14.4 (Chp16) Solubility and Solubility Product

Transcription

1 14.4 (Chp16) Solubility and Solubility Product What drives substances to dissolve and others to remain as a precipitate? Dr. Fred Omega Garces Chemistry 201 Miramar College 1 Solubility and Solubility Product

2 When a Substance Dissolve When a reaction occurs in which an insoluble product is produced, the K eq is called a solubility-product. BaSO 4 (s) D Ba +2 + SO 4 2- (aq) K eq = K sp = [Ba +2 ] [SO 4 2- ] Do not confuse solubility (s) with solubility product K sp. How water dissolves ionic compounds Solubility depends on Temp., conc., and ph Solubility Product depends on Temp. Only!!! 2 Solubility and Solubility Product

3 Solubility Solubility: What is the meaning of solubility? i. The ability of substance to dissolve in a solvent. ii. Quantity that dissolves to form a saturated solution. s g g per 100 ml or g /L (molar solubility, grams per liter saturated solution.) g mol g mol g Molarity g The greater the solubility, the smaller the amount of precipitation. 3 Solubility and Solubility Product

4 Factors Affecting Solubility (s). 1 Nature of Solute (Concentration). - Like dissolves like (IMF) 2 Temperature Factor - i) Solids/Liquids- Solubility increases with Temperature (mostly) Increase K.E. increases motion and collision between solute / solvent. ii) gas - Solubility decreases with Temperature Increase K.E. result in gas escaping to atmosphere. 3 Pressure Factor - i) Solids/Liquids - Very little effect Solids and Liquids are already lose together, extra pressure will not increase solubility. ii) Gas- Solubility increases with Pressure. Increase pressure squeezes gas solute into solvent. 4 Solubility and Solubility Product

5 In a Chemical Reaction - For a reaction to take place - Reactants should be soluble in solvent- This ensures that reactants combine (come in contact) with each other efficiently. When reaction proceeds- Products which are formed have different properties than that of the reactants. One such property is the solubility. When a precipitate forms upon mixing two solution it is a result of a new specie that is present in the solution. 5 Solubility and Solubility Product

6 Consider the reaction - Precipitation Reaction i AgNO 3 (aq) + HCl (aq) D AgCl (s) + HNO 3(aq) Net ionic equation: Ag + (aq) + Cl - (aq) D AgCl (s) Write the reverse, (standard convention) AgCl (s) D Ag + (aq) + Cl - (aq) Then K eq = K sp = [Ag + ] [Cl - ] If written in this form, then mass action is: K eq = 1 [Ag + ][Cl ] K sp - solubility product constant An indicator of the solubility of substance of interest. Yields info. on the amount of ions allowed in solution. (Must be determine experimentally) NaOCH 2 CH 3 (s) + H 2 O (l) D Na+ (aq) + OH- (aq) + CH 3 CH 2 OH (aq) K sp = [Na + ] [OH - ] [CH 3 CH 2 OH] 6 Solubility and Solubility Product

7 Solubility Equilibrium: Example What is the solubility product for the following reaction: Ag 3 PO 4 D 3Ag + (aq) + PO 4 3- (aq) [s] = M i Excess 0 0 Δ M 3( M ) M e Solid M M Solubility; K sp = [ ] 3 [ ] = No direct proportionality of K sp to solubility [s] K sp e # ions in solution but in general - the greater the solubility (s), the less the amount of precipitate (solid) in solution Unlike in homogeneous equilibrium, K sp isn t a good indicator of a substance solubility. The solubility is ultimately determine by the solving the equilibrium problem for solubility (s). 8 Solubility and Solubility Product

8 Solubility from Solubility Chart The diagram below shows the solubility of some salts as a function of temperature. This graph can be used to determine the K sp for any one of the salts shown. Consider KNO 3, the solubility of KNO 3 at 20 C according to the graph is approximately 30 g per 100ml H 2 O. The strategy is to find the molar concentration and then use this information to determine the K sp for the salt at the specified temperature. The solubility of KNO 3 at 20 C is 30 g per 100ml H 2 O. What is the K sp for KNO 3 at this temperature? MW KNO3 = g/mol mol KNO 3 = 30.0g mol g = 0.297mol Assume solution has ρ = 1.00 g/cc Mass solution = 100g H 2 O + 30g KNO 3 = 130 g Solution 130 g solution = L since ρ = 1.00 g/cc [s] = 0.297mol 0.130L = 2.285M K sp = [K + ][NO 3 - ] = [2.285] 2 = Solubility and Solubility Product

