General Chemistry II 2302102 Chemical Equilibrium for Gases and for Sparingly-Soluble Ionic Solids...

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General Chemistry IIGeneral Chemistry II23021022302102

Chemical Equilibrium for Gases and for Sparingly-Soluble Ionic

SolidsLecture 2

i.fraser@rmit.edu.auIan.Fraser@sci.monash.edu.au

Chemical EquilibriumChemical Equilibrium - 2 Lectures- 2 Lectures

• Equilibrium and Le Chatelier’s Principle (Completed)

• The Equilibrium Constant (Completed)

• Temperature and Pressure Effects (Completed)

• Sparingly-Soluble Ionic Compounds in Aqueous Solution

Outline - 4 Subtopics

By the end of this lecture AND completion of the set problems, you should be able to:

• Understand the concepts of: saturated and unsaturated solutions, freely soluble and sparingly soluble ionic compounds.

• Understand and know several examples of precipitation reactions.

• Understand the definition of the solubility product (Ksp) and its relationship to the solubility of sparingly soluble ionic compounds.

• Understand the definition of the Common Ion Effect.

• Calculate equilibrium concentrations in precipitation reactions.

Objectives - Lecture 2

Chemical EquilibriumChemical Equilibrium

Electrical ConductivityElectrical Conductivity

Open Circuit Closed circuit

ElectrolytesElectrolytesNonNon StrongStrong WeakWeak

NoNo ions ionsin solutionin solution

ManyMany ions ionsin solutionin solution

FewFew ions ionsin solutionin solution

ElectrolytesElectrolytesAn electrolyte is a substance that conducts an electric current when dissolved in water (or in the the molten state)

ElectrolytesElectrolytes

Examples: Most salts, Acids and bases

Strong electrolytes: their water solutions are good conductors (e.g. NaCl solution)

Weak electrolytes: their water solutions are poor conductors (e.g. vinegar - acetic acid, CH3COOH)

Non-electrolytes: their water solutions are nonconductors (e.g. sugar solution)

Sparingly Soluble SaltsSparingly Soluble Salts

• Some salts dissolve readily in water

e.g.NaCl(s) Na+(aq) + Cl-(aq)

NaCl dissociates in solution to form ions

• But other salts are only slightly soluble:

AgCl(s) Ag+(aq) + Cl-(aq)

Why are we interested in sparingly Why are we interested in sparingly soluble salts?soluble salts?

• Earth’s crust is dominated by these salts:– feldspars, gypsum, calcite, aluminosilicates, dolomite, and oxides

& sulfides of metals– control major geochemical processes

• Boiler “scale” - often iron & manganese oxides

• “Hard” water - – Waters originating from limestone areas contain Mg2+ and Ca2+

– these ions form soap “scum” precipitates

http://www.jenolancaves.org.au

CaCO3(s) + CO2(g) + H2O Ca2+(aq) + 2HCO3-(aq)

Caves are formedStalactites and stalagmites

are formed

TerminologyTerminology• Recall from an earlier lecture on phase equilibria, for solutions in

equilibrium with solids:

• The equilibrium concentration of a dissolved solid in solution is known as the solubility

• Such a solution is called “saturated”

• Undersaturated: More solute will dissolve.

• Supersaturated: Solution has an excess of solute.

Solubility TerminologySolubility TerminologyFreely soluble - several g (or more) dissolve in 100 g of water

– e.g. at 298K, 36 g of NaCl, 122 g of AgNO3

Sparingly soluble - << 1 g dissolves in 100 g of water– e.g. at 298K, 2.4 x 10-4 g of AgCl, 4.4 x 10-13 g of PbS, 9.3 x 10-4 g of

Intermediate solubility - ca. 1 g dissolves in 100 g of water (only a few of these)– e.g. 1.02 g of Ag(CH3COO) at 293K, 0.19 g of Ca(OH)2 at 273K

Net Ionic Equation: What is it?Net Ionic Equation: What is it?A reaction equation between two electrolytes, for example between silver nitrate and sodium chloride, can be written conventionally as:

AgNOAgNO33 (aq) + NaCl (aq) (aq) + NaCl (aq)

precipitateprecipitateAgCl (s) + NaNOAgCl (s) + NaNO33 (aq) (aq)

But AgNO3, NaCl, and NaNO3 are ionized in aqueous solution (AgCl is a solid - insoluble)

The conventional equation can be rewritten as:

Net Ionic Equation: What is it?Net Ionic Equation: What is it?

Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq)

AgCl (s) + Na+ (aq) + NO3- (aq)

Here, the NO3- and Na+ ions are unchanged. They are simply “spectator”

ions - they do not participate in the reaction.

It is known (observed) that the actual reaction occurs only between silver ions and chloride ions

Ag+ (aq) + Cl- (aq) AgCl(s)

Net Ionic Equation: What is it?Net Ionic Equation: What is it?

The above equation is a net ionic equation.

Spectator IonsSpectator Ions

Lead nitrate + Potassium chromateLead nitrate + Potassium chromate

Lead chromateLead chromate(s)(s) + Potassium nitrate+ Potassium nitrate

PbPb2+2+ + 2 NO + 2 NO33-- 2 K2 K++ + CrO + CrO44

PbCrOPbCrO44 (s)(s) 2 K2 K++ + 2 NO + 2 NO33--

Spectator IonsSpectator Ions

So Net Ionic Equation is:So Net Ionic Equation is:

PbPb2+2+ (aq) + CrO (aq) + CrO442- 2- (aq) (aq)

PbCrOPbCrO44 (s)(s)

The concentrations of all compounds are equal The concentrations of all compounds are equal at equilibriumat equilibrium

Forward and reverse rates of reaction are equal - Forward and reverse rates of reaction are equal - hence new products are formed at the same rate as hence new products are formed at the same rate as they are broken downthey are broken down

Equilibrium does Equilibrium does NOTNOT mean: mean:• Reactions are not occurring

The concentrations of all compounds are The concentrations of all compounds are determined by the position of the equilibrium - it determined by the position of the equilibrium - it may favour the may favour the productsproducts or the or the reactantsreactants

Equilibrium ConstantEquilibrium Constant

Generally for:Generally for:

aaA + A + bbB B ccC + C + ddDD

[C][C]cc [D] [D]dd

[A][A]aa [B] [B]bbKK = =

A Saturated Solution is at EquilibriumA Saturated Solution is at Equilibrium

The equilibrium constant expression The equilibrium constant expression is:is:

[Ag[Ag++] [Cl] [Cl--]]KK = =

AgCl(s) AgAgCl(s) Ag++(aq) + Cl(aq) + Cl--(aq)(aq)

The Solubility Product, KThe Solubility Product, Kspsp

AAbbBBaa(s) (s) bbAAa+a+(aq) + (aq) + aaBBb-b-(aq)(aq)

[A[Aa+a+]]bb [B [Bb-b-]]aa

[A[AbbBBaa]]KKcc = =

The concentration of [AThe concentration of [AbbBBaa] (a solid) is a ] (a solid) is a constant,constant,

and is by convention set and is by convention set 1 in the definition of K 1 in the definition of Kcc

KKcc = = [A[Aa+a+]]bb [B [Bb-b-]]aa = K= Kspsp

The Solubility Product, KThe Solubility Product, Kspsp

KKspsp (AgCl)(AgCl) = [Ag = [Ag++] [Cl] [Cl--]]

KKspsp (Fe(Fe22SS33)) = [Fe = [Fe3+3+]]22 [S [S2-2-]]33

KKspsp (BaSO(BaSO44)) = [Ba = [Ba2+2+] [SO] [SO442-2-]]

Examples:

Calculation of SolubilityCalculation of Solubility

KKspsp (AgCl) (AgCl) = [Ag = [Ag++] [Cl] [Cl--] ] = 1.8 x 10= 1.8 x 10-10-10 mol mol22 dm dm-6-6

