Solubility Silberberg 13.2, 13.3, 13.4, 19.3, 19.4 Problem book 8.63-8.82 ote on Units ncentrations of dissolved substances (1) mass % = (grams solute/grams solution) x 100 e.g., 5.0 g sugar (teaspoon) in a cup (250 ml) = 2.0 % 5.0 g salt in a cup = 2.0 % (2) molarity = moles solute / litre of solution [sugar] = 5.0 g / (342 g mol -1 x 0.25 L) = 0.058 M [a] = 5.0 g / ( 58.5 g mol -1 x 0.25 L) = 0.34 M (3) ppm = mass % x 10 000 Solubility Equilibria - how much dissolves? Solubility Equilibria - how much dissolves? Amount dissolved = Solubility Saturated solution: where Maximum amount of solute is dissolved Equilibrium between solution and some undissolved solid Molecular solutes Amount dissolvedequals solubility equilibrium constant e.g.,for sucrose at 25 o C:» 6 M = [products] / [reactants] = [sugar(aq)] / [sugar(s)] = [sugar(aq)] solute + solvent solution Ionic solutes Amount dissolveddoes T equal MX(s) M + (aq) + X (aq) = [M + ][X ] 1
Solubility Product - Example: How much Ag dissolves in water at 298 K? (Ag) = 1.6 x 10 10 M 2 every x moles of Ag produces x moles of Ag + and x moles of Solubility Product of 2:1 Salts Solubility product equilibrium constants differ between 1:1 and 2:1 electrolytes Example: Pb 2 (s) Pb 2+ (aq) + 2 (aq) for x moles dissolved: x 2x Ag(s) Ag + (aq) + (aq) x x = [Pb 2+ ] [ ] 2 = [x] [2x] 2 (units of are M 3) = [Ag + ][ ] = 1.6 x 10 10 M 2 = x 2 \ x = 1.3 x 10 5 M (solubility in water) Question: What is the solubility of AgI at 298 K? ( (AgI) = 8 x 10 17 M 2 ) (Pb 2 ) = 1.6 x 10 +5 M 3 = x. (2x) 2 = 4x 3 \ x = 1.6 x 10 2 M Solubility Product of 2:1 Salts Solubility product equilibrium constants differ between 1:1 and 2:1 electrolytes Example: A saturated solution of Pb 2 at 298 K has 3.2 x 10 2 M what is of Pb 2? = (1.6 x 10 2 )(3.2 x 10 2 ) 2 = 1.6 x 10 5 M 3 Question: What is the solubility of Ag 2 S? ( (Ag 2 S) = 6 x 10 50 M 3 ) Relative Solubilities Care must be taken when comparing the relative solubilities of a group of salts using the values. AgI = 1.5 x 10 16 x = ( ) 1/2 = 1.2 x 10 8 M CuI = 5.0 x 10 12 x = ( ) 1/2 = 2.2 x 10 6 M CaS 4 = 6.1 x 10 5 x = ( ) 1/2 = 7.8 x 10 3 M = [cation][anion] if x is the solubility in moles/litre then x = ( ) 1/2 CuS = 8.5 x 10 45 = x 2 x = 9.2 x 10 23 M Ag 2 S = 1.6 x 10 49 = (4x 3 ) x = 3.4 x 10 17 M Bi 2 S 3 = 1.1 x 10 73 = (4x 2 *27x 3 ) x = 1.0 x 10 15 M of CuS is larger, but more Bi 2 S 3 dissolves 2
mmon Ion Effect Example: How much Ag dissolves in 0.1 M H? Same equilibrium constant, but solubility is suppressed by the effect of common ion on equilibrium. Ag(s) Ag + (aq) + (aq) x (0.1+ x)» 0.1 = [Ag + ][ ] = 1.6 x 10 10 M 2 = [x][0.1 + x]» 0.1x x = 1.6 x 10 9 M (This is << than 10 5 M in pure H 2 ) ph and Solubility Mg(H) 2 (s) ( = 2 x 10 13 M 3 ) Mg 2+ (aq) + 2H (aq) What is the solubility in water at ph = 2 (~stomachacid)? in water, = 2 x 10-13 M 3 = 4x 3 x = 3 x 10 5 M at ph = 2, ph = 12, [H ] = 1 x 10 12 M = 2 x 10 