9 Solubility Vs. Ksp Consider ; La (IO 3 ) 3 Ba(IO 3 ) 2 AgIO 3 K sp = < K sp = < K sp = s = < s = > s = In general there is no direct proportionality of K sp to solubility because K sp depends on the number of ions in solution. But if same number of ions form between two chemicals, then greater the solubility, the greater the degree that a substance dissolves in solution (less precipitate). 12 Solubility and Solubility Product

10 Solubility Rules: What is it used for? How can one predict if a precipitate forms when mixing solutions? Solubility rules - Soluble Substances containing nitrates, (NO 3- ) chlorates (ClO 3- ) perchlorates (ClO 4- ) acetates (CH 3 COO - ) Exceptions Insoluble substances containing None carbonates (CO 3 2- ) phosphates (PO 4 3- ) Exceptions Slightly soluble halogens (X - ) X - = Cl -, Br -, I - sulfates (SO 4 2- ) alkali & NH 4 + Ag, Hg, Pb hydroxides (OH - ) alkali, NH 4+, Ca, Sr, Ba Ba, Hg, Pb None Solubility rule provides information on the specie that will form a precipitate. In general, a molar solubility of 0.01M or greater for a substance is considered soluble Solubility and Solubility Product

11 Precipitation Experiment Consider the reaction: 3 Ca PO 3-4 (aq) D Ca 3 (PO 4 ) 2 (s) According to the solubility rules, this should form a precipitate (ppt). The question is what concentration is required before a precipitate forms? To solve, It is always convenient to write the equation as a dissolution of solid Ca 3 (PO 4 ) 2 (s) D 3 Ca PO 3-4 (aq) Q = [Ca +2 ] 3 [PO 4 3- ] 2 next compare Q to K sp In solubility product problems, Q is the ion-product instead of reaction quotient 14 Solubility and Solubility Product

12 Ion Product; Q Q - indicator of direction of reaction for reaction not at equilb. MX (s) D M + (aq) + X - (aq) Direction Reaction Q (i ) Unsaturated. Q is small, system consist mostly of ion. No ppt. forms (s) g (aq) K sp Q Sat. Point Direction Reaction Q (h ) Supersaturated. Q is large, system will adjust to reduce high ion concentration. Solid forms (s) f (aq) 15 Solubility and Solubility Product

13 In Class: Precipitation Determination Question, (19.88 Sil): Will any solid Ba(IO 3 ) 2 form when 6.5 mg of BaCl 2 is dissolved in 0.500L of 0.033M NaIO 3? K sp Ba(IO 3 ) 2 = , BaCl 2 MW = g/mol Ba(IO 3 ) 2 : Ba(IO 3 ) 2 (s) D Ba 2+ (aq) + 2 IO 3 - (aq) [Ba +2 ] = 6.5 mg M [IO 3- ] = M Q = [Ba +2 ] [IO 3- ] 2 = M ( M) 2 Q = M > K sp Precipitate forms. 17 Solubility and Solubility Product

14 Common Ion Effect For a system containing a precipitate in equilibrium with its ions, solubility of solid can be reduced if an ionic compound is dissolved in the solution. Ag + I - Ag + I - Ag + I - AgI (s) AgI (s) Add I - via NaI AgI (s) AgI (s) D Ag + (aq) + I - (aq) add common ion i.e., NaI to solution Direction of Reaction Reduce solubility i.e., less AgI will dissolve in soln which means more ppt. forms in solution 18 Solubility and Solubility Product

15 ...according to LeChatelier AgI(s) Ag+ I- AgI (s) D Ag + + I - AgI(s) NaI(s) Q analogy: Q = [Ag + ] [I - ] Ag+ I- Ksp < Q Increase I - AgI(s) Ag+ I- Ksp Q More AgI forms and Solubility is lowered Direction of Rxn (ppt occur) 19 Solubility and Solubility Product

