Q vs Ksp

Coordination Number The atom of the ligand that supplies the nonbonding electrons for the metal-ligand bond is the donor atom. The number of these atoms is the coordination number.

Coordination Number Some metals, such as chromium(III) and cobalt(III), consistently have the same coordination number (6 in the case of these two metals). The most commonly encountered numbers are 4 and 6.

Geometries There are two common geometries for metals with a coordination number of four: – Tetrahedral – Square planar

Geometries By far the most- encountered geometry, when the coordination number is six, is octahedral.

Polydentate Ligands Some ligands have two or more donor atoms. These are called polydentate ligands or chelating agents. In ethylenediamine, NH 2 CH 2 CH 2 NH 2, represented here as en, each N is a donor atom. Therefore, en is bidentate.

Polydentate Ligands Ethylenediaminetetraacetate, mercifully abbreviated EDTA, has six donor atoms.

Polydentate Ligands Chelating agents generally form more stable complexes than do monodentate ligands.

Figure The spectrochemical series. For a given ligand, the color depends on the oxidation state of the metal ion. For a given metal ion, the color depends on the ligand. I - < Cl - < F - < OH - < H 2 O < SCN - < NH 3 < en < NO 2 - < CN - < CO WEAKER FIELD STRONGER FIELD LARGER  SMALLER  LONGER SHORTER

Complex Ion Equilibria Ag + + NH 3 ↔ Ag(NH 3 ) + K 1 = 2.1 x 10 3 Ag(NH 3 ) + + NH 3 ↔ Ag(NH 3 ) 2 + K 2 = 8.2 x 10 3 What is the net reaction? K> 1, products are “favored” Metal ions add ligands one at a time in steps symbolized by equilibrium constants called: formation constants/stability constants

Figure The stepwise exchange of NH 3 for H 2 O in M(H 2 O) 4 2+.

Figure Test for the presence of a carbonate.

Will a Precipitate Form? In a solution, – If Q = K sp, the system is at equilibrium and the solution is saturated. – If Q < K sp, more solid will dissolve until Q = K sp. – If Q > K sp, the salt will precipitate until Q = K sp.

Sample Problem 19.10Predicting Whether a Precipitate Will Form PROBLEM:A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when L of 0.30 M Ca(NO 3 ) 2 is mixed with L of M NaF? PLAN:Write out a reaction equation to see which salt would be formed. Look up the K sp values in a table (Appendix C). Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < K sp. Remember to consider the final diluted solution when calculating concentrations. SOLUTION:CaF 2 ( s ) Ca 2+ ( aq ) + 2F - ( aq ) K sp = 3.2x mol Ca 2+ = L(0.30 mol/L) = mol [Ca 2+ ] = mol/0.300 L = 0.10 M mol F - = L(0.060 mol/L) = mol [F - ] = mol/0.300 L = M Q = [Ca 2+ ][F - ] 2 =(0.10)(0.040) 2 = 1.6x10 -4 Q is >> K sp and the CaF 2 WILL precipitate.

Sample Problem 19.11Using Molecular Scenes to Predict Whether a Precipitate Will Form PROBLEM:Four ionic solutions are represented in these molecular scenes (silver is gray and carbonate black and red.) (a) Which scene best represents the solution at equilibrium with the solid? (b) In which, if any, other scene(s) will additional solid be formed? (c) Explain how, if at all, addition of a small volume of concentrated strong acid affects the [Ag + ] and the mass of solid Ag 2 CO 3. PLAN:(a) Count ions present in each scene. (b) Compare Q sp and K sp. (c) Strong acid reacts to remove carbonate from solution. SOLUTION: (a) Scene 3 is the only one with a 2:1 silver to carbonate ion ratio. [Ag + ]:[CO 3 2- ] ratios: scene 1, 2:4; scene 2, 3:3; scene 3, 4:2; and scene 4, 3:4.

Sample Problem 19.11Using Molecular Scenes to Predict Whether a Precipitate Will Form (b) Calculating the ion products: (c) Write the equations: SOLUTION: Ag 2 CO 3 ( s ) 2Ag + ( aq ) + CO 3 2- ( aq ) Q sp = [Ag + ] 2 [CO 3 2- ] Scene 1: Q sp = (2) 2 (4) = 16 From scene 3, K sp = 32; Q sp > K sp only in scene 4, precipitate forms. Scene 3: Q sp = (4) 2 (2) = 32 Scene 2: Q sp = (3) 2 (3) = 27 Scene 4: Q sp = (3) 2 (4) = 36 Ag 2 CO 3 ( s ) 2Ag + ( aq ) + CO 3 2- ( aq ) CO 3 2- ( aq ) + 2H 3 O + ( aq ) H 2 CO 3 ( aq ) + 2H 2 O( l ) 3H 2 O( l ) + CO 2 ( g ) Lowers [CO 3 2- ]. More CO 3 2- forms. [Ag + ] increases and mass of Ag 2 CO 3 decreases.

Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture.

