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19 Heterogeneous & complex equilibria 1 19 Heterogeneous and complex equilibria CaCO 3 exists as aragonite calcite Formation of crystals such as AgCl in.

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Presentation on theme: "19 Heterogeneous & complex equilibria 1 19 Heterogeneous and complex equilibria CaCO 3 exists as aragonite calcite Formation of crystals such as AgCl in."— Presentation transcript:

1 19 Heterogeneous & complex equilibria 1 19 Heterogeneous and complex equilibria CaCO 3 exists as aragonite calcite Formation of crystals such as AgCl in a solution is a heterogeneous equilibrium, because there are more than one phase, AgCl (s) = Ag + (aq) + Cl – (aq) Species such as Ag(NH 3 ) 2 + & Ag(CN) 2 – are complexes ( or complex ions ). Ag + (aq) + 2 NH 3 (aq) = Ag(NH 3 ) 2 + (aq)

2 19 Heterogeneous & complex equilibria 2 Beauty due to heterogeneous equilibria There are many natural heterogeneous equilibria. Please think of some!

3 19 Heterogeneous & complex equilibria 3 The solubility product For the dissolution, CaCO 3 (s) = Ca 2+ (aq) + CO 3 2- (aq) K sp = [Ca 2+ ] [CO 3 2- ] = 3.8e-9 (a constant) solubility product SubstanceFormula K sp Aluminum hydroxideAl(OH) 3 4.6e-35 Barium chromateBaCrO 3 1.2e-10 Calcium phosphateCa 3 (PO 4 ) 2 1e-26 Iron sulfiteFeS6e-18 Lead sulfitePbS2.5e-27 Mercury sulfiteHgS1.6e-52 K sp = [Al 3+ ] [OH - ] 3 K sp = [Ca 2+ ] 3 [PO 4 ] 2 Which is the most and least soluble?

4 19 Heterogeneous & complex equilibria 4 Solubility product constants CompoundK sp CompoundK sp AgBr5.0 x Fe(OH) x AgCl1.8 x FeS6.3 x AgI8.3 x HgS1.6 x AgIO x Mg(OH) x Ag 3 PO x MgC 2 O x Al(OH) x Mn(OH) x Ba(OH) x MnS2.5 x BaSO x NiS1.0 x Bi 2 S x PbCl x CaCO x PbSO x CaC 2 O x PbS8.0 x CaSO x SrSO x CdS8.0 x Zn(OH) x CoS2.0 x ZnS1.6 x CuS6.3 x Table like this is available in handbooks or data bases. Know where to find them when you need them. Write K sp expressions.

5 19 Heterogeneous & complex equilibria 5 K sp and solubility The K sp = 1.0e-6 for BaF 2, what are [Ba 2+ ] and [F¯]? Solution : BaF 2 = Ba F¯ assume x 2 x K sp = x (2x) 2 = 1.0e-6 x = 3 (1.0e-6 / 4) = 6.3e-3 M [Ba 2+ ] = x = 6.3e-3 M molar solubility = 6.3e-3 *175 = 1.1g/L [F¯] = 2 x = M Checking : 6.3e-3 * = 1e-6 = K sp Calculate solubility of BaCrO 3, K sp = 1.2e-10 Molar mass of BaF 2 =175 g/mol 19-2

6 19 Heterogeneous & complex equilibria 6 Concentrations of ions in solution CaF 2 (s) = Ca 2+ (aq) + 2 F - (aq), K sp know where to find K sp = 5.3e-9

7 19 Heterogeneous & complex equilibria 7 Perception of a saturate solution Ag 2 S = 2 Ag + + S 2– K sp = [Ag + ] 2 [ S 2– ]

8 19 Heterogeneous & complex equilibria 8 Calculate K sp from Solubility When 0.50 L saturated CaC 2 O 4 solution was dried, g dry salt was obtained. Evaluate the K sp. Solution : 3.0e-3 g 0.50 L 1 mol 128 g = 4.7e-5 mol / L = [Ca 2+ ] CaC 2 O 4 = Ca 2+ + C 2 O e-5 4.7e-5 K sp = 4.7e-5 * 4.7e-5 = 2.2e-9 Figure out how to calculate solubility from K sp. How many grams of HgS will dissolve in 1 L?

