Ch 15 Equilibria of Other Reaction Classes Dr

Ch. 15: Equilibria of Other Reaction Classes Dr. Namphol Sinkaset Chem 201: General Chemistry II

I. Chapter Outline I. III. IV. V. VI. Introduction Solubility Equilibria Predicting Precipitation The Common Ion Effect Lewis Acids and Bases Complex Ions

I. Last Aspects of Equilibria • In this chapter, we cover some final topics concerning equilibria. • Solubility can be reexamined from an equilibrium point of view. • Complex ions are introduced and their formation explained using equilibrium ideas.

II. Solubility • In 1 st semester G-chem, you memorized solubility rules and regarded compounds as either soluble or insoluble. • Reality is not as clear cut – there are degrees of solubility. • We examine solubility again from an equilibrium point of view.

II. Solubility Equilibrium • If we apply the equilibrium concept to the dissolution of Ca. F 2(s), we get: § Ca. F 2(s) Ca 2+(aq) + 2 F-(aq) • The equilibrium expression is then: § Ksp = [Ca 2+][F-]2 • Ksp is the solubility product constant, and just like any other K, it tells you how far the reaction goes towards products.

II. Some Ksp Values

II. Calculating Solubility • Recall that solubility is defined as the maximum possible concentration of a solute in a solution at a certain T and P (g solute/100 g water is common). • The molar solubility is the number of moles of a compound that dissolves in a liter of liquid. • Molar solubilities can easily be calculated using Ksp values.

II. Ksp Equilibrium Problems • Calculating molar solubility is essentially just another type of equilibrium problem. • You still set up an equilibrium chart and solve for an unknown. Pay attention to stoichiometry! Fe(OH)3(s) Fe 3+(aq) + 3 OH-(aq) Initial --- 0 0 Change --- +S +3 S Equil. --- S 3 S

II. Sample Problem • Which is more soluble: calcium carbonate (Ksp = 4. 96 x 10 -9) or magnesium fluoride (Ksp = 5. 16 x 10 -11)?

III. Precipitation • Ksp values can be used to predict when precipitation will occur. • Again, we use a Q calculation. § If Q < Ksp, solution is unsaturated. Solution dissolve additional solid. § If Q = Ksp, solution is saturated. No more solid will dissolve. § If Q > Ksp, solution is supersaturated, and precipitation is expected.

III. Sample Problems a) Will a precipitate form if 100. 0 m. L 0. 0010 M Pb(NO 3)2 is mixed with 100. 0 m. L 0. 0020 M Mg. SO 4? b) The concentration of Ag+ in a certain solution is 0. 025 M. What concentration of SO 42 - is needed to precipitate out the Ag+? Note that Ksp = 1. 2 x 10 -5 for silver(I) sulfate.

IV. The Common Ion Effect • The solubility of Fe(OH)2 is lower when the p. H is high. Why? * § Fe(OH)2(s) Fe 2+(aq) + 2 OH-(aq) • common ion effect: the solubility of an ionic compound is lowered in a solution containing a common ion than in pure water.

IV. Sample Problem • Calculate the molar solubility of lead(II) chloride (Ksp = 1. 2 x 10 -5) in pure water and in a solution of 0. 060 M Na. Cl.

IV. p. H and Solubility • As seen with Fe(OH)2, p. H can have an influence on solubility. • In acidic solutions, need to consider if H 3 O+ will react with cation or anion. • In basic solutions, need to consider if OH- will react with cation or anion.

IV. Sample Problems a) Will copper(I) cyanide be more soluble in acid or water? Why? b) In which type of solution is Ag. Cl most soluble: acidic or neutral?

V. A New Kind of Bond • How are ionic bonds formed? • Which atoms provide e-’s in a covalent bond? • There’s also a coordinate covalent bond (dative bond) whose formation is shown below.

