Chapter 17: Solubility & Complex Ion Equilibria

Transcription

1 Chapter 17: Solubility & Complex Ion Equilibria Solubility Equilibria equilibrium between a solid and its ions in solution ex. for calcium phosphate Ca3(PO4)2 (s) 3 Ca2+ (aq) + 2 PO43 (aq) this is a heterogeneous equilibrium equilibrium constant: Ksp solubility product constant for calcium phosphate: Ksp = [Ca2+]3[PO43 ]2 we can now refine our understanding of solubility rules from Chapter 4 solids that we classified as insoluble typically have small Ksp s and low molar solubilities slightly soluble or sparingly soluble Overview of the Chapter The goal of this chapter is to understand the equilibria that exist between ionic solids and their ions in solution, and factors that affect that equilibrium. write (heterogeneous) equilibrium equations & K expressions calculate and interpret Ksp Ksp is the solubility product constant using Ksp, calculate solubility of salts mol/l (molar solubility), and g/l factors that affect solubility common ion effect ph formation of complex ions calculations to determine whether precipitation of a solid will occur when 2 solutions are combined

2 Some Solution Terminology the solubility equilibrium can also be referred to as the dissolution precipitation equilibrium dissolution Ca3(PO4)2 (s) 3 Ca 2+ (aq) + 2 PO4 3 (aq) precipitation sol n may be saturated, unsaturated, or supersaturated unsaturated sol n: more solid can dissolve reaction continues in forward direction toward equilibrium (Q < Ksp) supersaturated sol n: ion [ ] s are too high; solid will precipitate out of the solution reaction continues in reverse direction toward equilibrium (Q > Ksp) saturated sol n: solution is at equilibrium (Q = Ksp) Ion Concentrations in a Saturated Solution consider 2 different preparations of a saturated sol n of CaF2 examine the relationships between [Ca 2+ ], [F ], Ksp CaF2 (s) Ca 2+ (aq) + 2 F (aq) Ksp = [Ca 2+ ][F ] 2 preparation 1: put solid CaF2 in a flask; add water; wait until equilibrium is established at equilibrium: [Ca 2+ ] = 2.0 x 10 4 M [F ] = 4.0 x 10 4 M Calculate Ksp. note: with this preparation, [F ] = 2 [Ca 2+ ] Why? Ion Concentrations in a Saturated Solution examine the relationships between [Ca 2+ ], [F ], KSP CaF2 (s) Ca 2+ (aq) + 2 F (aq) Ksp = [Ca 2+ ][F ] 2 preparation 2: a sol n with containing Ca 2+ is mixed with a solution containing F ; after some time, equilibrium is established at equilibrium: [Ca 2+ ] = M Ksp = 3.2 x Determine equilibrium [F ]. note: with this preparation, [F ] 2 [Ca 2+ ] Why not? example: A saturated solution of silver chromate is prepared by dissolving solid Ag2CrO4 in water, and allowing the solution to reach equilibrium. The saturated solution has [Ag + ] = 1.3 x 10 4 M. Determine [CrO4 2 ] and Ksp. Ag2CrO4 (s) 2 Ag + (aq) + CrO4 2 (aq) initial [ ] [ ] x + x equil [ ] --- 2x M x M

3 example: Using Ksp to Determine Solubility Calculate the solubility (in mol/l and g/l) of nickel (II) sulfide in water at 25 C. For NiS, Ksp = 3.0 x define: x = mol of solid that dissolve per L of sol n x is the molar solubility of the salt; units mol/l NiS (s) Ni 2+ (aq) + S 2 (aq) initial [ ] [ ] x + x equil [ ] --- x M x M Comparing Solubilities of Salts salts with greater solubility have higher [ions] in saturated solution solubility is related to equilibrium position the farther to the right the equilibrium position, the greater the solubility be careful about comparing Ksp s directly to determine relative solubilities of 2 salts... must consider the salt stoichiometry and relationship between Ksp & x Comparing Solubilities of Salts Which has greater solubility in water at 25 C? PbCl2 or PbF2 for PbCl2 Ksp = 1.6 x 10 5 for PbF2 Ksp = 4.0 x 10 8 Which has greater solubility in water at 25 C? PbCl2 or CaSO4 for PbCl2 Ksp = 1.6 x 10 5 for CaSO4 Ksp = 6.1 x 10 5 Factors that Affect Solubility the common ion effect The molar solubility of MgF2 in water at 25 C is 2.6 x 10 4 mol/l. Determine the molar solubility of MgF2 in 0.10 M NaF (aq). For MgF2, Ksp = 7.4 x MgF2 (s) Mg 2+ (aq) + 2 F (aq) initial [ ] [ ] x + 2x equil [ ] --- x M ( x) M solve x; x = molar solubility = 7.4 x 10 9 mol/l the presence of a common ion (here F ) reduces the solubility of a salt