KKspsp = s = s xx s s = 1.8 x 10= 1.8 x 10-10-10

01x 1.8 s 10-

s s = 1.34 x 10= 1.34 x 10-5-5 mol dm mol dm-3-3

Solubility Products at 25Solubility Products at 25°°CCCompound Ksp Compound Ksp

AgBr 5.0 x 10-13 Fe(OH)3 4.0 x 10-38

AgCl 1.8 x 10-10 FeS 6.3 x 10-18

AgI 8.3 x 10-17 HgS 1.6 x 10-52

AgIO3 3.1 x 10-8 Mg(OH)2 1.8 x 10-11

Ag3PO4 1.3 x 10-20 MgC2O4 8.6 x 10-5

Al(OH)3 2.0 x 10-32 Mn(OH)2 1.9 x 10-13

Ba(OH)2 5.0 x 10-3 MnS 2.5 x 10-13

BaSO4 1.1 x 10-10 NiS 1.0 x 10-24

Bi2S3 1.0 x 10-97 PbCl2 1.6 x 10-5

CaCO3 4.8 x 10-9 PbSO4 1.6 x 10-8

CaC2O4 4.0 x 10-9 PbS 8.0 x 10-28

CaSO4 1.2 x 10-6 SrSO4 3.2 x 10-7

CdS 8.0 x 10-27 Zn(OH)2 3.3 x 10-17

CoS 2.0 x 10-25 ZnS 1.6 x 10-23

CuS 6.3 x 10-36

Calculation of Solubility #2Calculation of Solubility #2KKspsp (PbF(PbF22) ) = [Pb = [Pb2+2+] [F] [F--]]22 = 3.7 x 10= 3.7 x 10-8-8 M M33

PbFPbF22(s) Pb(s) Pb2+2+(aq) + 2F(aq) + 2F--(aq)(aq)

KKspsp = s = s xx ( (22s)s)22 = 3.7 x 10= 3.7 x 10-8-8

3 8-01x 1.85 s

s s = 2.1 x 10= 2.1 x 10-3-3 M M [Pb[Pb2+2+] = 2.1 x 10] = 2.1 x 10-3-3 M, [F M, [F--] = 4.2 x 10] = 4.2 x 10-3-3 M M

The Common Ion EffectThe Common Ion Effect

[Pb[Pb2+2+] = 2.1 x 10] = 2.1 x 10-3-3 M, [F M, [F--] = 4.2 x 10] = 4.2 x 10-3-3 M M

The presence of an ion in solution which is common to the electrolyte will decrease the solubility:

We’ve just seen that if PbF2 is placed in water, the solubilities of the ions are:

What happens if PbF2 is placed in a solution of 0.02 M KF? (F- is the “common ion”)

The Common Ion EffectThe Common Ion Effect

KKspsp = s = s xx (2s + 0.02) (2s + 0.02)22 = 3.7 x 10= 3.7 x 10-8-8

3.7 x 103.7 x 10-8-8 = s x (0.02)= s x (0.02)22

s = [Pbs = [Pb2+2+] = 9.3 x 10] = 9.3 x 10-5-5 (c.f. 2.1 x 10(c.f. 2.1 x 10-3-3 M) M)

In this case, [F-] = 2s + 0.02 M

KKspsp = 4s = 4s33 ++ 0.08s 0.08s22 ++ 0.0004s 0.0004s = 3.7 x 10= 3.7 x 10-8-8

Solve as a cubic, or if 0.02 >> s

What happens if PbF2 is placed in a solution of 0.02 M KF? (F- is the “common ion”)

Application of Le Chatelier’s PrincipleApplication of Le Chatelier’s Principle

• Recall that:if possible, systems react to an imposed change by reducing the

impact of that change

PbFPbF22(s) Pb(s) Pb2+2+(aq) + 2F(aq) + 2F--(aq)(aq)