16 Factor Affecting Solubility: Solubility & Common Ion Calculate the solubility of copper(i) iodide, CuI (K sp = ) a) water b) 0.05 M sodium iodide CuI (s) D Cu + (aq) + I - (aq) a) i Lots 0 0 Δ -s +s +s e Solid +s +s CuI (s) D Cu + (aq) + I- (aq) b) i Lots Δ -s +s +s e Solid +s s a) K sp = =[Cu + ] [I - ] = s = s b) K sp = =[Cu + ] [I - ] = s (0.05M + s) = s (0.05M) = s Without common ion, solubility is, s = M but with common ion, solubility is, s = M 22 Solubility and Solubility Product

17 Calculation: Solubility, Iteration Method Reger Ex 12.19: What is the solubility of calcium fluoride in M sodium fluoride solution? Ksp = M CaF 2 (s) D Ca +2 (aq) + 2F- (aq) i Lots M Δ -s +s +2s e Solid +s 0.025M +2s K sp = = [Ca +2 ] [F - ] 2 = s (0.025M + 2s) = s (0.025M) = s = M assume s << = 0.4% Check assumption indeed << Solubility and Solubility Product

18 Calculation: Solubility of Two salts in same solution At 50 C, the solubility products, K sp, of PbSO 4 and SrSO 4 are M and M, respectively. What are the values of [SO 4 2- ], [Pb 2+ ], and [Sr 2+ ] in a solution at equilibrium with both substances? PbSO 4 (s) D Pb 2+ (aq) + SO (aq) ; K sp = = [Pb 2+ ][SO 4 2- ] SrSO 4 (s) D Sr + (aq) + SO 4 2- (aq) ; K sp = = [Sr 2+ ][SO 4 2- ] Let x = [Pb 2+ ], y = [Sr 2+ ], x + y = [SO 4 2- ] x(x+y) = = x = = ; x = 0.057y y(x+y) y y(0.057y+y) = ; y 2 ~ = ; y= = x = y; x = ( ) = x = [Pb 2+ ] = M, [Sr 2+ ] = M, [SO 4 2- ]= M 26 Solubility and Solubility Product

19 ][IO 3 - ] 2 n = n = m [n] = [Sr +2 ][IO 3 - ] 2 = n[2m+2n] 2 = n[2(11.818n)+2n] = n[(23.636n)+2n] 2 = n[(25.636n] 2 = 657.2n n= 3 = , m = n = [Sr +2 ] = M, [Zn +2 ] = M, [Zn +2 ] = M Calculation: Same type problem different data At 25 C, the solubility product of Zn(IO 3 ) 2 and Sr(IO 3 ) 2 is M and M, respectively. What are the values of [IO 3- ], [Zn +2 ], and [Sr 2+ ] in a solution at equilibrium with both substances? Zn(IO 3 ) 2 (s) D Zn 2+ (aq) + 2IO -2 3 (aq) i Lots 0 0 Δ -m +m +2m + 2n e Lots +m +2m + 2n K sp = [Zn 2+ ] [IO 3 - ] 2 K sp - [IO 3 ] 2 = [Zn 2+ ] Sr(IO 3 ) 2 (s) D Sr +2 (aq) + 2IO - 3 (aq) i Lots 0 0 Δ -n +n +2n + 2m e Lots +n +2n + 2m K sp =[Sr +2 ] [IO 3 - ] 2 [IO 4 - ] 2 = K sp [Sr +2 ] ][IO - =[Zn 3 ] [Sr +2 ][IO - 3 ] =[m][2m+2n] 2 [n][2m+2n - ] =[m] 2 [n] n = m 2 =[Zn -7 [Sr +2 ][IO - 3 ] =[m][2m+2n] 2 [n][2m+2n - ] =[m] 2... Answer: [Sr +2 ] = M [Zn +2 ] = M [IO 3 - ] = M 27 Solubility and Solubility Product

