Ch 16 Lecture 3 Solubility Equilibria I.Solubility Basics A.The Solubility Equilibrium 1)Dissolution of an ionic compound is an equilibrium process a)CaF 2 (s) Ca 2+ (aq) + 2 F - (aq) b)K sp = Solubility Product = [Ca 2+ ][F - ] 2 2)Remember, neither solids nor pure liquids (water) effect the equilibrium constant a)Dissolving and Reforming change proportionately to the amount of solid b)Solvent water is at such a high concentration as not to be effected 3)The Solubility Product is an equilibrium constant, so it has only one value at a given temperature 4)Solubility = the equilibrium position for a given set of conditions a)There are many different conditions that all must obey K sp b)Common ions effect the solubility much as they effect pH

5)K sp values of some slightly soluble ionic solids a)Most NO 3 - salts are soluble b)Most alkali metal and NH 4 + salts are soluble c)Most Cl -, Br -, and I - salts are soluble (except: Ag +, Pb 2+, and Hg 2 2+ ) d)Most SO 4 2- salts are soluble (except Ba 2+, Pb 2+, Hg 2 2+, and Ca 2+ ) e)Most OH- salts are insoluble (except NaOH, KOH) f)Most S 2-, CO 3 2-, CrO 4 2-, and PO 4 3- salts are insoluble Note: These values may differ from the ones in your text. Use the values from your text for all homework problems

6)Example: What is the K sp of a CuBr solution with a solubility of 2.0 x M? CuBr (s) Cu + (aq) + Br - (aq) K sp = [Cu + ][Br-] = [2.0 x ][2.0 x ] = 4 x )Example: K sp = ? for Bi 2 S 3 with solubility of 1.0 x M 8)Example: Find the solubility of Cu(IO 3 ) 2 (K sp = 1.4 x ) Cu (IO 3 ) 2 (s) Cu 2+ (aq) + 2 IO 3 - (aq) K sp = [Cu 2+ ][IO 3 - ] 2 = 1.4 x (x)(2x) 2 = 4x 3 = 1.4 x x = 3.3 x M B.Relative Solubilities 1)If the salts being compared produce the same number of ions, we can compare solubilities by comparing K sp values AgIK sp = 1.5 x CuIK sp = 5.0 x CaSO 4 K sp = 6.1 x K sp = [X n+ ] 1 [Y n- ] 1 for all of these, so we can directly compare them Solubility of CaSO 4 > solubility of CuI > solubility of AgI

2)If the salts being compared produce different numbers of ions, we must calculate the actual solubility values; we can’t use K sp values to compare. a)CuS (8.5 x ) > Ag 2 S (1.6 x ) > Bi 2 S 3 (1.1 x ) by K sp alone 2 ions 3 ions 5 ions b)Bi 2 S 3 (1.0 x ) > Ag 2 S (3.4 x ) > CuS (9.2 x ) in solubility C.The Common Ion Effect 1)Common Ion = any ion in the solid we are trying to dissolve that is present in solution from another source. 2)What is the solubility of Ag 2 CrO 4 (K sp = 9 x ) in 0.1 M AgNO 3 ? Ag 2 CrO 4 (s) 2 Ag + (aq) + CrO 4 2- (aq) Init Equil x x K sp = 9 x = [ x] 2 [x] ~~ (0.1) 2 (x) x = 9 x / 0.01 = 9 x = solubility 5% rule: 9 x / 0.1 < 5%, so the approximation is ok Solubility in pure water is 1.3 x M. Why does the solubility decrease?

3)Example: Find solubility of CaF 2 (K sp = 4.0 x ) in M NaF D.pH and Solubility 1)Many salts contain hydroxide anion: Mg(OH) 2 Mg OH - a)High pH means large OH - common ion concentration b)[OH - ] would shift the equilibrium to the left c)[H + ] would shift the equilibrium to the right by using up OH - ions 2)Any Basic Anion will be effected by pH a)OH -, S 2-, CO 3 2-, C 2 O 4 2-, CrO 4 2-, and PO 4 3- are all basic anions b)H + will increase the solubility of their salts by removing the anions c)Ag 3 PO 4 (s) 3 Ag + (aq) + PO 4 3- (aq) PO H + HPO )Acidic pH has no effect on non-basic anions or on most cations: Cl -, Br -, NO 3 - a)AgCl (s) Ag + (aq) + Cl - (aq) b)H + doesn’t react with either ion