13 M 3 = x(1 x 10 12 ) 2 x = 2 x 10 11 M Question: What is the solubility of AgBr in 0.05 M sodium bromide solution at 298 K? ( (AgBr) = 5 x 10 13 M 2 ) Question: Al(H) 3 is used to clarify drinking water - how much Al remains in water at ph 6.6? ( = 3 x 10 34 M 4 ) Ion Product, Q If Q > then a precipitate forms Ion Product, Q If Q > then a precipitate forms Ag(s) Ag + (aq) + (aq) Ag (s) Ag + (aq) + (aq) Example: Does a precipitate of Ag(s) form if equal volumes of 1 mm H and 1 mm Ag 3 solutions are mixed? (B: ncentration halved after mixing!) [Ag + ] = 5 x 10 4 M and [ ] = 5 x 10 4 M (after mixing) Q = [Ag + ][ ] = 2.5 x 10 7 M 2 > = 1.6 x 10 10 M 2 \ Ag(s) precipitates, i.e., reaction moves to left toward equilibrium. Example: What about equal volumes of 2.0 x 10 5 M solutions? [Ag + ] = 1.0 x 10-5 M and [ ] = 1.0 x 10 5 M (after mixing) Q = [Ag + ][ ] = 1.0 x 10 10 M 2 < = 1.6 x 10 10 M 2 \ Ag(s) does not precipitate. 3
Application: : Barium Sulfate for X-X Ray Imaging Ba 2+ is very toxic of BaS 4 = 1.1 x 10 10 How much is soluble: (a) in pure water? (b) in 0.1 M a 2 S 4? [Ba 2+ ] = (1.1 x 10 10 ) 1/2 ~ 1 x 10 5 M [Ba 2+ ][S 4 2 ] = 1.1 x 10 10 [Ba 2+ ] = 1.1 x 10 9 Application: : Calcium xalate in Kidney Stones of Ca(C 2 4 ) = 1.96 x 10 8 [Ca 2+ ][C 2 4 2 ] = 1.96 x 10 8 C 2 (H) 2 C 2 (H) (pk a = 1.23) C 2 (H) C 2 () 2 2 (pk a = 4.12) Drink lots of water Avoid dairy products (calcium) Avoid antacids (calcium) Don t take huge quantities of vitamin C Avoid foods rich in oxalic acid (tea, chocolate, rhubarb) Why is it so? Enthalpy of Hydration The smaller the ionic radius, the more energy is released by surrounding an ion with water molecules The more highly charged the ion, the more energy is released D H hyd - Always favours dissolution Lattice Energy (Silberberg 9.6) Energy released when crystal is formed from free ions DH lat cation charge x anion charge q 2 cation radius + anion radius r DH latt Always opposes dissolution Entropy of Hydration Solid Dissolved ions DG hyd = DH hyd - DH lat - TDS hyd Silberberg, p.499 DS hyd Almost Always favours dissolution 4
Trends in Ionic Heats of Hydration Trends in Ionic Heats of Hydration Ion Radius (pm) H hyd (kj/mol) Ion Radius (pm) H hyd (kj/mol) Li + a + 76 102 510 410 F 133 181 431 313 K + 138 336 Br 196 284 Cs + 167 282 I 220 247 Mg 2+ 72 1903 Ca 2+ 100 1591 Ba 2+ 135 1317 Important Things to Know: How to do solubility product calculations What the mmon Ion Effect is How to convert from one unit of solubility to another Why milk chocolate is worse for you than dark chocolate mplexes, aka ordination mpounds Silberberg 18.9, 19.4, 23.4 Problem book 8.83-8.91 Reaction of a metal ion (Lewis acid) with anions or molecules (Lewis base ~ ligands) forms complex species (coordination compounds) e.g., Ag + + 2 H 3 Ag(H 3 ) 2 + 5