20 Calculation: Same type problem different data At 25 C, the solubility product of AgI and PbI 2 is M and M, respectively. What are the values of [I - ], [Ag + ], and [Pb 2+ ] in a solution at equilibrium with both substances? (Answer behind) AgI (s) D Ag + (aq) + I - (aq) i Lots 0 2t Δ -s +s +s e Lots +s +s+2t K sp = [Ag + ] [I - ] [I - ] = K sp [Ag + ] PbI 2 (s) D Pb +2 (aq) + 2I - (aq) i Lots 0 s Δ -t +t +2t e Lots +t s+2t K sp = [Pb +2 ] [I - ] 2 [I - ] 2 = K sp [Pb +2 ] ][I - ] 2 =[Pb [Ag + ] [I - ] [Pb +2 ][I - ] 2 = =[t] [s+2t] 2 2 [t] [s+2t] = [s] [s+2t] = Assume s << t = [t] [2t] 2 = 4[t] 3 ; t = = [Pb +2 ] [Ag + ] [I - ] = = [s] [s+2t] Assume s << t = [s] [2t] = [s] [ ] ; [s] = M [Ag + ]= M, [Pb +2 ] = M, [I - ] = M 28 Solubility and Solubility Product

21 ][I - ] 2 [Ag + ] [I - ] [Pb +2 ][I - ] 2 = =[t] [s+2t] 2 Assume s << t = [t] [2t] 2 = 4[t] 3 ; t = = [Pb +2 ] = [Ag + ] [I - ] = [s] [s+2t], Assume s << t = [s] [2t] but t = Calculation: Same type problem different data At 25 C, the solubility product of AgI and PbI 2 is M and M, respectively. What are the values of [I - ], [Ag + ], and [Pb 2+ ] in a solution at equilibrium with both substances? (Answer behind) AgI (s) D Ag + (aq) + I - (aq) i Lots 0 2t Δ -s +s +s e Lots +s +s+2t K sp =[Ag + ] [I - ] PbI 2 (s) D Pb +2 (aq) + 2I - (aq) i Lots 0 s Δ -t +t +2t e Lots +t s+2t K sp =[Pb +2 ] [I - ] = s (2t) = t (2t) ][I - ] 2 =[Pb [Ag + ] [I - ] [Pb +2 ][I - ] 2 = =[t] [s+2t] 2 2 [t] [s+2t] = [s] [s+2t] = Assume s << t = [t] [2t] 2 = 4[t] 3 ; t = = [Pb +2 ] [Ag + ] = M [Pb +2 ] = M [I - ] = M = [Ag + ] [I - ] = [s] [s+2t], Assume s << t = [s] [2t] but t = = [s] [2( )] = s s = M= [Ag + ], [I - ] = s + 2t = = M -17 =[Pb [t] [s+2t] = [s] [s+2t] = Solubility and Solubility Product [Ag + ] [I - ] = = [s] [s+2t] Assume s << t = [s] [2t] = [s] [ ] ; [s] = M

22 Common Ion Effect B&L (7th Ed.): The K sp for cerium iodate, Ce(IO 3 ) 3, is a) Calculate the molar solubility of Ce(IO 3 ) 3 in pure water. b) What concentration of NaIO 3 in solution would be necessary to reduce the Ce 3+ concentration in a saturated solution of Ce(IO 3 ) 3 by a factor of 10 below that calculation in part (a). Ce(IO 3 ) 3 (s) D Ce +3 (aq) + 3 IO - 3 (aq) i Lots 0 0 Δ -s +s +3s e Lots +s +3s a) K sp = [Ce +3 - ] [IO 3 ] = s 27s 3 = 27s = s M = s b) reduce Ce 3+ by factor of 10: [Ce3+ ] = [IO 3 ] Total = [IO 3 ] Total M = [IO 3 ] Total 10 [NaIO 3 ] = M - [IO 3 - ] (from part previous) [NaIO 3 ] = M - (3 x [ [NaIO 3 ] = M **(same as Ce +3 ) ]) ** The IO 3 - in solution must be the same as Ce +3 for this condition 32 Solubility and Solubility Product