V. Lewis Acids/Bases • You probably recognized previous reactions as acid/base even as we were discussing them from the perspective of e- lone pairs. • Another acid/base definition centers on pairs of e-’s. • A Lewis acid is an e- pair acceptor; a Lewis base is an e- pair donor.

V. Lewis Acid-Base Adducts • A Lewis acid-base adduct is a compound that contains a coordinate covalent bond. • Boron compounds are Lewis acids because they are e- deficient:

V. More Examples

V. Lewis Displacements • “Stronger” Lewis bases can displace weaker Lewis bases.

VI. Complex Ions • In aqueous solution, transition metal cations are usually hydrated. § e. g. Ag+(aq) is really Ag(H 2 O)2+(aq). § The Lewis acid Ag+ reacts with the Lewis base H 2 O. • Ag(H 2 O)2+(aq) is a complex ion. § A complex ion has a central metal bound to one or more ligands. § A ligand is a neutral molecule or an ion that acts as a Lewis base with the central metal.

VI. Formation Constants • Stronger Lewis bases will replace weaker ones in a complex ion. § e. g. Ag(H 2 O)2+(aq) + 2 NH 3(aq) Ag(NH 3)2+(aq) + 2 H 2 O(l) § For simplicity, it’s common to write Ag+(aq) + 2 NH 3(aq) Ag(NH 3)2+(aq) § Since this is an equilibrium, we can write an equilibrium expression for it.

VI. Formation Constants Ag+(aq) + 2 NH 3(aq) Ag(NH 3)2+(aq) • Kf is called a formation constant. (Its inverse is Kd, a dissociation constant. ) • What does the large Kf mean?

VI. Sample Formation Constants

VI. Calculations w/ Kf’s • Since Kf’s are so large, calculations with them are slightly different. • We assume the equilibrium lies essentially all the way to the right, and we calculate the “leak back. ” • This changes how we set up our equilibrium chart.

VI. Illustrative Problem • Calculate the concentration of Ag+ ion in solution when 0. 085 g silver(I) nitrate is added to a 250. 0 m. L solution that is 0. 20 M in KCN.

VI. Illustrative Problem Solution 1) First, we must identify the complex ion. § § § In solution, we will have Ag+, NO 3 -, and K+, and CN-. The complex ion must be made from Ag+ and CN-. Looking at table of Kf’s, we find that Ag(CN)2 - has Kf = 1 x 1021.

VI. Illustrative Problem Solution 2) Next, we need initial concentrations. § § Already know that [CN-] = 0. 20 M. We calculate the [Ag+].

VI. Illustrative Problem Solution 3) Now we set up our equilibrium chart. § Since Kf is so big, we assume the reaction essentially goes to completion. Ag+(aq) + 2 CN-(aq) Ag(CN)2 -(aq) Initial 0. 00200 M 0. 20 M 0 Change ≈ -0. 00200 ≈ -0. 00400 ≈ +0. 00200 x 0. 196 0. 00200 Equil.

VI. Illustrative Problem Solution 4) Finally, we solve for x. Thus, [Ag+] = 5 x 10 -23 M. It is very small, so our approximation is valid. Note that book would use 0. 20 for [CN-] in the calculation.

VI. Sample Problem • A 125. 0 -m. L sample of a solution that is 0. 0117 M in Ni. Cl 2 is mixed with a 175. 0 m. L sample of a solution that is 0. 250 M in NH 3. After the solution reaches equilibrium, what concentration of Ni 2+(aq) remains?

VI. Complex Ions & Solubility • Formation of complex ions enhances the solubility of some normally insoluble ionic compounds. • Typically, Lewis bases will enhance solubility. • e. g. Adding NH 3 to a solution containing Ag. Cl(s) will cause more Ag. Cl(s) to dissolve. Why? * Ag. Cl(s) Ag+(aq) + Cl-(aq) Ksp = 1. 77 x 10 -10 Ag+(aq) + 2 NH 3(aq) Ag(NH 3)2+(aq) Kf = 1. 7 x 107

VI. Complex Ion Formation