Chapter 16 Solubility and Complex Ion Equilibria AP*

Chapter 16

Solubility and Complex Ion Equilibria

AP*

AP Learning Objectives

LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes. (Sec 16.1-16.3)

LO 6.21 The student can predict the solubility of a salt, or rank the solubility of salts, given the relevant Ksp values. (Sec 16.1)

LO 6.22 The student can interpret data regarding solubility of salts to determine, or rank, the relevant Ksp values. (Sec 16.1)

LO 6.23 The student can interpret data regarding the relative solubility of salts in terms of factors (common ions, pH) that influence the solubility. (Sec 16.1-16.2)

Section 16.1Solubility Equilibria and the Solubility Product

AP Learning Objectives, Margin Notes and References

Learning Objectives LO 6.1 The student is able to, given a set of experimental observations regarding physical, chemical, biological, or

environmental processes that are reversible, construct an explanation that connects the observations to the reversibility of the underlying chemical reactions or processes.

LO 6.21 The student can predict the solubility of a salt, or rank the solubility of salts, given the relevant Ksp values.

LO 6.22 The student can interpret data regarding solubility of salts to determine, or rank, the relevant Ksp values. LO 6.23 The student can interpret data regarding the relative solubility of salts in terms of factors (common ions,

pH) that influence the solubility.


Solubility Equilibria Because ionic compounds are strong

electrolytes, they dissociate completely to the extent that they dissolve.

When an equilibrium equation is written, the solid is the reactant and the ions in solution are the products.

The equilibrium constant expression is called the solubility-product constant. It is represented as Ksp.


Solubility ProductFor example:

BaSO4(s) Ba⇌ 2+(aq) + SO42–(aq)

The equilibrium constant expression is Ksp = [Ba2+][SO4

2]Another example:Ba3(PO4)2(s) 3 Ba⇌ 2+(aq) + 2 PO4

3–(aq)The equilibrium constant expression is

Ksp = [Ba2+]3[PO43]2


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Solubility Equilibria

Solubility product (Ksp) – equilibrium constant; has only one value for a given solid at a given temperature. (not affected by equilibrium position)

Solubility – an equilibrium position.

Bi2S3(s) 2Bi3+(aq) + 3S2–(aq)2 33+ 2

sp = Bi S K


Solubility vs. Solubility Product Ksp is not the same as solubility. Solubility is the quantity of a substance that dissolves

to form a saturated solution (reach equilibrium) Common units for solubility: Grams per liter (g/L) Moles per liter (mol/L)


Calculating Solubility from Ksp The Ksp for CaF2 is 3.9 × 10–11 at 25 °C. What

is its molar solubility? Follow the same format as before:1)CaF2(s) Ca⇌ 2+(aq) + 2 F–(aq)

2) Ksp = [Ca2+][F–]2 = 3.9 × 10–11

3) CaF2(s) [Ca2+](M) [F–](M)

Initial concentration

(M)

---

0

0

Change in concentration

(M)

---

+s

+2s

Equilibrium concentration

(M)

---

s

2s


Example (completed)4) Solve: Substitute the equilibrium

concentration values from the table into the solubility-product equation:3.9 × 10–11 = (s) (2s)2 = 4s3

s = 2.1 × 10–4 M(If you want the answer in g/L, multiply by molar mass; this would give 0.016 g/L.)



In comparing several different salts at a given temperature, does a higher Ksp value always mean a higher solubility? Explain. If yes, explain and verify. If no, provide a counter-example.

CONCEPT CHECK! Relative SolubilitiesCONCEPT CHECK! Relative Solubilities



EXERCISE!EXERCISE!

Which of these expressions correctly expresses the solubility-product constant for Ag3PO4 in water? (a) [Ag][PO4], (b) [Ag+][PO4

3–], (c) [Ag+]3[PO43–], (d)

[Ag+][PO43–]3, (e) [Ag+]3[PO4

3–]3.

Writing Solubility-Product (Ksp) Expressions



EXERCISE!EXERCISE!

Solid silver chromate is added to pure water at 25 °C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag2CrO4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 × 10–4 M. Assuming that the Ag2CrO4 solution is saturated and that there are no other important equilibria involving Ag+ or CrO4

2– ions in the solution, calculate Ksp for this compound.

Calculating Ksp from Solubility



Calculate the solubility of silver chloride in water. Ksp = 1.6 × 10–10

1.3×10-5 M

Calculate the solubility of silver phosphate in water. Ksp = 1.8 × 10–18

1.6×10-5 M

EXERCISE!EXERCISE!


Factors Affecting Solubility The Common-Ion Effect

If one of the ions in a solution equilibrium is already dissolved in the solution, the solubility of the salt will decrease.

If either calcium ions or fluoride ions are present, then calcium fluoride will be less soluble.


Calculating Solubility with a Common Ion

What is the molar solubility of CaF2 in 0.010 M Ca(NO3)2?

