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1 Chapter 16 Precipitation Equilibria Copyright 2001 by Harcourt, Inc. All rights reserved. Requests for permission to make copies of any part of the work should be mailed to the following address: Permissions Department, Harcourt, Inc Sea Harbor Drive, Copyright Orlando, Florida by Harcourt, Inc. All rights reserved.

2 10 (10.1) (10.2),,

3 CaCO 2 (s) Ca 2+ (aq)+co 3 2- (aq) CO 3 2- (aq) +H 3 O + (aq) HCO 3- (aq) + H 2 O(l) CO 2, CO 2

4 16.1 Precipitate Formation : Solubility Product Constant(K sp ) K sp PbC1 2 (s) Pb 2+ (aq)+2c1 - (aq) K sp K sp

5 K sp Expression PbC1 2 (s) Pb 2+ (aq)+2c1 - (aq) K sp = [Pb 2+ ] [Cl ] 2 = PbI 2 (s) Pb 2+ (aq) + 2I (aq) K sp = [Pb 2+ ] [I ] 2 = See, Table 16.1, vs.

6 K sp and The Equilibrium Concentration of Ions Ex) [Cl ] = M, (PbCl 2 ) [Pb 2+ ]? PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) K sp = [Pb 2+ ] [Cl ] 2 = [Pb 2+ ] = ( ) / ( ) 2 = M

7 K sp and Precipitate Formation : Q(ion product) Q < Ksp : no precipitate ( ) Q>K sp Q=K sp Q > Ksp : precipitate forms until P = Ksp Q<K sp

8 K sp and Precipitate Formation Ex1) M CrO 2 4 Ag + [Ag + ]= M, (K sp Ag 2 CrO 4 = )? P = (orig. conc. Ag + ) 2 (orig. conc. CrO 2 4 ) = ( ) 2 ( )= > K sp ; Ex16.4) M Sr(NO 3 ) L M K 2 CrO 4 C0.100 L. SrCrO 4 (K sp = )? [Sr 2+ ]= (0.200L mol/L)/0.300L= M [CrO 2 4 ]=(0.100L 0.015mol./L)/0.300L= M P=( ) ( ) < Ksp ;

9 Selective Precipitation K sp (BaSO 4 )= , K sp (CaSO 4 )= [Ba 2+ ]=[Ca 2+ ]=0.10M, SO 2-4? For BaSO 4 [SO 4 2- ]= /0.1= For CaSO 4 [SO 4 2- ]= /0.1=

10 K sp and Water Solubility PbC1 2 (s) Pb 2+ (aq)+2c1 - (aq) s 2s [Pb 2+ ] = s, [Cl ] = 2s : K sp = s (2s) 2 = s 3 = ; s = M Ex) 100 ml g? mol g 1 L 10 2 ml = 0.44 g L 1 mol 10 3 ml vs

11 K sp and the Common Ion Effect PbC1 2 (s) Pb 2+ (aq)+2c1 - (aq) K sp = s (0.5+2s) 2 = C ini s (0.5) 2 = C +s +2s S= mol/l C fin. s s

12 K sp and the Common Ion Effect Ex2) BaSO 4 (s) Ba 2+ (aq) + SO 4 2- (aq) Ksp = M Na 2 SO 4 BaSO 4 (mol/l)? BaSO 4 (s) Ba 2+ (aq) + SO 2-4 (aq) C ini. C +s +s C fin. s 0.10+s K sp = s(0.10+s) = : 0.10+s 0.10 s = , mol/l(m) cf) M in water

13 Μ Μ 0.10 Μ

14 16.2 Dissolving Precipitate (works with basic anions) CaCO 3 (s) + 2H + (aq) Ca 2+ (aq) + H 2 CO 3 (aq) Zn(OH) 2, ZnCO 3, ZnS Driving force? CaCO 3 (s) Ca 2+ (aq) + CO 2-3 (aq) K sp = CO 2-3 (aq) + 2H + (aq) H 2 CO 3 (aq) K =1/(K a1 K a2 ) = CaCO 3 (s) + 2H + (aq) Ca 2+ (aq) + H 2 CO 3 (aq) K= K sp K >10 7

15 Dissolving Precipitate NH 3 or OH (works with cation that forms complexes) AgCl(s) + 2NH 3 (aq) Ag(NH 3 ) 2+ (aq) + Cl (aq) K AgCl(s) Ag + (aq) + Cl (aq) K sp = Ag + (aq) + 2NH 3 (aq) Ag(NH 3 ) 2+ (aq) K f = K = K sp K f = ( )( ) = 3.1 ex) Solubility of AgCl in 1.00 M NH 3? Ag(NH 3 ) 2+ (aq) + Cl (aq) s 2 s 1.00 AgCl(s) + 2NH 3 (aq) = ; s = mol / L 1.00 s