• Increase [F-] (by adding KF - soluble)• Force equilibrium to left, and therefore reduce

[Pb2+]

Precipitation ReactionsPrecipitation Reactions• A solid will form (precipitate) if the concentrations of the

ions exceed the solubility product.

bbAAa+a+(aq) + (aq) + aaBBb-b-(aq) A(aq) AbbBBaa(s) (s)

First calculate the initial “reaction quotient”, Qo :

QQoo = = [A[Aa+a+]]oobb [B [Bb-b-]]oo

Then compare Qo with Ksp

Precipitation Reactions #2Precipitation Reactions #2First calculate the initial “reaction quotient”, Qo :

QQoo = = [A[Aa+a+]]oobb [B [Bb-b-]]oo

Then compare Qo with Ksp :

If Qo > Ksp Precipitation Occurs

If Qo < Ksp No Precipitation

Precipitation Reactions #3Precipitation Reactions #3

First calculate the initial “reaction quotient”, Qo :

QQoo = = [Ag[Ag++]]oo [Cl [Cl--]]oo = = 22 x 10 x 10-4-4 xx 11 x 10 x 10-5-5

= 2 x 10= 2 x 10-9-9 M M2 2

Then compare Qo with Ksp (1.8 x 101.8 x 10-10-10 M M22):

Is Qo > Ksp Yes! Precipitation Occurs

Example 1. 4 x 10-4 M AgNO3 is mixed with an equal volume of 2 x 10-5 M NaCl. Will a precipitate form?

Precipitation Reactions #4Precipitation Reactions #4Example 2. Electroplating with cadmium is a common industrial process. Before waste solutions can be discharged, [Cd2+] must be reduced to < 1 x 10-6 M.

This may be achieved by adding sodium hydroxide to precipitate the cadmium as the hydroxide salt: Ksp(Cd(OH)2) = 5.3 x 10-15 M3

What concentration of NaOH is required?

CdCd2+2+(aq) + (aq) + 22OHOH--(aq) Cd(OH)(aq) Cd(OH)22(s) (s)

Precipitation Reactions #5Precipitation Reactions #5

Example 2 cont...

Ksp(Cd(OH)2) = 5.3 x 10-15 M3

CdCd2+2+(aq) + 2OH(aq) + 2OH--(aq) Cd(OH)(aq) Cd(OH)22(s)(s)

KKspsp = = 1 1 xx 10 10-6-6 xx s s22 = 5.3 x 10= 5.3 x 10-15-15

01x 5.3 s 9-

s = 7.3 x 10-5

Hence to reduce [Cd2+] < 1 x 10-6 M, [OH-] must be > 7.3 x 10-5 M

Selective PrecipitationSelective Precipitation

Problem: Ag+(aq) as silver nitrate is added to a mixture containing 0.001 M Cl- and 0.001 M CrO4

2-. Ksp (AgCl) = 2.8 x 10-10, Ksp (Ag2CrO4) = 1.9 x 10-12. What will happen as [Ag+] is increased?

1/ Precipitation of AgCl(s) will begin when [Ag+] > 2.8 x 10-7 M. (Qo > Ksp)

2/ Precipitation of Ag2CrO4(s) will begin when [Ag+] > 4.4 x 10-5 M. (Qo > Ksp)

Selective PrecipitationSelective Precipitation

1/ Precipitation of AgCl(s) will begin when [Ag+] > 2.8 x 10-7 M. (Qo > Ksp)

2/ Precipitation of Ag2CrO4(s) will begin when [Ag+] > 4.4 x 10-5 M. (Qo > Ksp)

But in order for the value of [Ag+] to exceed 4.4 x 10-5 it is necessary that [Cl-] < 6.4 x 10-6

Thus on addition of silver nitrate, AgCl will precipitate until the concentration of the chloride ion remaining is less than 6.4 x 10-6 M, at which stage precipitation of silver chromate will occur.

Chemical Equilibrium - End of Lecture 2 - End of Lecture 2

After studying this lecture should be able to: