Please Pick Up Solubility Products Constants Data Sheet Heterogeneous Equilibria Problem Set.

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Solubility Products Constants Data Sheet

Heterogeneous Equilibria Problem Set

Heterogeneous Equilibria

Edward A. Mottel

Department of Chemistry

Rose-Hulman Institute of Technology

04/18/23

Heterogeneous Equilibria

• Reading assignment: Fine, Beall & Stuehr, Chapter 10.5

• Sparingly soluble salts• Solubility product• Common ion effect• Molar solubility• Applications

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Some Salts are Soluble

• Soluble salts

• Group IA, NH4+, NO3

–, ClO3–, ClO4

–

• Most halides (Cl–, Br–, I–)

• except Ag+, Cu+, Hg22+, Pb2+

• Most sulfates

• except CaSO4, BaSO4, Ag2SO4, HgSO4, PbSO4

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Some Salts are Not Very Soluble

• Sparingly soluble salts• Most hydroxides

• except Group IA, Ba(OH)2

• Carbonates, phosphates, sulfides• except Group IA, NH4

+

• Solubility depends on• Temperature and solvent

• Strong electrolyte

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Sparingly Soluble Salts

AgCl (s) Ag+(aq) + Cl–(aq)

The solubility product of silver chloride is 1.8 x 10-10, how can it be a strong electrolyte?

AgCl (s) AgCl (aq)

This indicates that AgClisn’t very soluble.(sparingly soluble)

Ag+(aq) + Cl–(aq)

This indicates that theamount of AgCl which dissolves

ionizes extensively.(strong electrolyte)

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Solubility Product, Ksp

• Products over reactants with simplifications

[Ag+] [Cl–]

[AgCl(s)]Kc

[Ag+] [Cl–] = Kc · [AgCl(s)]

[Ag+] [Cl–] = Ksp

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For Each of the Following Determine the Solubility Product Expression

• lead(II) carbonate• calcium fluoride• iron(III) hydroxide

• iron(II) sulfide• lead(II) iodide• silver sulfate

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For Each of the Following Determine the Solubility Product Expression

• [Pb2+] [CO32–]

• [Ca2+] [F–]2 • [Fe3+] [OH–]3

• [Fe2+] [S2–]• [Pb2+] [I–]2 • [Ag+]2 [SO4

2–]

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How to Determine If a Compound is Soluble or Not?

• Solubility Rules• Table of Solubility Product Constants• Handbook of Chemistry and Physics

• Physical Constants of Inorganic Compounds

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Physical Constants of Inorganic Compounds

• Name• Formula• Molecular Weight• Crystalline Form

• Color• Density

• Melting Point• Boiling Point• Solubility

• Cold Water• Hot Water• Other Solvents

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Copper(II) Formate

• Formula and Molecular Weight

• Cu(HCOO)2, 153.55 amu

• Color• blue, monoclinic crystals

• Is copper(II) formate soluble in water?• yes, 12.5 grams per 100 mL of water• greater than 1 gram per 100 mL water

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Group Activity

• An aqueous solution of copper(II) nitrate is mixed with aqueous sodium iodate.• Write an equation for this process and use

appropriate tables to determine if a reaction occurs.

• What observations are expected to be made?

• Is Cu(NO3)2 soluble?• Is NaIO3 soluble?

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Copper(II) Iodate

Cu2+(aq) + 2 NO3–(aq) + 2 Na+(aq) + 2 IO3

–(aq)

Cu(IO3)2(s) + 2 NO3–(aq) + 2 Na+(aq)

Ionic Equation:

Cu2+(aq) + 2 IO3–(aq) Cu(IO3)2(s)

Net Ionic Equation:

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Copper(II) Iodate

• Sparingly soluble• less than 1 gram per 100 mL of water• 0.136 grams per 100 mL of water at 15 °C

Cu2+(aq) + 2 IO3–(aq) Cu(IO3)2(s)

• Green, monoclinic crystals

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Pre-laboratory Assignment

• Experiment Q: Qualitative Analysis• Complete the Predicted Reactions grid on

page Q-3 in laboratory manual • Predict double displacement reactions, use

• Solubility Rules• Table of Solubility Product Constants• Handbook of Chemistry and Physics

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Al2(SO4)3

ZnCl2

Al2(SO4)3 + ZnCl2 AlCl3 + ZnSO4

predicted to be solublepredicted to be soluble

NR

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Al2(SO4)3

ZnCl2 NR

BaCl2

Al2(SO4)3 + BaCl2 AlCl3 + BaSO4

predicted to be solublepredicted to be an insoluble, white ppt

BaSO4

white ppt

Molar Solubility

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Two Beakers Each Contain500 mL of Water

50 grams of PbF2 is added to Beaker 1

400 grams of PbF2 is added to Beaker 2

Some of the solid dissolves.

In which beaker will morelead(II) fluoride dissolve?

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Solubility Product

• What is the solubility of lead(II) fluoride in pure water?• Write the chemical equation for the

dissolution process.• Let +x be the moles per liter of lead(II) ion

that is produced.• Write the mass-action expression.• Substitute variables and solve the solubility

product mass-action expression.

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Lead(II) Fluoride

-x +x +2x

x = 2.1 x 10–3 M

• What does 2.1 x 10–3 M represent?