4 example: Determine the solubility of lead (II) hydroxide in a solution with ph = For Pb(OH)2, Ksp = 1.2 x sol n with ph = has [OH ] = 1.0 x 10 4 M ph of solution Factors that Affect Solubility In general, for an ionic compound with a basic anion, solubility will increase as the ph of the solution decreases. H + present reacts with the basic anion [anion] decreases Le Chatelier s principle predicts that equilibrium will shift to the right (in the direction of greater dissolution of solid). example: CaCO3 (s) Ca 2+ (aq) + CO3 2 (aq) calcium carbonate will be more soluble in acidic sol n because the following reaction results in decreased [CO3 2 ]: CO3 2 (aq) + H + (aq) HCO3 (aq) some common examples of basic anions: CO3 2, OH, PO4 3, SO4 2, C2O4 2, CN, F, S 2 remember that the following anions are neutral: Cl, Br, I, NO3, ClO4 the solubility of salts with these anions is not affected by lowering the ph of the solution complex ion formation Factors that Affect Solubility The solubility of an ionic compound may increase dramatically if a solution containing a Lewis base is added. added Lewis base may react with a metal cation to form a Lewis acid-base adduct called a complex ion formation of complex ion is an equilibrium with equilibrium constant, Kf formation constant

5 example: AgCl has very limited solubility in water and acidic solution, but will dissolve in NH3(aq): AgCl (s) Ag+ (aq) + Cl (aq) Ksp = 1.6 x Ag+ (aq) + 2 NH3 (aq) [Ag(NH3)2]+ (aq) Kf = 1.7 x 107 AgCl (s) + 2 NH3 (aq) [Ag(NH3)2]+ (aq) + Cl (aq) KC = Ag+ ions in solution react with NH3 to form the complex ion [Ag(NH3)2]+ as the complex ion forms, [Ag+] in sol n decreases as [Ag+] decreases, the solubility equilibrium position to shifts to the right more AgCl dissolves another example of complex ion formation and effect on solubility: NH3 (aq) added to a solution of CuSO4 (aq) initially a precipitate of Cu(OH)2 (s) forms as solution becomes basic then the dissolution of Cu(OH)2 is observed after further addition of NH3 (aq) as Cu(NH3)22+ forms Zn(OH)2 (s)! Zn2+ (aq) + 2 OH- (aq); Ksp = 2.1 x Zn2+ (aq) + 4 OH- (aq)! Zn(OH)42-(aq); Kf = 2.8 x 1015 Zn(OH)2 (s) + 2 OH- (aq)! Zn(OH)42- (aq); KC = 0.59

6 Calculation of Solubility After Complex Ion Formation example (see example in text): Calculate the molar solubility of AgBr in 2.25 M Na2S2O3 (aq). For AgBr, Ksp = 5.0 x ; for the complex ion [Ag(S2O3)2] 3, Kf = 2.9 x Calculation of Solubility After Complex Ion Formation example (see example in text): Calculate the molar solubility of CuI in 0.88 M KCN (aq). For CuI, Ksp = 1.1 x ; for the complex ion [Cu(CN)2], Kf = 1.0 x Precipitation of Ionic Solids For an ionic solid, MX the solubility equilibrium is given by: MX (s) M n+ (aq) + X n (aq) When 2 solutions are combined one sol n containing M n+, and one sol n containing X n will a precipitate of MX form? Is the system at equilibrium? If not, in what direction does the reaction proceed to reach equilibrium? Q vs K calculation Precipitation of Ionic Solids determine concentrations of ions after solutions are combined: [ion] = mol ion total sol n volume calculate the ion product, Q compare Q to Ksp: if Q = Ksp, the solution is saturated; solution is at equilibrium if Q > Ksp, the solution is supersaturated; precipitation of solid will occur if Q < Ksp, the solution is unsaturated; precipitation will not occur - more solid will dissolve

7 example: ml of M AgNO3 (aq) and ml of M Na2CO3 are combined. Will a precipitate of Ag2CO3 form? For Ag2CO3, Ksp = 8.1 x Ag2CO3 (s) 2 Ag + (aq) + CO3 2 (aq) Q = [Ag + ] 2 [CO3 2 ] example: Determine the minimum concentration of carbonate ion required to cause the precipitation of silver carbonate from a 5.8 x 10 4 M solution of AgNO3. For Ag2CO3, Ksp = 8.1 x Ag2CO3 (s) 2 Ag + (aq) + CO3 2 (aq) Ksp = [Ag + ] 2 [CO3 2 ] solve for [CO3 2 ] present in a saturated solution of Ag2CO3; [CO3 2 ] at equilibrium OR when Q = Ksp any greater [CO3 2 ] will result in Q > Ksp and precipitation of solid