9 19 Heterogeneous & complex equilibria 9 Precipitation of AgCl Goto URL: dl.clackamas.cc.or.us/ch105-05/solubili.htm for a demonstration if you have not seen the experiment Ag + (aq) + Cl - (aq) = AgCl(s)

10 19 Heterogeneous & complex equilibria 10 Common-ion effect on solubility Like acid-base equilibria, presence of common ions from more than one electrolyte affects the solubility, since K sp remains constant. For example, the K sp = 1.8e-10 for AgCl. The maximum [Ag + ] is governed by the condition, NaCl = Na + + Cl AgCl = Ag + + Cl - [Ag + ]0.10 Thus[Ag + ] * 0.10 = 1.8e-10 [Ag + ] = 1.8e-9 M Solubility of AgCl = 1.8e-10 = 1.3e-5 M in pure water is 7,454 times more. ? Should we consider AgCl = Ag + + Cl - x0.10+x 19-3

11 19 Heterogeneous & complex equilibria 11 Graph the common ion effect AgCl = Ag + + Cl - NaCl = Na + + Cl -

12 19 Heterogeneous & complex equilibria 12 Condition for precipitation Recall: Predicting reaction directions by comparing Q c and K c. Same principle applies to precipitation (ppt) Q K sp, super-saturated (unstable, often needs a seed to start the ppt) 19-5

13 19 Heterogeneous & complex equilibria 13 Separation by precipitation A 10-mL solution contains 0.10 M each of Cl –, Br –, and I – ions. Micro amounts of 0.10 M AgNO 3 solution is added to the system. What are the Ag+ concentrations before AgI, AgBr, and AgCl precipitate? Solution : Data sheet K sp :AgCl 1.8e-10, AgBr 5.0e-13, and AgI 8.3e-17. [Ag + ] for AgI, AgBr, & AgCl solids to form: [I – ] = 0.10 M[Ag + ] = 8.3e-17 / 0.10 = 8.3e-16 AgI(s) appears [Br – ] = 0.10 M[Ag + ] = 5.0e-13 / 0.10 = 5.0e-12 AgBr(s) appears [Cl – ] = 0.10 M[Ag + ] = 1.8e-10 / 0.10 = 1.8e-9 AgCl(s) appears As [Ag + ] increases, how do [Cl - ], [Br - ] and [I - ] vary ? [Ag + ] [Cl – ] [Br – ] [I – ] 0 8.3e AgI(s) 5.0e AgBr(s) 1.7e-5 1.8e AgCl(s) 2.8e-4 4.6e-8 1e-7 1.8e-3 5.0e-6 8.3e ____?____?____? 19-6

14 19 Heterogeneous & complex equilibria 14 pH H 2 S and solubility of metal ions The pH affects the equilibrium of many species, for example: H 2 S = H + + HS – K a1 = 1e-9 HS – = H + + S 2 – K a2 = 1e-14 (doubtful but adopt) H 2 S = 2 H + + S 2 – K overall = 1e-23 [S 2 – ] = [H 2 S] *1e-23 / [H + ] 2 = [H 2 S] *1e(2*pH-23), strongly affected by pH If [H 2 S] = 1.0 M pH [S 2 – ] K sp of MS [S 2 – ] for [M 2+ ] = e e-52 HgS 1.6e-49 (ppt) 21e e-27 PbS 2.5e-24 (ppt) e-181.1e-21 ZnS 1.1e-18 (no ppt pH < 2.5) 4.396e-156e-18 FeS6e-15 (no ppt pH < 4.4) 71e-9 2.5e-13 MnS_____ (no ppt pH < ____) 101e Recalculate [s2-] at various pH if [H 2 S] = 0.10 M