II.Precipitation and Qualitative Analysis A.We can use the solubility product to predict precipitation 1)If Q > K sp, precipitation occurs until K sp is reached 2)If Q < K sp, no precipitation will occur 3)Example: 750 ml M Ce(NO 3 ) 3 is added to 300 ml 0.02 M KIO 3. Will Ce(IO 3 ) 3 (K sp = 1.9 x ) precipitate? K sp = [Ce 3+ ][IO 3 - ] 3 We need to know concentrations. Q = (2.86 x )(5.71 x ) 3 = 5.32 x > K sp Precipitate 4)We can also calculate the equilibrium concentrations after precipitation. a)Examine the stoichiometry of the precipitation reaction allowed to go to completion b)Becomes a K sp problem with a common ion (ion in excess)

3)Calculate the equilibrium concentrations after precipitation when 100 ml of 0.05 M Pb(NO 3 ) 2 is added to 200 ml 0.10 M NaI. K sp for PbI 2 = 1.4 x 10 –8. a)PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) K sp = [Pb 2+ ][I - ] 2 b)[Pb 2+ ] = 1.67 x 10 -2, [I - ] = 6.67 x 10 -2, Q = 7.43 x > K sp precipitate c)Stoichiometry: Pb I - PbI 2 Initial5mmol20 mmol---- Completion0 10 mmol---- d)Equilibrium: some PbI 2 redissolves, with I - common ion present i.[I - ] common ion = 10mmol / 300ml = M PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) K sp = [Pb 2+ ][I - ] 2 Initial M Equilibrium---- x x i.K sp = 1.4 x 10 –8 = [Pb 2 + ][I - ] 2 = (x)( x) 2 ~~(x)(0.033) 2 x = 1.3 x M = [Pb 2 + ], [I - ] = M 4)Example: 150 ml 0.01 M Mg(NO 3 ) ml 0.1 M NaF. Find [Mg 2+ ] and [F - ] at equilibrium. K sp for MgF 2 = 6.4 x 10 -9

B.Qualitative Analysis 1.Selective Precipitation = addition of an ion that causes only one of a mixture of ions to precipitate Add NaCl to Ag + (aq) + Ba 2+ (aq) AgCl(s) + Na + (aq) + Ba 2+ (aq) 2.Example: Which ion will precipitate first? I - is added to a solution of M Cu + and M Pb 2+ ? K sp CuI = 5.3 x K sp PbI 2 = 1.4 x a)Use K sp CuI to find [I - ] that will just start precipitation b)Use K sp PbI 2 to find [I - ] that will just start precipitation c)Whichever has the lowest [I - ] will precipitate first 3.Sulfide Ion (S 2- ) is particularly useful for selective cation precipitation a)Metal sulfides have very different solubilities i.FeS K sp = 2.3 x ii.MnS K sp = 3.7 x iii.Mn would precipitate first from an equal mixture of Fe 2+ and Mn 2+ b)[S 2- ] can be controlled by pH i.H 2 S H + + HS - K a1 = 1 x ii.HS - H + + S 2- K a2 = 1 x

iii.S 2- is quite basic. At low pH, there will be very little S 2- iv.At high pH, there is much more S 2- c)We can selectively precipitate metal ions by adding S 2- in acidic solution, and then slowly adding base. CuSK sp = 8.5 x HgSK sp = 1.6 x MnSK sp = 2.3 x NiSK sp = 3.0 x Hg, then Cu, then Ni, then Mn would precipitate as we raise pH Solution of Mn 2+, Ni 2+, Cu 2+, Hg 2+ Solution of Mn 2+, Ni 2+ Precip. Of MnS, NiS Precipitate of CuS, HgS Add H 2 S, pH = 2 Add OH - to pH = 8

4)Qualitative Analysis = scheme to separate and identify mixtures of cations by precipitation a)Group I Insoluble Chlorides: Add HCl. AgCl, PbCl 2, Hg 2 Cl 2 precipitate b)Group II Sulfides Insoluble in Acid Solution: Add H 2 S. Low [S 2- ] due to [H + ]. HgS, CdS, Bi 2 S 3, CuS, and SnS 2 precip. c)Group III Sulfides Insoluble in Basic Solution: Add OH -. CoS, ZnS, MnS, NiS, FeS, Cr(OH) 3, and Al(OH) 3 precipitate. d)Group IV Insoluble Carbonates: Add CO BaCO 3, CaCO 3, and MgCO 3 precipitate. e)Group V Alkali Metal and Ammonium These ions (Na +, K +, NH 4 + ) are soluble with common ions f)Further tests would tell us which specific ions in each group are present

