Chapter 18 Solubility and Complex- Ionic Equilibria 1
The common ion effect Le Chatelier Why is AgCl less soluble in sea water than in fresh water? AgCl(s) Ag + + Cl Seawater contains NaCl 2
Problem: The solubility of AgCl in pure water is 1.3 x 10 5 M. What is its solubility in seawater where the [Cl ] = 0.55 M? (K sp of AgCl = 1.8 x 10 10 ) I. C. E. AgCl(s) Ag + + Cl N/A 0 0.55 N/A +x +x N/A +x 0.55 + x K sp = [Ag + ][Cl ] K sp = [x][0.55 + x] try dropping this x K sp = 0.55x 1.8 x 10 10 = 0.55x x = 3.3 x 10 10 = [Ag + ]=[AgCl] AgCl is much less soluble in seawater 3
more Common ion effect: a. What is the solubility of CaF 2 in 0.010 M Ca(NO 3 ) 2? K sp (CaF 2 ) = 3.9 x 10 11 CaF 2 (s) Ca 2+ + 2F I. C. E. [Ca 2+ ] [F ] 0.010 0 +x +2x 0.010 + x 2x K sp =[Ca 2+ ][F - ] 2 K sp = [0.010 + x][2x] 2 [0.010][2x] 2 = 0.010(4x 2 ) 3.9 x 10 11 = 0.010(4x 2 ) x = 3.1 x 10 5 M Ca 2+ from CaF 2 so = M of CaF 2 Now YOU determine the solubility of CaF 2 in 0.010 M NaF. 4
Answer: 3.9 x 10 7 M Ca 2+ CaF 2 (s) Ca 2+ + 2F 0 0.010 +x 2x x 0.010 + 2x K sp = [x][0.010 + 2x] 2 x(0.010) 2 3.9 x 10-11 =x(0.010) 2 x = 3.9 x 10-7 What does x tell us 5
Predicting Formation of Precipitate Q sp = K sp saturated solution, but no precipitate Q sp > K sp saturated solution, with precipitate Q sp < K sp unsaturated solution, Q sp is ion product expressed in the same way as K sp for a particular system.
Predicting Precipitation Consider the following case: 20.0 ml of 0.025 M Pb(NO 3 ) 2 is added to 30.0 ml of 0.10 M NaCl. Predict if precipitate of PbCl 2 will form. (Ksp for PbCl 2 = 1.6 x 10-5)
Predicting Precipitation Calculation: [Pb 2+ ] =(20.0 ml x 0.025 M)/(50.0 ml)= 0.010 M [Cl - ] = (30.0 ml x 0.10 M)/(50.0 ml) = 0.060 M Q sp = [Pb 2+ ][Cl - ] 2 = (0.010 M)(0.060 M) 2 = 3.6 x 10-5 Q sp > K sp precipitate of PbCl 2 will form.
Practical Applications of Solubility Equilibria Qualitative Analyses Isolation and identification of cations and/or anions in unknown samples Synthesis of Ionic Solids of commercial interest Selective Precipitation based on K sp
Reaction Quotient (Q): will a ppt. occur? Use Q (also called ion product) and compare to K sp Q < K sp Q = K sp Q > K sp reaction goes Equilibrium reaction goes 10
Practical Applications of Solubility Equilibria Qualitative Analyses Isolation and identification of cations and/or anions in unknown samples Synthesis of Ionic Solids of commercial interest Selective Precipitation based on K sp
Qualitative Analysis Separation and identification of cations, such as Ag +, Ba 2+, Cr 3+, Fe 3+, Cu 2+, etc. can be carried out based on their different solubility and their ability to form complex ions with specific reagents, such as HCl, H 2 SO 4, NaOH, NH 3, and others. Separation and identification of anions, such as Cl -, Br -, I -, SO 4 2-, CO 3 2-, PO 4 3-, etc., can be accomplished using reagents such as AgNO 3, Ba(NO 3 ) 2 under neutral or acidic conditions.
Selective Precipitation (Mixtures of Metal Ions) Use a reagent whose anion forms a precipitate with only one or a few of the metal ions in the mixture. Example: Solution contains Ba 2+ and Ag + ions. Adding NaCl will form a precipitate with Ag + (AgCl), while still leaving Ba 2+ in solution.
Separation of Cu 2+ and Hg 2+ from Ni 2+ and Mn 2+ using H 2 S At a low ph, [S 2 ] is relatively low and only the very insoluble HgS and CuS precipitate. When OH is added to lower [H + ], the value of [S 2 ] increases, and MnS and NiS precipitate.
Separation of Cu 2+ and Hg 2+ from Ni 2+ and Mn 2+ using H 2 S
Separating the Common Cations by Selective Precipitation
Synthesis of Ionic Solids Chemicals such as AgCl, AgBr, and AgI that are important in photography are prepared by precipitation method. AgNO 3 (aq) + KBr(aq) AgBr(s) + KNO 3 (aq)
Selective Precipitation Compounds with different solubility can be selectively precipitated by adjusting the concentration of the precipitating reagents. For example, AgCl has a much lower K sp than PbCl 2 If Ag + and Pb 2+ are present in the same solution, the Ag + ion can be selectively precipitated as AgCl, leaving Pb 2+ in solution.
Problem: A solution is 1.5 x 10 6 M in Ni 2+. Na 2 CO 3 is added to make the solution 6.0 x 10 4 M in CO 2 3. K sp (NiCO 3 ) = 6.6 x 10 9. Will NiCO 3 ppt? We must compare Q to K sp. NiCO 3 Ni 2+ + CO 2 3 K sp = [Ni 2+ ][CO 2 3 ] Q = [Ni 2+ ][CO 2 3 ] Q = [1.5 x 10 6 ][6.0 x 10 4 ] = 9.0 x 10 10 Q < K sp no ppt. 19
Problem: 0.50 L of 1.0 x 10 5 M Pb(OAc) 2 is combined with 0.50 L of 1.0 x 10 3 M K 2 CrO 4. a. Will a ppt occur? K sp (PbCrO 4 ) = 1.8 x 10 14 Pb(OAc) 2 (aq) + K 2 CrO 4 (aq) PbCrO 4 (s) + 2KOAc(aq) then: PbCrO 4 (s) Pb 2+ + CrO 4 2 K sp = [Pb 2+ ][CrO 4 2 ] [Pb 2+ ]: 0.50 L -5 1.0 x 10 mol Pb(OAc) 2 L 2 1 Pb 1 Pb(OAc) 2 1 L 5.0 x 10 6 mol Pb L 2 [CrO 4 2- ]: 0.50 L -3 1.0 x 10 mol K 2CrO4 L 2-1CrO4 1 K CrO 2 4 1 L 5.0 x 10-4 mol CrO L 2-4 Q = [5.0 x 10-6 ][5.0 x 10-4 ] = 2.5 x 10-9 compare to K sp : Q > K sp so a ppt. will occur 20
b. find the Eq. conc. of Pb 2+ remaining in solution after the PbCrO 4 precipitates. K sp (PbCrO 4 ) = 1.8 x 10 14 Since [Pb 2+ ] = 5.0 x 10-6 and [CrO 2-4 ] = 5.0 x 10-4 and there is a 1:1 stoichiometry, Pb 2+ is the limiting reactant. PbCrO 4 (s) Pb 2+ + CrO 2 4 I. (after ppt.) 0 5.0 x 10-4 - 5.0 x 10-6 = 5.0 x 10-4 C. +x +x E. x 5.0 x 10 4 + x K sp = [x][5.0 x 10 4 + x] Try dropping the + x term. 1.8 x 10 14 = x(5.0 x 10-4 ) x = [Pb 2+ ] = 3.6 x 10 11 This is the concentration of Pb 2+ remaining in solution. 21
Complex ion formation: AgCl(s) Ag + + Cl Ag + (aq) + NH 3 (aq) Ag(NH 3 ) + (aq) Ag(NH 3 ) + (aq) + NH 3 (aq) Ag(NH 3 ) 2+ (aq) K sp = 1.8 x 10 10 Ag + (aq) + 2NH 3 (aq) {Ag(NH 3 ) 2 } + (aq) formation or stability constant: Ag(NH3) 2 K f 1.7 x 10 Ag NH 3 2 7 For Cu 2+ : Cu 2+ + 4NH 3 [Cu(NH 3 ) 4 ] 2+ (aq) K 1 x K 2 x K 3 x K 4 = K f = 6.8 x 10 12 22
K Solubility and complex ions: Problem: How many moles of NH 3 must be added to dissolve 0.050 mol of AgCl in 1.0 L of H 2 O? (K sp AgCl = 1.8 x 10 10 ; K f [Ag(NH 3 ) 2 ] + = 1.6 x 10 7 ) AgCl(s) Ag + + Cl Ag + (aq) + 2NH 3 (aq) Ag(NH 3 ) 2+ (aq) sum of RXNS: AgCl(s) + 2NH 3 Ag(NH 3 ) 2+ (aq) + Cl overall Ag(NH3) 2 Cl 2 NH Now use K overall to solve the problem: 3 K sp x K f = 2.9 x 10 3 K overall = 2.9 x 10 3 = Ag(NH ) Cl 0.050 0.050 3 2 2 NH NH 2 3 3 [NH 3 ] eq = 0.93 but... How much NH 3 must we add? [NH 3 ] total = 0.93 + (2 x 0.050) = 1.03 M 2 ammonia s for each Ag + 23
Fractional Precipitation: ppting one ion at a time. Compounds must have different K sp values (i.e. different solubilities) Example: K sp CdS = 3.9 x 10 29 and K sp NiS = 3.0 x 10 21? Which will ppt. first? least soluble Problem: A solution is 0.020 M in both Cd 2+ and Ni 2+. Just before NiS begins to ppt., what conc. of Cd 2+ will be left in solution? Approach: Find conc. of S 2 ion when Ni 2+ just begins to ppt. since Cd 2+ will already be ppting. Then use this S 2 conc. to find Cd 2+. NiS(s) Ni 2+ + S 2 K sp = 3.0 x 10 21 = [0.020][S 2 ] [S 2 ] = 1.5 x 10 19 M when Ni 2+ just begins to ppt. So: CdS(s) Cd 2+ + S 2 K sp = 3.9 x 10 29 = [Cd 2+ ][1.5 x 10 19 ] [Cd 2+ ] = 2.6 x 10 10 M when NiS starts to ppt. 24