mplexes, aka ordination mpounds ordination umber and Geometry Reaction of a metal ion with anions or molecules (ligand(s)) forms complex species (coordination compound) e.g., Ag + + 2 H 3 Ag(H 3 ) 2 + A complex is a chemical species whose components are capable of separate existence: e.g. Ag(H 3 ) 2 + distinct from Ag + and H 3, (not C 3 2 because 2 and C 4+ do not exist independently) Metals are usually transition metals (but remember we had Mg 2+ ) Ligands have at least one atom with a lone (loan) pair: e.g, H 2, H 3,, C The number of ligands attached to the metal is called the coordination number CRD.. GEMETRY CMMETS / Examples 2 LIEAR uncommon [Ag(H3)2]+ 4 PLAAR electronic cause* [Cu(H3)4]2+ 4 TETRAHEDRAL common [Zn(H3)4]2+ 6 CTAHEDRAL common [Cr(H3)6]3+ d8 Ligands Some ligands have more than one atom with lone pairs that can be bonded to the metal ion these are CHELATES ( cela = aw) Monodentate ligands ( denta = tooth) e.g. H 2, H 3, Bidentate ligands can form 2 bonds e.g. ethylenediamine (en) Polydentate ligands can form more than two bonds some can form as many as 6, e.g. EDTA ethylenediaminetetraacetate (hexadentate) - forms very stable complexes with most metal ions and is used as a scavenger to remove TXIC heavy metals such as lead from the human body. H2 H2 H2 H2 Fe H2 H2 3+ H2 H2 H2 H2 Fe H 2 H2 3+ Fe - Isomers ordination isomerism (which ligands are in coordination sphere and which outside) Example: Cr 3. 6H 2 [Cr(H 2 ) 6 ] 3 hexaaquachromium(iii) chloride [Cr(H 2 ) 5 ] 2. H 2 pentaaquachlorochromium(iii) chloride water (1/1) [Cr(H 2 ) 4 2 ]. 2H 2 tetraaquadichlorochromium(iii) chloride water (1/2) [Cr(H 2 ) 3 3 ]. 3H 2 triaquatrichlorochromium(iii) water (1/3) 6
Isomers Geometric isomers: Four coordinate: cis- and trans- [Pt(H 3 ) 2 2 ] H 3 Pt H 3 Pt H 3 H 3 Isomers ptical isomers: compare cis-[cr(h 3 ) 4 2 ] + and cis-[cr(en) 2 2 ] + H 3 + H 3 H 3 H 3 H 2 + H 2 H 2 H 2 Six coordinate: cis- and trans- [(H 3 ) 4 2 ] + [M(en) 3 ] n+ complexes have optical isomers H 3 H 3 + H 3 + H 3 H H 3 3 H 3 H 3 H 2 H 2 H 2 3+ H 2 H 2 3+ H 2 H H H 2 2 2 H 2 H H 2 2 Types of Isomerism ordination Geometric ptical Know what they are! Know at least one example of each! Know how to name them (more on naming to come...) 7
Stability nstant (K( stab ) K stab = [mplex] [Metal Ion] [Ligand] n e.g., [AlF 6 ] 3 Kstab = [AlF6 ]3 = 4 x 10 19 [Al 3+ ][F ] 6 Stability nstant (K( stab ) Example: A solution is prepared by dissolving Ag 3 (0.01 mol) in a 1.00 M water solution of KC (500 ml) and adding enough water to make 1.00 L of solution. Calculate the equilibrium [Ag + ] given K stab [Ag(C) 2 ] = 10 20.0 M 2. Ag + + 2C [Ag(C) 2 ] initial / M 0.01 0.500 change ~ 0.01 0.02 0.01 equilibrium / M x 0.480 0.01 [Ag(C) 2] K 10 20.0 stab = = + 2 [Ag ][C] [Ag(C) ] 0.01 [Ag + ] = 2 = = 4 10 22 M 10 20.0 [C] 2 10 20.0 (0.48) 2 Question A solution is prepared by dissolving Zn( 3 ) 2 (0.10 mol) and ammonia (3.00 mol) in water and making 1.00 L of solution. Calculate the equilibrium concentration of Zn 2+ (aq) given K stab [Zn(H 3 ) 4 ] 2+ = 1 x 10 9 M 4. initial / M 0.10 3.00 Zn 2+ + 4H 3 [Zn(H) 4 ] 2+ change ~ 0. 