116 PLTL Activity sheet / Solubility Equilibrium Set 11

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1 Predicting Solubility Solubility problems are equilibrium problems. The reactant in a solubility equilibrium is a slightly soluble salt and the equilibrium constant for the reaction is the solubility product constant, K sp. Note that since the reactant is a solid, its concentration does not appear in the K sp expression. For example, the solubility equilibrium and K sp for the salt SrF 2 (s) are SrF 2 (s) Sr 2+ (aq) + 2 F (aq) K sp = [Sr 2+ ][F ] 2 = The molar solubility of a salt in water can be formed by setting up an equilibrium table and solving for x. The solubility is the same quantity expressed in g/l (rather than M = mol/l). Example 1: Calculate (a) the molar solubility and (b) the solubility of SrF 2 (s) in water. Solution: We first set up an equilibrium table: Balanced Equation SrF 2 (s) Sr 2+ (aq) + 2 F (aq) Initial Concentrations (M) c o 0 0 Change (M) - x x 2x Equilibrium Concentrations (M) (c o - x) x 2x To determine the solubility we use the equilibrium concentrations from the table and the K sp value given above, and solve for x: K sp = [Sr 2+ ][F ] 2 = (x)( 2x) 2 = 4x 3 = x 3 = x = = molar solubility To find the solubility we use the molar mass to convert the molar solubility to g/l: mol SrF g SrF 2 2 2? g / L = = SrF2 /L 1L 1molSrF 2 Predicting Precipitation Away from equilibrium, the ion product Q sp may be defined. As always, Q has the same form as K, but typically involves non-equilibrium concentrations. Precipitation can be predicted by comparing Q sp with K sp : Relationship Solution Type Result Q sp < K sp Unsaturated More salt can dissolve without ppt forming Q sp = K sp Saturated No more salt can dissolve Q sp > K sp Supersaturated Salt will precipitate until Q sp = K sp Chemical Separations If two precipitates are possible, the least soluble salt will precipitate first (the one with the smaller K sp ). For example, if you add NaOH(aq) to a solution containing equal amounts of Ca 2+ and Mg 2+, Mg(OH) 2 (s) will precipitate before Ca(OH) 2 (s) since K sp (Mg(OH) 2 ) = is less than 1

2 K sp (Ca(OH) 2 ) = The reason is that for Mg(OH) 2, Q sp > K sp will occur at lower [OH ] since the K sp is smaller. If the K sp values for two salts differ widely, the salts may be separated by fractional precipitation. Reduction in Solubility due to the Common Ion Effect Addition of an ion common to a solubility equilibrium will reduce solubility. This can be predicted in a qualitative way using Le Chatelier s Principle. For example, adding fluoride ion, F (aq), to the SrF 2 (s) equilibrium above will shift it left. The shift will increase the amount of SrF 2 (s) in solid form, and thus decrease solubility. The new solubility can be calculated as illustrated in the following example. Example 2. What would be the molar solubility of SrF 2 (s) in a 0.10 M NaF(aq) solution? Again we set up an equilibrium table, but now we have an initial concentration of fluoride ion: Balanced Equation SrF 2 (s) Sr 2+ (aq) + 2 F (aq) Initial Concentrations (M) c o Change (M) - x x 2x Equilibrium Concentrations (M) (c o - x) x ( x) We next use the equilibrium concentrations in the table and the K sp value given above, and solve for x. Note that since x is small, ( x) K sp = [Sr 2+ ][F ] 2 = (x)( x) 2 (x)( 0.10) 2 = x = 2.0 x x = = molar solubility Note that as predicted by the common ion effect, this solubility is much lower than what we calculated in Example 1 for pure water! Increase in Solubility due to addition of a Species that Reacts with an Ion The solubility of a salt will increase if a species is added which reacts with one of its ions. Once again this is an example of shifts predicted by Le Chatelier s Principle. For example if the anion reacts with the added substance, the concentration of the anion will be reduced. Thus the equilibrium will shift right to undo, in part, the disturbance. The shift right will reduce the amount of salt in solid form, and thereby increase its solubility. There are two common examples of this phenomena: (1) Reaction of the basic anion of a salt with a strong acid. Salts such as Mg(OH) 2 (s), BaCO 3 (s), and NiS(s) are much more soluble in a strong acid solution than in water. On the other hand, the solubility of AgCl(s) does not increase when 6 M HNO 3 is added. (2) Reaction of the acidic cation (Lewis acid) of a salt with a Lewis base to form a soluble complex ion. Salts such as AgCl(s) and Cu(OH) 2 (s) are much more soluble in 6 M NH 3 (aq) solution than in pure water due to the formation of Ag(NH 3 ) 2 + (aq) and Cu(NH 3 ) 4 2+ (aq) complex ions respectively. Likewise, amphoteric hydroxides such as Al(OH) 3 (s) and Zn(OH) 2 (s) are soluble in 6 M NaOH(aq) due to the formation of Al(OH) 4 - (aq) and Zn(OH) 4 2- (aq) ions. 2

3 Calcium chromate, CaCrO 4, K sp of What happens when calcium and chromate solutions are mixed to give M Ca 2+ 2 and M CrO 4 Q=6.00x10-4, no ppt since Q < Ksp What concentration of the lead ion, Pb 2+, must be exceeded to precipitate PbCl 2 from a solution that is M in the chloride ion, Cl? K sp for lead chloride is [Pb 2+ ] = M 3

4 AgI, BaF 2, and AgBr are all sparingly soluble salts. Which of these salts will be more soluble in an acidic solution than in water? As a chemist for an agricultural products company, you have just developed a new herbicide,"herbigon," that you think has the potential to kill weeds effectively. A sparingly soluble salt, Herbigon is dissolved in 1M acetic acid for technical reasons having to do with its production. You have determined that the solubility product Ksp of Herbigon is Although the formula of this new chemical is a trade secret, it can be revealed that the formula for Herbigon is X-acetate (XCH 3 COO, where "X" represents the top-secret cation of the salt). It is this cation that kills weeds. Since it is critical to have Herbigon dissolved (it won't kill weeds as a suspension), you are working on adjusting the ph so Herbigon will be soluble at the concentration needed to kill weeds. What ph must the solution have to yield a solution in which the concentration of X + is M? The pka of acetic acid is ph=2.10 4

5 Cu 2+ and Pb 2+ are both present in an aqueous solution. To precipitate one of the ions and leave the other in solution, add a. H 2 S (aq) b. H 2 SO 4 (aq) c. HNO 3 (aq) d. NH 4 NO 3 (aq) What is the minimum ph at which Cd(OH) 2 will precipitate from a solution that is M in Cd 2+ (aq)? ph=8.32 5

6 What [I ] should be maintained in KI(aq) to produce a solubility of Mol PbI 2 /L when PbI 2 is added? [I ] = M 6