Solubility Equilibria These are associated with ionic solids dissolving in water to form aqueous solutions Chapter 16 Solubility Equilibria It is assumed that when an ionic compound dissolves in water, it dissociates completely into hydrated ions: A a B b (s) aa + (aq) + bb - (aq) Initially, when the solid is added to the water, there are no ions present. However, as the dissolution proceeds, the ion concentrations build up and the reverse reaction starts to re-form the solid: aa + (aq) + bb - (aq) A a B b (s) Eventually, an equilibrium is reached when the solution becomes saturated and no more solid can dissolve: A a B b (s) aa + (aq) + bb - (aq) KCl(s) K + (aq) + Cl - (aq) Solubility Product (Constant), K sp For the general equilibrium: A a B b (s) aa + (aq) + bb - (aq) We can write the equilibrium expression using the Law of Mass Action: K sp = [A + ] a [B - ] b where [A + ] and [B - ] are expressed in moll -1 Since A a B b (s) is a pure solid it is not included in the equilibrium expression It might be thought that the amount of excess solid would affect the position of the equilibrium since more solid means more surface area exposed to the solvent. However, it does not since even though the rate of dissolution increases with increasing surface area so does the rate of re-formation of the solid Solubility and the Solubility Product Solubility and the Solubility Product are not the same thing! The Solubility Product is an equilibrium constant and has only one value for a given solid at a given temperature Solubility is an equilibrium position at a given temperature and can vary with the presence of a common ion In all cases the product of ion concentrations must satisfy the K sp expression 1
Calculating K sp from Solubility Calculating Solubility from K sp 2
Shortcut For the general solubility equilibrium: A x B y (s) xa(aq) + yb(aq) K sp = [A] x [B] y The solubility, S is given by the expression: S x y K sp y x x. y Relative Solubilities An ionic compound s K sp value gives information about its solubility. However, we must be careful when using K sp values to predict relative solubilities since we need to take into account the different numbers of ions produced Relative solubilities can only be predicted using K sp values directly in the case of ionic compounds that produce the same number of ions: Compounds that produces more ions can sometimes be more soluble than a compounds that produces fewer ions, even though their K sp values are smaller: Salt K sp Calculated Solubility (moll -1 ) AgI(s) 1.5 x 10-16 1.2 x 10-8 CuI(s) 5.0 x 10-12 2.2 x 10-6 CaSO 4 (s) 6.1 x 10-5 7.8 x 10-3 3
The Common Ion Effect and Solubility The solubility of a solid is lowered if the solution contains ions common to the solid since the presence of the common ion shifts the solubility equilibrium to the left ph and Solubility Solution ph can greatly affect the solubility of an ionic compound Example: Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) Addition of OH - decreases the solubility since the equilibrium shifts to the left Addition of H + increases solubility since OH - is removed and the equilibrium shifts to the right Example: Ag 3 PO 4 (s) 3Ag + (aq) + PO 4 3- (aq) Addition of H + increases solubility since it reacts with the strong base PO 4 3- forming the HPO 4 2- and shifting the equilibrium to the right 4
In general, if the anion, X -, in an ionic compound is a strong base (i.e. HX is a weak acid) then its salt MX will show increased solubility in an acidic solution Examples: OH -, S 2-, CO 3 2-, C 2 O 4 2- and CrO 4 - In general, if the anion, X -, in an ionic compound is a weak base (i.e. HX is a strong acid) then its salt MX will show no change in solubility in an acidic solution Examples: Ag 3 PO 4 CaCO 3 Hg 2 Cl 2 PbI 2 CdCO 3 Sr 3 (PO 4 ) 2 Cl -, Br -, I -, NO 3- and ClO 4 - Precipitation and Qualitative Analysis When solutions of two ionic compounds are mixed it is possible to use K sp to predict whether a precipitate will form The ion product, Q of an ionic compound is defined like its K sp expression, except that initial rather than equilibrium concentrations are used Example, for CaF 2 : Q = [Ca 2+ ] 0 [F - ] 0 2 If we add a solution containing Ca 2+ ions to a solution containing F - ions: A precipitation will occur if Q > K sp and will continue until the concentrations are reduced to the point that Q = K sp A precipitation will not occur if Q < K sp 5
It is also possible to calculate the equilibrium concentrations in the solution after precipitation occurs Procedure: 1. Find initial concentrations 2. Calculate Q 3. Compare Q and K sp to see if precipitation will occur 4. Run reaction to completion 5. Calculate concentration of excess ion 6. Determine concentrations of ions at equilibrium Selective Precipitation Mixtures of metal ions in solution are often separated by selective precipitation by using a reagent whose anion forms a precipitate with one of the metal ions in the mixture Example: If NaCl is added to a solution containing both Ag + and Ba 2+ ions, the Ag + will selectively precipitate out of solution as insoluble AgCl while the soluble BaCl 2 will stay in solution 6
Qualitative Analysis Qualitative analysis of a mixture of involves first separating them into five major groups (these groups are not based on groups in the periodic table) based on solubilities. Each group is then treated further to separate and identify the individual ions 7