15 19 Heterogeneous & complex equilibria 15 pH, and CO 2 on CaCO 3 solubility The [CO 3 2– ] in a M H 2 CO 3 solution is determined by, H 2 CO 3 = H + + HCO 3 – K a1 = 4.3e-7 HCO 3 – = H + + CO 3 2– K a2 = 5.6e-11 H 2 CO 3 = 2 H + + CO 3 2– K overall = K a1 * K a2 = 2.4e-17 = [H + ] 2 [CO 3 2– ] / [H 2 CO 3 ] [CO 3 2– ] = (0.0010*2.4e-17) [H + ] –2 = 2.4e-20 * 1e(2*pH) = 2.4e(2*pH-20) [CO 3 2– ] affects [Ca 2+ ] due to equilibrium CaCO 3 = Ca 2+ + CO 3 2– K sp = 8.7e–9 [Ca 2+ ] = 8.7e–9 [CO 3 2– ] –1 = 8.7e–9 / 2.4e(2*pH–20) = 3.6e(20 – 9–2*pH) = 3.6e(11-2*pH) Decrease pH by 1 increases [Ca 2+ ] by 2 order of magnitude pH [Ca 2+ ] 83.6e e

16 19 Heterogeneous & complex equilibria 16 Stalactites and stalagmites stalactites stalagmites Rain dissolves limestone and when water drops form stalactites and stalagmites in caves. They grew about 2 cm per 1000 years. The slightly acidic rain dissolves lime stone: CaCO 3 (s) + H + = Ca 2+ (aq) + HCO 3 - (aq) When acidity is reduced, solid forms: Ca 2+ (aq) + HCO 3 - (aq) + OH - (aq) = CaCO 3 (s) + H 2 O

17 19 Heterogeneous & complex equilibria 17 Stalactites hanging down from the ceiling formed over hundreds of years in Mercer Cavern

18 19 Heterogeneous & complex equilibria 18 Aragonite formations found at Mercer Caverns, California The "Angel Wings" are two delicate and translucent crystalline formations, over 9 feet long and 2.5 feet wide in Mercer Caverns

19 19 Heterogeneous & complex equilibria 19 Equilibrium of complexes Metal ions tend to attract Lewis bases forming coordinated complexes, or complex ions. These formation is governed by equilibrium, metal ion ligand Step-wise formation constant Ag + (aq) + NH 3 (aq) = Ag(NH 3 ) + (aq) K 1 Ag(NH 3 ) + (aq) + NH 3 (aq) = Ag(NH 3 ) 2 + (aq) K 2 formation constant Ag + (aq) + 2 NH 3 (aq) = Ag(NH 3 ) 2 + (aq) K f = K 1 K 2 = 1.7e7 Ag + (aq) + 2 S 2 O 3 2 – (aq) = Ag(S 2 O 3 ) 2 3 – (aq) K f = 2.9e13 Ag + (aq) + 2 CN – (aq) = Ag(CN) 2 – (aq) K f = 5.6e18 Dissociation constant Ag(NH 3 ) 2 + (aq) = Ag + (aq) + 2 NH 3 (aq) K d = 1 / K f = 5.9e-8 Ag(S 2 O 3 ) 2 3 – (aq) = Ag + (aq) + 2 S 2 O 3 2 – (aq) K d = 1 / 2.9e13 = 3.4e

20 19 Heterogeneous & complex equilibria 20 Logical, but impractical method!!! Evaluate [Ag+] in a solution containing 1.0 M NH 3 and 0.10 M AgNO 3. Solution : Ag + (aq) + 2 NH 3 (aq) = Ag(NH 3 ) 2 + (aq) K f = 1.7e7 (find data) 0.1-y1-2yy(think way) y = 1.7e7 (1-2y) 2 (0.10-y) y = (1-2y) 2 (0.1-y) By trial an error, y = e7 [Ag + ] = 0.10 – y ~ 0 (cannot evaluate [Ag + ]) 0.10-y = 6e-9, too small, impractical! 5.9e-8y = 0.1 –0.8y –2y 2