C.Complex Ion Equilibria 1)A complex ion = Lewis Acid—Lewis Base complex of a metal ion a)Ligand = generic name for the Lewis Base bonded to a metal ion b)H 2 O, NH 3, Cl -, CN - all have lone pairs to donate; can be ligands 2)Coordination Number (CN) = the number of ligands bonded to a metal ion a)CN = 6 is common: Co(H 2 O) 6 2+, Ni(NH 3 ) 6 2+ b)CN = 4 CoCl 4 2-, Cu(NH 3 ) 4 2+ c)CN = 2 Ag(NH 3 ) 2 + d)Coordination number depends on size and properties of the metal ion 3)Formation (or Stability) Constants a)Ligand addition to the metal ion is stepwise Ag + + NH 3 Ag(NH 3 ) + K 1 = 2.1 x 10 3 Ag(NH 3 ) + + NH 3 Ag(NH 3 ) 2 + K 2 = 8.2 x 10 3 Ag + + 2NH 3 Ag(NH 3 ) 2 + K f = 1.7 x 10 7 b)Usually, [Ligand] >> [M n+ ] to force the equilibria to the right c)What are the equilibrium conditions of 100 ml 2.0 M NH 3 added to 100 ml of M AgNO 3 ? i.K 1 x K 2 = K f is large, favoring complete reaction ii.[NH 3 ] is much larger than [Ag + ], so assume constant

iii.Do stoichiometry of the reaction: Ag + +2 NH 3 Ag(NH 3 ) 2 + Initial 5 x M1 M 0 Equilib. 01 M 5 x M iv.Then use equilibrium expressions to find concentrations

4)Example: Find [Ag + ], [Ag(S 2 O 3 ) - ], [Ag(S 2 O 3 ) 2 3- ] for 150 ml M AgNO ml 5.0 M Na 2 S 2 O 3. K 1 = 7.4 x 10 8, K 2 = 3.9 x )Complex Ions and Solubility a)How do we dissolve AgCl(s)? K sp = 1.6 x Ag + + NH 3 Ag(NH 3 ) + K 1 = 2.1 x 10 3 Ag(NH 3 ) + + NH 3 Ag(NH 3 ) 2+ K 2 = 8.2 x 10 3 b)Adding NH 3 to a suspension of AgCl, forces more of it to dissolve. The NH 3, removes Ag + from solution, forcing the equilibrium to the right. c)[Ag + ] T = [Ag + ] + [Ag(NH 3 ) + ] + [Ag(NH 3 ) 2+ ] d)The overall reaction is the sum of three individual reactions: AgCl Ag + + Cl - K sp = 1.6 x Ag + + NH 3 Ag(NH 3 ) + K 1 = 2.1 x 10 3 Ag(NH 3 ) + + NH 3 Ag(NH 3 ) 2+ K 2 = 8.2 x 10 3 AgCl + 2 NH 3 Ag(NH 3 ) Cl - K = ??? K = K sp x K 1 x K 2 = (1.6 x )(2.1 x 10 3 )(8.2 x 10 3 ) = 2.8 x 10 -3

e)Example: Find the solubility of AgCl(s) in 10 M NH 3. AgCl + 2 NH 3 Ag(NH 3 ) Cl - Initial Equil – 2xx x Solve using the expression for K as in a normal equilibrium problem. 6)Strategies for Dissolving Insoluble Solids a)If the anion is basic, add acid b)If the cation will form a complex, add ligand c)Heat often increases solubility (temp. effects all equil. constants) d)Example: HgS Hg 2+ + S 2- K sp = 1 x i.S 2- is a basic anion, so add HCl S 2- + H + HS - ii.Hg 2+ will form a chloride complex Hg Cl - HgCl 4 2-

7)The qualitative analysis of the Group I cations illustrates complex ion equilibria Solution of Ag +, Hg 2 2+, Pb 2+ Solution of Pb 2+ Precipitate of AgCl, Hg 2 Cl 2 Precipitate of PbCrO 4 Solution of Ag(NH 3 ) 2+, Cl - Precip. of AgCl Precipitate of Hg, HgNH 2 Cl Precipitation of AgCl, Hg 2 Cl 2, PbCl 2 Add cold HCl Heat Add CrO 4 2- Add H + Add NH 3