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# Chapter 18 - PowerPoint PPT Presentation

Chapter 18. Solubility and Complex-Ion Equilibria. Overview. Solubility Equilibria Solubility Product Constant Solubility and Common Ion Effect Precipitation Calculations Effect of pH on Solubility Complex-Ion Equilibria Complex Ion Formation Complex Ions and Solubility

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### Chapter 18

Solubility and Complex-Ion Equilibria

• Solubility Equilibria

• Solubility Product Constant

• Solubility and Common Ion Effect

• Precipitation Calculations

• Effect of pH on Solubility

• Complex-Ion Equilibria

• Complex Ion Formation

• Complex Ions and Solubility

• Application of Solubility Equilibria

• Qualitative analysis of metal ions

• Solubility of a solid treated as with other equilibria. Solution is saturated. No more solid will dissolve since dynamic equilibrium.

AgCl(s)  Ag+(aq) + Cl(aq) Ksp = [Ag+][Cl]

• Solid not included in the equilibrium expression.

• MyXz(s)  yM+p(aq)+zXq(aq) Ksp=[M+p]y[Xq]z

where Ksp = solubility product.

E.g. determine the equilibrium expression for each: PbCl2, Ag2SO4,Al(OH)3.

• Ksp can be determined if the solubility is known.

E.g. Determine Ksp for silver chromate (Ag2CrO4) if its solubility in water is 0.0290 g/L at 25C.

• Determine molar solubility.

• Determine Ksp.

E.g. 2 Determine Ksp of CaF2 if its solubility is 2.20x104M.

• Can be determined by using stoichiometry to express all quantities in terms of one variable- solubility, x. i.e. for the reaction. Use equilibrium table to write concentration of each in terms of the compound dissolving

E.g. determine the solubility of PbCl2 if its Ksp = 1.2x105

PbCl2(s)  Pb2+(aq) + 2Cl(aq)

E.g.1 Determine solubility of AgCl if its Ksp = 1.8x1010M2.

E.g.2 Determine solubility of Ag2CO3 if its Ksp = 8.1x1012M3.

E.g.3 Determine solubility of Fe(OH)3 if its Ksp = 4x1038M4

• The common–Ion effect(Remember LeChatelier’s Principle)

E.g. Determine solubility of PbCl2 (Ksp = 1.2x105)in 0.100M NaCl.

• Write equilibrium table in terms of x and [Cl]

a common–ion reduces the solubility of the compound.

• Assume that [Cl]NaCl >>x

• Solve for x.

• E.g. determine the solubility of CaF2 in a solution of CaCl2. Ksp = 3.9x1011.

• Starting with two solutions, Qsp used to predict precipitation and even the extent of it.

• Precipitation = reverse of dissolution

• Precipitation occurs when Qsp > Ksp until Qsp = Ksp

• If Qsp < Ksp, precipitation won’t occur.

E.g. determine if precipitation occurs after mixing 50.00 mL 3.00x103 M BaCl2 and 50.00 mL 3.00x103 M Na2CO3.

Solution:

• CBaCl2 = 1.50x103 M; CNa2CO3 = 1.50x103 M

• Qsp = 1.50x103 M1.50x103 M = 2.25x106

• Qsp >1.1x1010.= Ksp precipitation.

E.g. 2 determine equilibrium concentration of each after precipitation occurs.

Solution:

• assume complete precipitation occurs;

• set up equilibrium table; and solve for equilibrium concentration of barium and carbonate ion concentrations.

Eg. 3 determine the fraction of Ba2+ that has precipitated.

Solution:

• Use the amount remaining in solution (results of E.g. 2) divided by starting concentration to determine the fraction of barium that is left in solution.

• Subtract from above.

E.g.4 determine the Br concentration when AgCl starts to precipitate if the initial concentration of bromide and chloride are 0.100 M. Ksp(AgBr) = 5.0x1013; Ksp(AgCl) = 1.8x1010.

• pH of the Solution: LeChatelier’s Principle again.

E.g. determine the solubility of CaF2 at a pH of 2.00. Ksp = 3.9x1011. Ka(HF) = 6.6x104.

Strategy:

• Determine the ratio of [F] and [HF] from the pH and Ka.

• Write an expression for solubility in terms of Ka and pH and

• Substitute into solubility equation to determine the solubility.

Solution:

• Ksp = 3.9x1011 = x[F]2 (pH changes the amount of free Fluoride.)

• Let x = solubility. Then 2x = [F] + [HF]

• From equilibrium equation:

• 2x = [F](1+1/0.066) = 16.15*[F] or

• [F] = 2*x/16.15 = 0.124*x

• 3.9x1011 = x(0.0124*x)2

• x = 1.36x103 M vs. 2.13x104 M (normal solubility)

• Metal ions with very different Ksp can be separated.

• Divalent metal ions are often separated using solubility variations for the metal sulfides.

• Solution is saturated with H2S at 0.100 M; pH adjusted to keep one component soluble and the other insoluble.

• H2S is diprotic acid; the overall reaction to get to sulfide is:

• Combine with solubility equilibrium reaction to get the overall equilibrium expression and constant.

E.g. determine the solubility of 0.00500 M Zn2+ in 0.100 M H2S at pH = 1. Ksp = 1.10x1021.

• Formation of Complex Ions (Coordination Complexation ) = an ion formed from a metal ion with a Lewis base attached to it by a coordinate covalent bond.

Ag+(aq) + 2NH3(aq)  Ag(NH3)2(aq) Kf = 1.7x107

• Large equilibrium constant indicates that “free” metal is completely converted to the complex.

Eg. What is the concentration of the silver amine complex above in a solution that is originally 0.100 M Ag+ and 1.00 M NH3?

E.g. determine the [Ag+] (free silver concentration) in 0.100 M AgNO3 that is also 1.00 M NaCN.

• Free metal ion concentration in solution is reduced when complexing agent added to it;

• Free metal ion concentration needed in solubility expression.

E.g. determine if precipitation will occur in a solution containing 0.010 M AgNO3 and 0.0100 M Nal in 1.00 M NaCN. Recall Kf = 5.6x1018

Agl(s)Ag+ + l Ksp = 8.5.x1017

Strategy:

• Determine the free metal concentration in the solution.

• Use free metal concentration with iodide concentration to get Qsp

• If Qsp < Ksp, no precipitation

• If Qsp > Ksp, precipitation

• If Qsp = Ksp, precipitation is starting.

E.g. Determine the solubility of AgI in 1.00 M NaCN. Recall Kf = 5.6x1018

Agl(s)Ag+ + l Ksp = 8.5.x1017

Strategy:

• Combine to equilibria equations to find a single equation describing the equilibrium.

the presence of a complexing agent increases the solubility

• Setup equilibrium table and solve.