21 19 Heterogeneous & complex equilibria 21 Concentration of ions in presence of ligands Evaluate [Ag+] in a solution containing 1.0 M NH 3 and 0.1 M AgNO 3. Solution : Ag + (aq) + 2 NH 3 (aq) = Ag(NH 3 ) 2 + (aq) K f = 1.7e7 (find data) x0.8+2x0.1-x(think way) 0.1-x (treated as dissociation) = 1.7e7 (0.8+2x) 2 x x ~ 0.1; 0.8+2x ~ 0.8 (x is very small) then = 1.7e7 x = = 9.2e-9 = [Ag + ] x *1.7e7 3.4e7 x e7 x – 0.1 = 0 x =[ -1.7e7+ {(1.7e7) 2 + 4*0.1}] / 2 = 0 (cannot solve) Small indeed

22 19 Heterogeneous & complex equilibria 22 Complex and precipitate formation-1 What is the maximum [Cl - ] before AgCl(s) forms in a solution containing 1.0 M NH 3 and 0.1 M AgNO 3 ? Solution : Previous slide showed [Ag+] = 9.2e-9 M in a solution when [NH 3 ] = 1.0 M K sp = 1.8e-10 for AgCl (know where to look up) Thus, the max. [Cl - ] = 1.8e-10 / 9.2e-9 = 0.02 M When no NH 3 is present, max. [Cl-] = 1.8e-10 / 0.1 = 1.8e-9 What is the maximum [Br - ] before AgBr(s) forms in a solution containing 1.0 M NH 3 and 0.1 M AgNO 3 ? K sp = 5e-13 for AgBr, Thus, the max. [Br - ] = 5e-13 / 9.2e-9 = 5.4e-5 M = 368 times 5.4e-5

23 19 Heterogeneous & complex equilibria 23 Complex and precipitate formation-2 What is the maximum [Br - ] in a solution suppose to containing 0.10 M AgNO 3 and 0.20 M Na 2 S 2 O 3 before AgBr(s) ( K sp = 5e-13) forms? Solution : K f = 2.9e13 for Ag(S 2 O 3 ) 2 3 – ; Ag(S 2 O 3 ) 2 3 – = Ag S 2 O 3 2 –, K = 1 / 2.9e13 = 3.4e xx2x 4x 3 = 3.4e-14; x = (3.4e-14*0.1) 1 / 3 = 1.5e-5 = [Ag + ] (0.1–x) small max [Br – ] = K sp / [Ag+] = 5e-13 / 1.5e-5 = 3.3e-8 (very low) What is [Br – ] = ? If [Na 2 S 2 O 3 ] is increased to 1.0 M? See next page

24 19 Heterogeneous & complex equilibria 24 Complex and precipitate formation-3 What is the maximum [Br - ] in a solution suppose to containing 0.10 M AgNO 3 and 1.0 M Na 2 S 2 O 3 before AgBr(s) ( K sp = 5e-13) forms? Solution : K f = 2.9e13 for Ag(S 2 O 3 ) 2 3 – K = 1 / 2.9e13 = 3.4e-14 Ag(S 2 O 3 ) 2 3 – = Ag S 2 O 3 2– 0.1-xx2x +0.8 (0.8+2x) 2 x = 3.4e-14, x = 3.4e-14*0.1/0.8 2 = 5.3e-14 = [Ag + ] (0.1-x) small max [Br – ] = 5e-13 / 5.3e-14 = 94 M, unrealistically large What is the maximum [I – ] in a solution containing 0.10 M AgNO 3 and 1.0 M Na 2 S 2 O 3 before AgI(s) ( K sp = 8e-17) forms? [I - ] = 8e-17 / 5.3e-14 = M; small compare to 0.1 M