10 0.40 0.10 equilibrium / M x 2.60 0.10 K stab = [Zn(H 3 ) 4 2+ ] [Zn 2+ ][H 3 ] 4 = 1 109 Biologically Important mplexes We have already had haemoglobin (Fe 2+ ) and chlorophyll (Mg 2+ ) Also: Zinc many enzymes including carbonic anhydrase, carboxypeptidase A, alcohol dehydrogenase balt xylose isomerase, cobalamin (vitamin B 12 ) Manganese, Chromium, Vanadium... [Zn] 2+ = [Zn(H 3 ) 2+ 4 ] 1 10 9 [H 3 ] 4 = 0.1 45.7 10 9` = 2.2 10 11 M 8
Biologically Important mplexes omenclature ame cation, then anion, as separate words Ligands, then metal, in same word Purple Acid Phosphatase umber of ligands as Greek prefixes (di-, tri-, tetra-, penta-, hexa-) xidation state in Roman numeral in parentheses after name of metal e.g. [Ag(H 3 ) 2 ]( 3 ) diamminesilver(i) nitrate Anionic ligands end in '-o'; e.g. chloro, hydroxo, harpo eutral ligands named as molecule, except H 2 (aqua), H 3 (ammine), C (carbonyl) and (nitrosyl) omenclature (pp.1016-10171017 Silberberg) Ligands named in alphabetical order e.g. [(H 3 ) 5 ](S 4 ) pentaamminechlorocobalt(iii) sulfate Anionic complexes end in '-ate e.g. K 3 [Cr 6 ] potassium hexachlorochromate(iii) Some metals in anionic complexes use Latinate names: Fe (ferrate), Cu (cuprate), Ag (argentate), Pb (plumbate, Au (aureate), Sn (stannate) mplex ligands require Latin prefixes bis, tris, e.g. bis(ethylenediamine) for (en) 2 Examples: [(H 2 ) 6 ]C 3 [Cu(H 3 ) 4 ]S 4 (H 4 ) 3 [FeF 6 ] K 4 [Mn(C) 6 ] [(en) 2 2 ] 3 hexaaquacobalt(ii) carbonate tetraamminecopper(ii) sulfate ammonium hexafluoroferrate(iii) potassium hexacyanomanganate(ii) dichlorobis(ethylenediamine)cobalt (III) nitrate 9
ame: K 2 [ 4 ] [Cu (H 3 ) 2 2 ] [Ag(H 3 ) 2 ] + [Fe(H 2 ) 6 ] 3 ot on exam Incoming ligands avoid electrons Ligand : d-orbital repulsion Half-filled orbitals repel more than filled orbitals Big, electron rich ligands repelled more than little ones Fig. 23.17 Fig. 23.18 10
Hybrid rbitals and Bonding in the ctahedral [Cr(H 3 ) 6 ] 3+ Ion Hybrid rbitals and Bonding in the Square Planar [i(c) 4 ] 2- Ion Fig. 23.13 unpaired electrons in d xy, d xz, d yz orbitals ligands approach along d z 2, d x 2 -y 2 axes Fig. 23.14 unpaired electrons on d z 2, d x 2 -y 2 axes ligands approach along x-y plane d x 2 -y 2 becomes less favourable; d z 2 becomes more - electrons pair up in d z 2 orbital Hybrid rbitals and Bonding in the Square Planar [i(c) 4 ] 2- Ion But what do you really need to know? Fig. 23.14 unpaired electrons on d z 2, d x 2 -y 2 axes ligands approach along x-y plane d x 2 -y 2 becomes less favourable; d z 2 more - electrons pair up in d z 2 orbital doesn t happen for - - [i() 4 ] 2- tetrahedral K stab calculations The types of isomerism ordination number and geometry, and what metal usually has what coordination number How to name coordination compounds How inorganic chemists can weasel their way into Biochemistry departments 11