23 Qualitative Analysis Ions are precipitated selectively by adding a precipitation ion until the K sp of one compound is exceeded without exceeding the K sp of the others. An extension of this approach is to control the equilibrium of the slightly soluble compound by simultaneously controlling an equilibrium system that contains the precipitating ion. Qualitative analysis of ion mixtures involves adding precipitating ions to separate the unknown ions into ion groups. The groups are then analyzed further through precipitation and complex ion formation. 33 Solubility and Solubility Product

24 Selective Precipitation Qualitative Analysis Q < K sp, solid dissolve until Q=K sp Q = K sp, equilib Q > K sp, ppt until Q=K sp Grp1: Insoluble Chlorides Ag +, Pb 2+, Hg 2+ 2 Grp2: Acid-Insoluble sulfides Cu 2+, Bi 3+, Cd 2+, Pb 2+, Hg 2+, As 3+, Sb 3+, Sn 4+ Grp3: Base-Insoluble sulfides Cu 2+, Al 3+, Fe 2+, Fe 3+, Co 2+, Ni 2+, Cr 3+, Zn 2+, Mn 2+ Grp4: Insoluble Phosphates Ba 2+, Ca 2+, Mg 2+ Grp5: Alkali Metals and NH + 4 Na +, K +, NH Solubility and Solubility Product

25 Lab Practical, Qualitative Analysis Qualitative Analysis of Household Chemicals 36 Solubility and Solubility Product

26 Selective Precipitation(2) A solution contains M Ag + and M Pb 2+. If NaI is added, will AgI (K sp = ) or PbI 2 (K sp = ) precipitate first? Specify the concentration of I - needed to begin precipitation. This is a Q problem PbI 2 (s)! Pb 2+ (aq) + 2I - (aq ) AgI (s)! Ag + + I - i Excess y (aq) (aq ) i Excess! x Δ -x x 2x [e] Excess Δ -y y y -3 +x y + 2x [e] Excess y 2x+y Assume y << x, furthermore assume x << Assume y << x and < [Ag + ] [I - ] = [ y] [2x + y] < [ ] [2x] 2x > = , x > M [I - ] = 2x+y = 2x, y << x, [I - ] = M < [Pb 2+ ] [I - ] 2 = [ x ] [y + 2x] = [ ] [2x] x = = M = x, Note x = therefore need second iteration for x. [Pb 2+ ] = = Second iteration, = [ ] [2x] 2 2x = M [I - ] = 2x = M for PbI 2 (s) to precipitate AgI will precipitate first. The [I - ] concentration needs to be M, PbI 2 will not precipitate until the [I - ] concentration reaches M 38 Solubility and Solubility Product

27 Complex Ion Process Metals forming Complexes: Metals by virtue of electron pairs will act as Lewis acids and bind a substrate to form a Complex Consider the following: Ag(NH 3 ) 2 + (aq) Complex ion AgCl (s) D Ag + (aq) + Cl- (aq) Ag + (aq) + 2NH 3(aq) D Ag(NH 3 ) 2 + (aq) AgCl (s) + 2NH 3(aq) D Ag(NH 3 ) 2 + (aq) + Cl - (aq) Normally insoluble AgCl can be made soluble By the addition of NH 3. The presence of NH 3 drives the top reaction to the right and increase the solubility of AgCl An assembly of metal ion and the Lewis base (NH 3 ) is called a complex ion. The formation of this complex is describe by Formation Constant, K f Ag + (aq) + 2NH 3(aq) D Ag(NH 3 ) 2 + (aq) K f = [Ag(NH 3 ) 2 + ] [Ag + = ] [NH 3 ] Solubility of metal salts affected presence of Lewis Base: e.g., NH 3, CN -, OH - 39 Solubility and Solubility Product