Follow the same format as before:1) CaF2(s) Ca⇌ 2+(aq) + 2 F–(aq)

2) Ksp = [Ca2+][F–]2 = 3.9 × 10–11

3) CaF2(s) [Ca2+](M) [F–](M)

Initial concentration

(M)

---

0.010

0

Change in concentration

(M)

---

+s

+2s

Equilibrium concentration

(M)

---

0.010 + s

2s


Example (completed)

4) Solve: Substitute the equilibrium concentration values from the table into the solubility-product equation:3.9 × 10–11 = (0.010 + s)(2s)2 (We assume that s << 0.010, so that 0.010 + s = 0.010!)3.9 × 10–11 = (0.010)(2s)2 s = 3.1 × 10–5 M



Calculate the solubility of AgCl in: Ksp = 1.6 × 10–10

a) 100.0 mL of 4.00 x 10-3 M calcium chloride. 2.0×10-8 M

b) 100.0 mL of 4.00 x 10-3 M calcium nitrate. 1.3×10-5 M

EXERCISE!EXERCISE!


© 2015 Pearson Education, Inc.

Factors Affecting Solubility pH: If a substance has a basic anion, it will be

more soluble in an acidic solution.



How does the solubility of silver chloride in water compare to that of silver chloride in an acidic solution (made by adding nitric acid to the solution)?

Explain.

The solubilities are the same.

CONCEPT CHECK!CONCEPT CHECK!



How does the solubility of silver phosphate in water compare to that of silver phosphate in an acidic solution (made by adding nitric acid to the solution)?

Explain.

The silver phosphate is more soluble in an acidic solution.




How does the Ksp of silver phosphate in water compare to that of silver phosphate in an acidic solution (made by adding nitric acid to the solution)?

Explain.

The Ksp values are the same.


Section 16.2Precipitation and Qualitative Analysis




LO 6.23 The student can interpret data regarding the relative solubility of salts in terms of factors (common ions, pH) that influence the solubility.


Precipitation (Mixing Two Solutions of Ions)


Will a Precipitate Form? To decide, we calculate the reaction quotient, Q,

and compare it to the solubility product constant, Ksp.

If Q = Ksp, the system is at equilibrium and the solution is saturated.

If Q < Ksp, more solid can dissolve, so no precipitate forms.

If Q > Ksp, a precipitate will form.


Does a precipitate form when 0.10 L of 8.0 × 10–3 M Pb(NO3)2 is added to 0.40 L of 5.0 × 10–3 M Na2SO4? (Ksp of 6.3 × 10–7 )

Predicting Whether a Precipitate Forms



What are the equilibrium concentrations of the remaining ions once precipitation has occurred?


Selective Precipitation (Mixtures of Metal Ions)

Use a reagent whose anion forms a precipitate with only one or a few of the metal ions in the mixture.

Example: Solution contains Ba2+ and Ag+ ions. Adding NaCl will form a precipitate with Ag+ (AgCl),

while still leaving Ba2+ in solution.



A solution contains 1.0 × 10–2 M Ag+ and 2.0 × 10–2 M Pb2+. When Cl– is added, both AgCl(Ksp = 1.8 × 10–10) and PbCl2(Ksp = 1.7 × 10–5) can precipitate. What concentration of Cl– is necessary to begin the precipitation of each salt? Which salt precipitates first?

Selective Precipitation


Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S

At a low pH, [S2–] is relatively low and only the very insoluble HgS and CuS precipitate.

When OH– is added to lower [H+], the value of [S2–] increases, and MnS and NiS precipitate.



Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S



Separating the Common Cations by Selective Precipitation


Ksp values can be compare directly for ions that produce the same number of ions in solution.


31

potassium. sodium lithium

Section 16.3Equilibria Involving Complex Ions





Complex Ion Equilibria

Charged species consisting of a metal ion surrounded by ligands. Ligand: Lewis base (a molecule with a lone pair that

can be donated to an empty orbital on the metal form a covalent bond)

The number of ligands attached to the metal is called the coordination number. (commonly 6,4,or 2)




Formation (stability) constant. Equilibrium constant for each step of the formation of

a complex ion by the addition of an individual ligand to a metal ion or complex ion in aqueous solution.




Be2+(aq) + F–(aq) BeF+(aq) K1 = 7.9 × 104

BeF+(aq) + F–(aq) BeF2(aq) K2 = 5.8 × 103

BeF2(aq) + F–(aq) BeF3–

(aq) K3 = 6.1 × 102

BeF3–

(aq) + F–(aq) BeF42–

(aq) K4 = 2.7 × 101





How Complex Ion Formation Affects Solubility

Silver chloride is insoluble. It has a Ksp of 1.6 × 10–10.

In the presence of NH3, the solubility greatly increases because Ag+ will form complex ions with NH3.


Complex Ions and Solubility

Two strategies for dissolving a water–insoluble ionic solid. If the anion of the solid is a good base, the solubility

is greatly increased by acidifying the solution. In cases where the anion is not sufficiently basic, the

ionic solid often can be dissolved in a solution containing a ligand that forms stable complex ions with its cation.




Amphoterism and Solubility• Amphoteric oxides and hydroxides are soluble in strong

acids or base, because they can act either as acids or bases.

• Examples are oxides and hydroxides of Al3+, Cr3+, Zn2+, and Sn2+.


Calculate the solubility of silver chloride in 10.0 M ammonia given the following information:

Ksp (AgCl) = 1.6 × 10–10

Ag+ + NH3 AgNH3+ K = 2.1 × 103

AgNH3+ + NH3 Ag(NH3)2

+ K = 8.2 × 103

0.48 MCalculate the concentration of NH3 in the final equilibrium mixture.

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Chapter 16 Solubility and Complex Ion Equilibria AP*

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