16 Qualitative Analysis Group I : Ag +, Pb 2+, Hg 2+ 2 : AgCl, PbC1 2,Hg 2 C1 2 Add hot water : PbC1 2 (s) Pb 2+ (aq) + 2C1 - (aq) Add NH 3 : AgC1(s) + 2NH 3 (aq) Ag(NH 3 ) 2+ (aq) + C1 - (aq) Group II : Cu 2+, Bi 3+, Hg 2+, Cd 2+, Sn 4+, Sb 3+ Cu 2+ (aq) + H 2 S(aq) CuS(s) + 2H + (aq) at ph=0.5 See, fig. 16.4

17 Group III : at ph=9.0 Al 3+, Cr 3+, Co 2+, Fe 2+, Mn 2+, Ni 2+, Zn 2+ Al(OH) 3, Cr(OH) 3 K sp (CuS )= , K sp (NiS )= Group IV : Ba 2+ (Ca 2+, Mg 2+ )+ CO 2-3 (aq) Ba CO 3 (s) Na +, K +, NH + 4

18 Step1 : Cu 2+, Ag +, Pb 2+ + Cl - (aq) Cu 2+ (aq), AgCl (s), PbC1 2 (s) Step 2 : Add hot water : AgCl (s), PbC1 2 (s) AgCl (s), Pb 2+ (aq) + 2C1 - (aq) Step 3 : Pb 2+ (aq) + CrO 4 2- (aq) PbCrO 4 (s)

19 Metals + NaCl, precipitate? Yes AgCl, PbC1 2, Hg 2 C1 2 + H 2 S (ph=0.5), precipitate? Yes Cu 2+, Bi 3+, Hg 2+, Cd 2+, Sn 4+, Sb 3+ + H 2 S (ph=9.0), precipitate? Yes Al 3+, Cr 3+, Co 2+, Fe 2+, Mn 2+, Ni 2+, Zn 2+ + CO 3 2- (aq) Yes Ba 2+, Ca 2+, Mg 2+ Na +, K +, NH 4 +

20 Solutility Equilibrium; K sp Expression for K sp : K sp represents a particular type of equilibrium constant known as the solubility product constant Uses of K sp Calculation of concentration of one ion, knowing that of the other Determination of whether precipitate will form Water solubility

21 Expression for K sp PbC1 2 (s) Pb 2+ (aq)+2c1 - (aq) K sp = [Pb 2+ ] [Cl ] 2 PbI 2 (s) Pb 2+ (aq) + 2I (aq) K sp = [Pb 2+ ] [I ] 2 =

22 Uses of K sp Calculation of concentration of one ion, knowing that of the other What is [Pb 2+ ] in a solution in equilibrium with lead chloride if [Cl ] = M? [Pb 2+ ] = ( ) / ( ) 2 = M

23 Uses of K sp Determination of whether precipitate will form. If P < K sp no precipitate (equilibrium not established) If P > K sp precipitate forms until P becomes equal to K sp Suppose enough Ag + is added to a solution M in CrO 2 4 to make [Ag + ]= M. Will silver chromate precipitate (K sp Ag 2 CrO 4 = )? P = (orig. conc. Ag + ) 2 (orig. conc. CrO 2 4 ) = ( ) 2 ( )= > K sp ; precipitate forms

24 Uses of K sp Water solubility PbC1 2 (s) Pb 2+ (aq)+2c1 - (aq) [Pb 2+ ] = s ; [Cl ] = 2s 4s 3 = K sp = ; s = M Solubility in grams per 100 ml? mol g 1 L 10 2 ml = 0.44 g L 1 mol 10 3 ml

25 Dissolving Precipitates Strong acid (works with basic anions) CaCO 3 (s) + 2H + (aq) Ca 2+ (aq) + H 2 CO 3 (aq) NH 3 or OH (works with cation that forms complexes) AgCl(s) + 2NH 3 (aq) Ag(NH 3 ) 2+ (aq) + Cl (aq) K = K sp K f = ( )( ) = Solubility of AgCl in 1.00 M NH 3? s = ; s = mol / L

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untitled Chapter 4 Reactions in Aqueous Solution 3 4.1 Pb(NO 3 ) 2 (aq) + 2KI (aq) HCl (aq) + NaOH (aq) Alka-Selter/Water Cu wire/ag(no 3 ) (aq) 3 4.2 (Solution) KMnO 4 (s) K + (aq)+ MnO 4- (aq) 3 4.3 (Molarity)