A complex ion is a polyatomic cation or anion composed of a central metal ion to which ligands are bonded.

Transcription

1 Equilibria Involving Complex Ions Complex Ion Equilibria and Solubility A complex ion is a polyatomic cation or anion composed of a central metal ion to which ligands are bonded. Coordination compounds are substances containing complex ions Coordination complex ions or complexes are often called Werner complexes 1

2 In 1893 Alfred Werner first proposed the structure of [Co(NH 3 ) 6 ]Cl 3 as we know it today. Alfred Werner Nobel Prize (1913) "In recognition of his work on the linkage of atoms in molecules by which he has thrown new light on earlier investigations and opened up new fields of research especially in inorganic chemistry".

3 Equilibria Involving Complex Ions AgCl(s) + NH 3 (aq) [Ag(NH 3 ) ] + (aq) + Cl (aq) Complex ion formation: dissolution of AgCl(s) in NH 3 Complex Ion Equilibria and Solubility The formation constant or stability constant (K f ) is the equilibrium constant for the complex ion formation. Co + (aq) + 4Cl - (aq) - CoCl 4 (aq) Co(H O) 6 + K f = - CoCl 4 - [CoCl 4 ] [Co + ][Cl - ] 4 HCl K f stability of complex 3

4 Two sample problems follow that were done on the class whiteboard. 4

5 A00mole 0.0-mole quantity of CuSO 4 is added to a liter of 1.0 M NH 3 solution. What is the concentration of Cu + ions at equilibrium? Strategy The addition of CuSO 4 to the NH 3 solution results in complex ion formation K f = 5.0x10 13 Cu + (aq) +4NH 3 (aq) Cu(NH ) (aq) 3 4 Reaction lies mostly to the right. Concentration of Cu + will be very small at equilibrium. As a good approximation, we can assume that essentially all the dissolved Cu + ions end up as Cu(NH ) ions

6 SLOW DOWN AND ASK AND DO THE FOLLOWING STEPS: How many moles of NH 3 will react with 0.0 mole of Cu +? How many moles of Cu(NH 3 ) 4 + will be produced? Recall: a very small amount of Cu + will be present at equilibrium. Set up the K f expression for the equilibrium Cu + (aq) + 4NH 3 (aq) Cu(NH 3 ) 4 + (aq) to solve for [Cu + ]. Solution Strategy The amount of NH 3 consumed in forming the complex ion is 4 x 0.0 mol, or 0.80 mol. (Note that 0.0 mol Cu + is initially present in solution and four NH 3 molecules are needed to form a complex ion with one Cu + ion.). The concentration of NH 3 at equilibrium is therefore ( ) mol/l soln or 0.40 M, and that of Cu(NH 3 ) + 4 is 0.0 mol/l soln or 0.0 M, the same as the initial concentration of Cu +. [There is a 1:1 mole ratio between Cu + and Cu(NH + 3 ) 4 ]. Because Cu(NH 3 ) 4 + does dissociate to a slight extent, we call the concentration of Cu + at equilibrium x and write the following set of formation constant equation: 6

7 Given that x = [Cu + ] Solving for x and keeping in mind that the volume of the solution is 1 L, we obtain Check x = [Cu + ] = 1.6 x M The small value of [Cu+] at equilibrium, compared with 0.0 M, certainly justifies our approximation. Calculate the molar solubility of AgCl in a 1.0 M NH 3 solution. Recall that AgCl(s) + NH 3 (aq) [Ag(NH 3 ) ] + (aq) + Cl (aq) 7

8 Strategy AgCl is only slightly soluble in water AgCl(s) Ag + (aq) +Cl - (aq) The Ag + ions form a complex ion with NH 3 (see Table 16.4) Ag + (aq) + NH 3 (aq) Ag(NH ) 3 Combining these two equilibria will give the overall equilibrium for the process. Solution Step 1: Initially, the species in solution are Ag + and Cl - ions and NH 3. The reaction between Ag + and NH 3 produces the complex ion Ag(NH ). 3 Step : The equilibrium reactions are AgCl(s) Ag + (aq) + Cl - (aq) K sp = [Ag + ][Cl - ] = 1.6 x Ag + (aq) + NH Ag(NH 3 (aq) ) (aq) Overall: AgCl(s) + NH 3 (aq) 3 Kf + 3 [Ag(NH ) ] 7 = = [Ag ][NH 3] Ag(NH ) (aq) + Cl - (aq) 3 8

9 The equilibrium constant K for the overall reaction is the product of the equilibrium constants of the individual reactions. [Ag(NH 3) ][Cl ] K= KspKf = [NH ] = (16 10 ( )(1.5 )( ) = Let s be the molar solubility of AgCl (mol/l). We summarize the changes in concentrations that result from formation of the complex ion as follows: AgCl(s) + NH 3 (aq) Ag(NH ) (aq) + Cl-(aq) Initial (M): Change (M): -s -s +s +s Equilibrium (M): (1.0 s) s s 3 The formation constant for Ag(NH 3) is quite large, so most of the silver ions exist in the complexed form. In the absence of ammonia we have, at equilibrium, [Ag + ] = [Cl - ]. As a result of complex ion formation, however, we can write[ Ag(NH ) 3 ] = [Cl - ]. 9

10 Step 3: ()() s s K= (1.0 - s ) 3 s.4 10 = (1.0 - s) Taking the square root of both sides, we obtain s = s s = Μ Step 4: At equilibrium, mole of AgCl dissolves in 1 L of 1.0 M NH 3 solution. Check The molar solubility of AgCl in pure water is 1.3 x 10-5 M. Thus, the formation of the complex ion Ag(NH 3) enhances the solubility of AgCl. 10

11 Effect of Complexation on Solubility AgNO 3 + NaCl Add NH 3 AgCl Ag(NH 3 ) + 11

12 Qualitative Analysis of Cations 1

13 Outline of a qualitative cation analysis 13