PbF2(s) Pb2+(aq) + 2 F–(aq)

• Molar solubility of PbF2 in pure water

• Molarity of lead(II) ion in solution• What is the molarity of the fluoride ion?

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How many grams of lead(II) fluoride will dissolve in 500 mL of

water?

• Molar solubility of PbF2 (FW 245.2 amu) in pure water is 2.1 x 10–3 M

2.1 10–3moles PbF2

liter

0.500 liter

245.2 grams PbF2

moles PbF2

0.26 g PbF2

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Each beaker contains 500 mL of water.

50 grams of PbF2 is added to Beaker 1

400 grams of PbF2 is added to Beaker 2

If there is excess solid at equilibrium,the same amount dissolves!

In which beaker will more lead(II) fluoride dissolve?

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Common Ion Effect

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Two beakers each contain500 mL of solution

• 50 grams of PbF2 is added to pure water

• 50 grams of PbF2 is added to 0.0200 M sodium fluoride

Into which solution will morelead(II) fluoride dissolve?

H2OH2O

Na+F-

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Common Ion Effect

• Equilibrium concentrations of each ion depend on the number of those ions in solution, regardless of the source.

What is the molar solubility of lead(II) fluoridein a solution containing 0.0200 M sodium fluoride?

Assume one liter of solution

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The Molar Solubility of Lead(II) Fluoridein 0.0200 M Sodium Fluoride

PbF2(s) Pb2+(aq) + 2 F–(aq)

0.0200

-x +x +2x

Ksp = [Pb2+] [F–]2 = 3.7 x 10–8 M3

Ksp = (x) (0.0200 + 2x)2 = 3.7 x 10–8 M3

x = 9.2 x 10–5 Mx = 0.023 g in one liter

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Group Activity

Why is the molar solubility oflead(II) fluoride different in a solution

containing sodium fluoride?

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Group Activity

PbF2(s) Pb2+(aq) + 2 F–(aq)

Why is the molar solubility of lead(II) fluoridedifferent in a solution containing sodium fluoride?

Explain in terms of Le Châtelier’s Principle

Applications

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Selective Precipitation

A solution contains0.100 M cadmium ion

0.050 M iron(III) ion

0.200 M nickel(II) ion

If sodium hydroxide is slowly added to

the solution

What precipitates will form?In what order will the precipitates form?

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Selective Precipitation Strategy

Determine the hydroxide ion concentrationneeded to precipitate each metal.

Determine the order in which thosehydroxide ion concentrations will be reached.

0.100 M cadmium ion

0.050 M iron(III) ion

0.200 M nickel(II) ion

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Hydroxide Ion Concentration Neededto Precipitate 0.100 M Cd2+

Cd(OH)2(s) Cd2+(aq) + 2 OH–(aq)

[Cd2+] [OH–]2 = Ksp

[0.100] [OH–]2 = 2.0 x 10–14 M3

[OH–] =2.0 x 10–14 M3

[0.100]

1/2

[OH–] = 4.5 x 10–7 M

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Hydroxide Ion Concentration Neededto Precipitate

• 0.100 M Cd2+

• [OH–] = 4.5 x 10–7 M• 0.050 M Fe3+

• [OH–] = 1.1 x 10–12 M• 0.200 M Ni2+

• [OH–] = 2.8 x 10–8 M

As the solution is slowly made basic,which precipitates first?

10–5

10–6

10–7

10–8

10–9

10–10

10–11

10–12

10–13

10–14

[OH–]

Cd(OH)2

Fe(OH)3

Ni(OH)2Bas

ic

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Selective Precipitation

At what pH will the last metal ion

begin to precipitate?

9876543210

pH

7.65

When the cadmium ion begins to

precipitate how much of the iron(III)

ion is still in solution?

10–5

10–6

10–7

10–8

10–9

10–10

10–11

10–12

10–13

10–14

[OH–]

Cd(OH)2

Ni(OH)2

Fe(OH)3

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Selective Precipitation

• 0.100 M cadmium ion begins to precipitate when the hydroxide ion concentration is 4.5 x 10–7 M.

• What is the maximum iron(III) ion concentration under these conditions?

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Selective Precipitation

Fe(OH)3(s) Fe3+(aq) + 3 OH–(aq)

[Fe3+] [OH–]3 = Ksp

[Fe3+] (4.5 x 10–7 M)3 = 6.0 x 10–38 M4

[Fe3+] =6.0 x 10–38 M4

(4.5 x 10–7 M )3

[Fe3+] = 6.6 x 10–19 M

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Determination ofMolar Solubility and Ksp

• The Handbook of Chemistry and Physics indicates that the solubility of zinc fluoride (FW 103.37) is 1.62 grams per 100 mL of water.• What is the molar solubility of zinc

fluoride?• What is the solubility product of zinc

fluoride?

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Zinc FluorideMolar Solubility

1.62 g /100 mL water

16.2 g /1000 mL water

16.2 g /1000 mL solution

16.2 g /103.37 g·mol–1 = 0.157 mol ZnF2

The molar solubility of zinc fluoride is 0.157 M.

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Zinc FluorideSolubility Product

ZnF2(s) Zn2+(aq) + 2 F–(aq)

Ksp = [Zn2+] [F–]2

Ksp = (0.157 M) (2 x 0.157 M)2

Two moles of fluoride ionform for every zinc ion.

Ksp = (0.157 M) (0.314 M)2

Ksp = 1.54 x 10–2 M3

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or

or

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