Sample Problem 19.12Separating Ions by Selective Precipitation SOLUTION: PROBLEM:A solution consists of 0.20 M MgCl 2 and 0.10 M CuCl 2. Calculate the [OH - ] that would separate the metal ions as their hydroxides. K sp of Mg(OH) 2 = is 6.3x ; K sp of Cu(OH) 2 is 2.2x PLAN:Both precipitates are of the same ion ratio, 1:2, so we can compare their K sp values to determine which has the greater solubility. It is obvious that Cu(OH) 2 will precipitate first so we calculate the [OH - ] needed for a saturated solution of Mg(OH) 2. This should ensure that we do not precipitate Mg(OH) 2. Then we can check how much Cu 2+ remains in solution. Mg(OH) 2 ( s ) Mg 2+ ( aq ) + 2OH - ( aq ) K sp = 6.3x Cu(OH) 2 ( s ) Cu 2+ ( aq ) + 2OH - ( aq ) K sp = 2.2x [OH - ] needed for a saturated Mg(OH) 2 solution = = 5.6x10 -5 M

Sample Problem 19.12Separating Ions by Selective Precipitation Use the K sp for Cu(OH) 2 to find the amount of Cu remaining. [Cu 2+ ] = K sp /[OH - ] 2 = 2.2x /(5.6x10 -5 ) 2 =7.0x M Since the solution was 0.10 M CuCl 2, virtually none of the Cu 2+ remains in solution.

Sample Problem 19.13Calculating the Concentration of a Complex Ion SOLUTION: PROBLEM:An industrial chemist converts Zn(H 2 O) 4 2+ to the more stable Zn(NH 3 ) 4 2+ by mixing 50.0 L of M Zn(H 2 O) 4 2+ and 25.0 L of 0.15 M NH 3. What is the final [Zn(H 2 O) 4 2+ ]? K f of Zn(NH 3 ) 4 2+ is 7.8x10 8. PLAN:Write the reaction equation and K f expression. Use a reaction table to list various concentrations. Remember that components will be diluted when mixed as you calculate final concentrations. It is obvious that there is a huge excess of NH 3 and therefore it will drive the reaction to completion. Zn(H 2 O) 4 2+ ( aq ) + 4NH 3 ( aq ) Zn(NH 3 ) 4 2+ ( aq ) + 4H 2 O( l ) K f = [Zn(NH 3 ) 4 2+ ] [Zn(H 2 O) 4 2+ ][NH 3 ] 4 [Zn(H 2 O) 4 2+ ] initial = (50.0 L)( M) 75.0 L = 1.3x10 -3 M [NH 3 ] initial =(25.0 L)(0.15 M) 75.0 L = 5.0x10 -2 M

Sample Problem 19.13Calculating the Concentration of a Complex Ion Zn(H 2 O) 4 2+ ( aq ) + 4NH 3 ( aq ) Zn(NH 3 ) 4 2+ ( aq ) + 4H 2 O( l ) Concentration (M) Initial Change Equilibrium 1.3x x ~(-1.3x10 -3 )~(-5.2x10 -3 ) Since we assume that all of the Zn(H 2 O) 4 2+ has reacted, it would use 4 times its amount in NH 3. [NH 3 ] used = 4(1.3x10 -3 M) = 5.2x10 -3 M ~(+1.3x10 -3 )- x4.5x x10 -3 [Zn(H 2 O) 4 2+ ] remaining = x(a very small amount) - K f = [Zn(NH 3 ) 4 2+ ] [Zn(H 2 O) 4 2+ ][NH 3 ] 4 = 7.8x10 8 = (1.3x10 -3 ) x(4.5x10 -2 ) 4 x = 4.1x10 -7 M

Sample Problem 19.14Calculating the Effect of Complex-Ion Formation on Solubility SOLUTION: PROBLEM:In black-and-white film developing, excess AgBr is removed from the film negative by “hypo”, an aqueous solution of sodium thiosulfate (Na 2 S 2 O 3 ), through formation of the complex ion Ag(S 2 O 3 ) Calculate the solubility of AgBr in (a) H 2 O; (b) 1.0 M hypo. K f of Ag(S 2 O 3 ) 2 3- is 4.7x10 13 and K sp AgBr is 5.0x PLAN:Write equations for the reactions involved. Use K sp to find S, the molar solubility. Consider the shifts in equilibria upon the addition of the complexing agent. AgBr( s ) Ag + ( aq ) + Br - ( aq ) K sp = [Ag + ][Br - ] = 5.0x S = [AgBr] dissolved = [Ag + ] = [Br - ]K sp = S 2 = 5.0x ; S = 7.1x10 -7 M(a) (b)AgBr( s ) Ag + ( aq ) + Br - ( aq ) Ag + ( aq ) + 2S 2 O 3 2- ( aq ) Ag(S 2 O 3 ) 2 3- ( aq ) AgBr( s ) + 2S 2 O 3 2- ( aq ) Br - ( aq ) + Ag(S 2 O 3 ) 2 3- ( aq )

Sample Problem 19.14Calculating the Effect of Complex-Ion Formation on Solubility K overall = K sp x K f = [Br - ][Ag(S 2 O 3 ) 2 3- ] [S 2 O 3 2- ] 2 = (5.0x )(4.7x10 13 )= 24 AgBr( s ) + 2S 2 O 3 2- ( aq ) Br - ( aq ) + Ag(S 2 O 3 ) 2 3- ( aq )Concentration (M) Initial Change Equilibrium S S 00 +S+S+S+S SS K overall = S2S2 ( S) 2 = 24; S S = (24) 1/2 S = [Ag(S 2 O 3 ) 2 3- ] = 0.45 M