28 Formation Constants Table Complex Kf Complex Kf Complex Kf [Ag(CN)2] 5.6e18 [Cr(OH)4] 8e29 [HgI4]2 6.8e29 [Ag(EDTA)]3 2.1e7 [CuCl3]2 5e5 [Hg(ox)2]2 9.5e6 [Ag(en)2]+ 5.0e7 [Cu(CN)2] 1.0e16 [Ni(CN)4]2 2e31 [Ag(NH3)2]+ 1.6e7 [Cu(CN)4]3 2.0e30 [Ni(EDTA)]2 3.6e18 [Ag(SCN)4]3 1.2e10 [Cu(EDTA)]2 5e18 [Ni(en)3]2+ 2.1e18 [Ag(S2O3)2]3 1.7e13 [Cu(en)2]2+ 1e20 [Ni(NH3)6]2+ 5.5e8 [Al(EDTA)] 1.3e16 [Cu(CN)4]2 1e25 [Ni(ox)3]4 3e8 [Al(OH)4] 1.1e33 [Cu(NH3)4]2+ 1.1e13 [PbCl3] 2.4e1 [Al(Ox)3]3 2e16 [Cu(ox)2]2 3e8 [Pb(EDTA)]2 2e18 [Cd(CN)4]2 6.0e18 [Fe(CN)6]4 1e37 [PbI4]2 3.0e4 [Cd(en)3]2+ 1.2e12 [Fe(EDTA)]2 2.1e14 [Pb(OH)3] 3.8e14 [Cd(NH3)4]2+ 1.3e7 [Fe(en)3]2+ 5.0e9 [Pb(ox)2]2 3.5e6 [Co(EDTA)]2 2.0e16 [Fe(ox)3]4 1.7e5 [Pb(S2O3)3]4 2.2e6 [Co(en)3]2+ 8.7e13 [Fe(CN)6]3 1e42 [PtCl4]2 1e16 [Co(NH3)6]2+ 1.3e5 [Fe(EDTA)] 1.7e24 [Pt(NH3)6]2+ 2e35 [Co(ox)3]4 5e9 [Fe(ox)3]3 2e20 [Zn(CN)4]2 1e18 [Co(SCN)4]2 1.0e3 [Fe(SCN)]2+ 8.9e2 [Zn(EDTA)]2 3e16 [Co(EDTA)] 1e36 [HgCl4]2 1.2e15 [Zn(en)3]2+ 1.3e14 [Co(en)3]3+ 4.9e48 [Hg(CN)4]2 3e41 [Zn(NH3)4]2+ 2.8e9 [Co(NH3)6]3+ 4.5e33 [Hg(EDTA)]2 6.3e21 [Zn(OH)4]2 4.6e17 [Co(ox)3]3 1e20 [Hg(en)2]2+ 2e23 [Zn(ox)3]4 1.4e8 [Cr(EDTA)] 1e23 40 Solubility and Solubility Product

29 Calculations: Formation of Complex Ion Tro, End of chapter Q 109 (2 nd Edition) A solution is made that is M Zn(NO 3 ) 2 and M NH 3. After the solution reaches equilibrium, what concentration of Zn 2+ (aq) remains. Answer = M Zn 2+ (aq) + 4NH 3 (aq) D Zn(NH 3 ) 2+ 4 (aq) Zn 2+ (aq) + NH 3 (aq)! Zn(NH 3 ) 4 2+ i Δ -x [e] x (aq) Assume reaction goes to completion Zn(NH 3 ) 4 2+ = M = [Zn(NH 3 ) 4 2+ ] [Zn +2 ] [NH 3 ] 4 = x [ x] [ ] 4 = [ x] [0.1456] x = [0.1456] 4 = , [Zn 2+ ] eq = M 41 Solubility and Solubility Product

30 Calculations: Formation of Complex Ion By using the values of K sp for AgI and K f for Ag(CN) 2-, calculate the equilibrium constant for the following reaction: AgI (s) + 2CN - (aq) D Ag(CN) 2 - (aq) + I - (aq) Note that this equation is the sum of AgI (s) D Ag + (aq) + I - (aq) ; K sp = = [Ag + ][I - ] Ag + (aq) + 2CN - (aq) D Ag(CN) - 2 (aq) ; K f = = [Ag(CN) 2- ]/[Ag + ][CN - ] 2 AgI (s) + 2CN - (aq) D Ag(CN) - 2 (aq) + I - (aq) K = K sp K f = ( [Ag + ][I - ] ) ( [Ag(CN) 2 ]/[Ag + ][CN - ] 2 ) K = K sp K f = K = Solubility and Solubility Product