25 19 Heterogeneous & complex equilibria 25 Chemical casserole Experiment AgNO 3 solution | Cl – AgCl (s) | 6.0 M NH 3 Ag(NH 3 ) Cl – | Br – AgBr (s) + NH 3, Cl – | S 2 O 3 2– Ag(S 2 O 3 ) 2 3– + Br – + NH 3, Cl – | I – AgI (s) Data Solid K sp AgCl1.8e-10 AgBr5.0e-13 AgI8.0e-17 Complex K f Ag(NH 3 ) e7 Ag(S 2 O 3 ) 2 3– 2.9e13 white ivory brown June

26 19 Heterogeneous & complex equilibria 26 Some complexes Ligands (Lewis bases): Negative ions, F -, Cl -, Br -, I -, OH -, CN -, SO 4 2-, RCOO -, … Neutral molecules: H 2 O, NH 3, NR 3, ROR, CO, C 5 H 5, PR 3, C 5 H 5 N, … Multidentate ligands: H 2 NCH 2 CH 2 NH 2, H 2 NCH 2 CH 2 NHCH 2 CH 2 NH 2, Ag(NH 3 ) 2 +, Cu(CN) 3 2- (Cu(I)), Cu(PPh 3 ) 2 Br, (NH 3 ) 2 PtCl 2 (cis & trans), Ni(CN) 5 3-,

27 19 Heterogeneous & complex equilibria 27 The heme and hemoglobin (a) The structure of heme is a planar porphoryin ring with iron at the center. (b) Four heme units and four coiled polypeptide chains are bonded together in a molecule of hemoglobin.

28 19 Heterogeneous & complex equilibria 28 Amphoteric hydroxides Amphoteric metal hydroxides react with both acids and bases. Examples: (M = Fe, Zn) M(H 2 O) 4 (OH) 2 + H + = M(H 2 O) 5 (OH) + M(H 2 O) 5 (OH) + + H + = M(H 2 O) 6 2+ M(H 2 O) 4 (OH) 2 + OH – = M(H 2 O) 3 (OH) 3 – M(H 2 O) 3 (OH) 3 – + OH – = M(H 2 O) 2 (OH) 4 2 – Aluminum hydroxide behave similarly: Al(H 2 O) 3 (OH) 3 + H + = Al(H 2 O) 4 (OH) 2 + Al(H 2 O) 4 (OH) H + = Al(H 2 O) 5 (OH) 2+ Al(H 2 O) 5 (OH) 2+ + H + = Al(H 2 O) 6 3+ Al(H 2 O) 3 (OH) 3 + OH – = Al(H 2 O) 2 (OH) 4 – Al(H 2 O) 2 (OH) 4 – + OH – = Al(H 2 O)(OH) 5 2– Al(H 2 O)(OH) 5 2– + OH – = Al(OH) 6 3 –

29 19 Heterogeneous & complex equilibria 29 Qualitative analysis of metal ions Mixture of metal ions add HCl group I Ag +, Hg 2 2+, Pb 2+ filtrate of soluble chloride chloride | add 0.3 M H + & H 2 S group II Cu 2+, Cd 2+, Hg 2+, Pb 2+ filtrate of soluble metal sulfide As 3+, Sb 3+, Bi 3+, Sn 4+ | | add OH - & H 2 S acid insoluble sulfide | (NH 4 OH) group III Mn 2+, Fe 2+, Co 2+, Ni 2+, Zn 2+ filtrate of soluble metal ions base insoluble sulfide | | add CO 3 2- or PO 4 3- Al(OH) 3, Cr (OH) 3 | | hydroxide | group IV Mg 2+, Ca 2+, Sr 2+, Ba 2+ filtrate of soluble metal ions ppt as carbonates or phosphates |K +, Na + group V

30 19 Heterogeneous & complex equilibria 30 Heterogeneous and complex ion equilibria Summary Calculate K sp Evaluate molar solubility (or in other units) from K sp Discuss concentrations of species in solution with two or more electrolytes (common ion effect, pH effect, separation by ppt) Predicts ppt (heterogeneous equilibria) Describe ligands, metal ions, complexes (ions), and formation constants, and dissociation constant Apply complex formation to explain solubility Be able to get out of traps (think in both directions)


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