Lecture 11. Solubility of Ionic Compounds in Water

Transcription

1 Lecture 11 Solubility of Ionic Compounds in Water Soluble Compounds Exceptions Acetates, Nitrates, Chlorates Na +, K +, NH 4 + compounds None None Halides ( Cl -, Br -, I - ) Ag +, Hg 2 2+, Cu +, Pb 2+ Sulfates ( SO 4 2- ) Ca 2+, Sr 2+, Ba 2+, Pb 2+, Ag +, Hg 2 2+ Insoluble Compounds Exceptions Sulfides ( S 2- ) Carbonates ( CO 2-3 ) Phosphates ( PO 3-4 ) Hydroxides ( OH - ) NH + 4, groups IA and IIA ions NH + 4 and group IA ions NH + 4 and group IA ions NH + 4, Ba 2+ and group IA ions Week 6-1

2 Molar Solubility The molar solubility s, of a saturated solution (s), of a precipitate is defined as the number of moles of that precipitate(dissolved) per litre of solution. (s) signifies that the solution is saturated i.e. solid ppt. is visible on the bottom of the container s moles of dissolved ppt. per litre e.g. Silver Chloride ppt. AgCl (s) º Ag + + Cl - s s and the solubility product, K sp = [Ag + ] [Cl - ] solid AgCl is treated as unity Week 6-2

3 General expression for solubility of a salt M x A y, where the ions do not react with water. M x A y º xm y+ + ya x- mass balance gives y[m y+ ] = x[a x- ] if solubility = s then from the equilibrium equation, [M y+ ] = xs and [A x- ] = ys now K sp = [M y+ ] x [A x- ] y = [xs] x [ys] y e.g. for PbI 2, K sp = [Pb 2+ ] 1 [I - ] 2 = [1s] 1 [2s] 2 for Hg 2 Cl 2, [Hg 2 2+ ] is treated as a monomer i.e. the two Hg + are held together by sigma bonds Week 6-3

4 Solubility Equilibria, K sp - the solubility product is the equilibrium constant for the reaction in which a sparingly soluble solid salt dissolves to give its constituent ions. e.g. for a saturated solution of PbI 2, K sp = 7.9 x 10-9 PbI 2 (s) º Pb I - K sp = [Pb 2+ ] [I - ] 2 This is straight forward since Pb 2+ & I - do not react with water. ˆ [H + ] = [OH - ] and ph = 7 The mass balance and K sp are used to obtain solubility i.e. - mass balance: 2 [Pb 2+ ] = [I - ] - and K sp = 7.9 x 10-9 = [Pb 2+ ] ( 2 [Pb 2+ ] ) 2 = 4 [Pb 2+ ] 3 [Pb 2+ ] = 1.25 x 10-3 M aside PbI 2 (s) º Pb I - s 2s Week 6-4

5 Another example where there is no reaction of the ions with water: Hg 2 Cl 2 (s) º Hg 2 2+ (aq) + 2Cl - (aq) charge balance: 2 [Hg 2 2+ ] + [H + ] = [Cl - ] + [OH - ] mass balance: [Cl - ] = 2 [Hg 2 2+ ] Week 6-5

6 Now consider a saturated solution of silver phosphate, Ag 3 PO 4 (s) Ag 3 PO 4 (s) º 3 Ag + + PO 4 3- mass balance:?? previously asked to determined PO 4 3- reacts with water to give HPO H 2 PO 4! + H 3 PO 4 and the correct mass balance is charge balance: [Ag + ] + [H + ] = [OH - ] + 3[PO 4 3- ] + 2[HPO 4 2- ] + [H 2 PO 4! ] The mass balance equation is critical in solubility calculations. Week 6-6

7 Lecture 12 Factors affecting solubility 1. Temperature 2. Effect of other ions in solution 3. Nature of the solvent 1. The solubility of most of the inorganic salts that we are interested in, increase with temperature. 2a. The Common Ion Effect - Le Chatelier's principle - If Cl - is added to the following reaction, the solubility of Hg 2 Cl 2 (s) MUST decrease. Hg 2 Cl 2 (s) º Hg 2 2+ (aq) + 2Cl - (aq) e.g. in water, s = 6.7 x 10-7 M in M NaCl, s = 1.3 x M Week 6-7

8 2b. Separation by Precipitation Is it possible to "completely" separate Pb 2+ and Hg 2 2+, each at a concentration of 0.010M, by selective precipitation with I -? i. Define "complete" separation as 0.01% of removed ion left in solution, i.e % of one ion removed leaving the other ion at 0.01 M. Thus we want one ion concentration to be: and the other ion = M. Which is which? ii. K sp values are: PbI 2 (s) º Pb I x 10-9 Hg 2 I 2 (s) º Hg I x 10-28?? Week 6-8

9 iii. How much I - (concentration) should we add?? Hg 2 I 2 (s) º Hg I - initial equilibrium 1.0 x 10-6 x iv. Will any Pb 2+ precipitate? Q = (0.010) (1.0 x ) 2 = 1.0 x Q < K sp ( 7.9 x 10-9 ) ˆ No precipitation. Week 6-9

10 2c Effect of Complex Formation The solubility of a slightly soluble salt depends on whether the ions of the salt can form soluble complexes e.g. i. silver halides are much more soluble in ammonia solutions than in water Ag + + NH 3 º Ag(NH 3 ) + K 1 = 2.0 x 10 3 Ag(NH 3 ) + + NH 3 º Ag(NH 3 ) 2 + K 2 = 6.3 x 10 3 ii. only a small excess of precipitating agent is used in AgCl precipitations, because Ag + + Cl - º AgCl (aq) AgCl (aq) + Cl -! º AgCl 2 AgCl! 2 + Cl - 2- º AgCl 3 s = [Ag + ] + [AgCl] + [AgCl! 2 ] + [AgCl 2-3 ] K sp complex formation Cumulative formation constant, $ i i.e. $ 1 = K 1 $ 2 = K 1 K 2 $ 3 = K 1 K 2 K 3 Week 6-10

11 Log [AgCl Total] -2.0E -3.0E -4.0E -5.0E -6.0E -7.0E -8.0E -9.0E Complex Formation Ksp alone -1.0E E -5.0E -4.0E -3.0E -2.0E -1.0E 0.0E+ 00 Log [Cl-] Solubility of AgCl in solutions containing excess Cl -. The curved line include complex formation species. The straight line represent solubility without the complex species. AgCl (s) º Ag + + Cl - K sp Ag + + Cl - º AgCl (aq) K 1 AgCl(aq) + Cl - º AgCl 2! K 2 K 3 AgCl! 2 + Cl - 2- º AgCl 3 s = [Ag + ] + [AgCl] + [AgCl 2! ] + [AgCl 3 2- ] Week 6-11