of complex ions Solubility of a solid increases if there is, in the solution, a Lewis base that is capable of forming a covalent bond with the metal cation of the solid to form a complex ion. An example of a complex ion is Ag(NH 3 ) 2+ where two ammonia molecules act as Lewis bases to the Lewis acid Ag + ion to form the complex. Such complexes affect solubility in the exact same way that ph does, by reducing in solution the concentration of the cation so that the dissolution reaction must shift to the products to replace the cation concentration to reestablish equilibrium. Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) + 2 (aq) K f = 1.7 x 10 7 (note how large K f is ) (formation constant) AgCl (s) Ag + (aq) + Cl - (aq) K sp = 1.8 x

5 Therefore, in the presence of ammonia, the dissolution of AgCl can be expressed by the sum of these two reactions Ag + (aq) + 2 NH 3 (aq) Ag(NH 3 ) 2+ (aq) K f =1.7 x AgCl (s) Ag + (aq) + Cl - (aq) K sp = 1.8 x Ag + (aq) + 2 NH 3 (aq) + AgCl (s) Ag(NH 3 ) 2+ (aq) + Ag + (aq) + Cl - (aq) or 2 NH 3 (aq) + AgCl (s) Ag(NH 3 ) 2+ (aq) + Cl - (aq) K = K f x K sp = (1.7 x 10 7 ) (1.8 x ) = 3.1 x 10-3 We see the dissolution of AgCl occurs much more completely in the presence of ammonia (3.1 x 10-3 ) than it does in pure water (1.8 x ). Amphoterism Will not be covered in the course this year Precipitation of Ionic Compounds Now, when we mix two solutions, the system is most likely not at equilibrium. With other reactions, we saw we could calculate the reaction quotient Q c to see if the system was at equilibrium, and which direction the reaction would proceed in. For solid dissolution / precipitation reactions, we use the same procedure to define the ion product (IP), which, like Q c, has the exact same form as the equilibrium constant expression (K sp in this case) it is compared to. So for a solution containing Ca 2+ and F - IP = [Ca 2+ ][ F - ] 2 When we mix the solutions, then, we can calculate IP and determine what will happen. If IP = Ksp, the solution is saturated If IP > Ksp, the solution is supersaturated, and equilibrium does not exist Precipitation to occur. If IP < Ksp, the solution is unsaturated, equilibrium does not exist. So the reaction shifts from the solid towards the ions 5

6 Problem Will a precipitate form on mixing equal volumes of the following solutions? First note we are mixing equal volumes. Since the final volume after mixing is doubled, all concentrations will need to be halved! A) 100 ml each of 3.0 x 10-3 mol/l BaCl 2 and of 2.0 x 10-3 mol/l Na 2 CO 3 K sp ( BaCO 3 ) = 2.6 x 10-9 Solids will form from the cations of one compound, and the anions from the other, so there are two possible solids, BaCO 3, and NaCl. We know NaCl is very soluble in water, so we want to see if BaCO 3 will precipitate. IP = [Ba 2+ ][CO 3 ] The concentrations of these ions right after mixing are [Ba 2+ ] = (3.0 x 10-3 mol/l / 2) = 1.5 x 10-3 mol/l, and [CO 3 ] = (2.0 x 10-3 mol/l / 2) = 1.0 x 10-3 mol/l, so IP = (1.5 x 10-3 ) (1.0 x 10-3 ) = 1.5 x 10-6 K sp of BaCO 3 is 2.6 x 10-9, and IP > K sp, so we expect a precipitate to form. Problem X What [OH - ] would be required to initiate precipitation of a Fe(III) solution containing 5.0 x 10-6 M FeCl 3? Given: K sp (Fe(OH) 3 = 3 x To just initiate precipitation IP = K sp IP=K sp = [Fe 3+ ] [OH - ] 3 [OH - ] 3 = IP / [Fe 3+ ] [OH - ] = (IP / [Fe 3+ ]) 1/3 = (3.0 x10-39 / 5.0x10-6 ) 1/3 = 8 x M Finally If more than one species can be precipitated from solution on addition of the appropriate reagent, the species for which the Ksp is exceeded precipitates and the others remain in solution 6

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