Transcript ENG1071 S1 2005 - BSAC (Bachelor of Science in Applied

General Chemistry II
2302102
Chemical Equilibrium for Gases
and for Sparingly-Soluble Ionic
Solids
Lecture 2
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Chemical Equilibrium
- 2 Lectures
Outline - 4 Subtopics
• Equilibrium and Le Chatelier’s Principle (Completed)
• The Equilibrium Constant (Completed)
• Temperature and Pressure Effects (Completed)
• Sparingly-Soluble Ionic Compounds in Aqueous
Solution
Chemical Equilibrium
Objectives - Lecture 2
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.
Electrical Conductivity
Open Circuit
Closed circuit
Electrolytes
Non
No ions
in solution
Strong
Weak
Many ions
in solution
Few ions
in solution
Electrolytes
An electrolyte is a substance that conducts an electric current when dissolved in water (or in
the the molten state)
Electrolytes
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 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
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
Caves are formed
Ca2+(aq) + 2HCO3-(aq)
Stalactites and stalagmites
are formed
Terminology
• 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 Terminology
Freely 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 CaCO3
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?
A reaction equation between two electrolytes, for example between silver
nitrate and sodium chloride, can be written conventionally as:
AgNO3 (aq) + NaCl (aq)
AgCl (s) + NaNO3 (aq)
precipitate
But AgNO3, NaCl, and NaNO3 are ionized in aqueous solution (AgCl is a
solid - insoluble)
Net Ionic Equation: What is it?
The conventional equation can be rewritten as:
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.
Net Ionic Equation: What is it?
It is known (observed) that the actual reaction occurs only between
silver ions and chloride ions
Ag+ (aq) + Cl- (aq)
The above equation is a net ionic equation.
AgCl(s)
Spectator Ions
Lead nitrate + Potassium chromate
Pb2+ + 2 NO3-
2 K+ + CrO42-
Lead chromate(s) + Potassium nitrate
PbCrO4 (s)
2 K+ + 2 NO3-
Spectator Ions
So Net Ionic Equation is:
Pb2+ (aq) + CrO42- (aq)
PbCrO4 (s)
Equilibrium does NOT mean:
• Reactions are not occurring
Forward and reverse rates of reaction are equal hence new products are formed at the same rate as
they are broken down

The concentrations of all compounds are equal
at equilibrium
The concentrations of all compounds are
determined by the position of the equilibrium - it
may favour the products or the reactants
Equilibrium Constant
Generally for:
aA + bB
K =
cC + dD
[C]c [D]d
[A]a [B]b
A Saturated Solution is at Equilibrium
AgCl(s)
Ag+(aq) + Cl-(aq)
The equilibrium constant expression
is:
K =
[Ag+] [Cl-]
The Solubility Product, Ksp
AgCl(s)
Ag+(aq) + Cl-(aq)
AbBa(s)
bAa+(aq) + aBb-(aq)
Kc =
[Aa+]b [Bb-]a
[AbBa]
The concentration of [AbBa] (a solid) is a constant,
and is by convention set  1 in the definition of Kc
 Kc =
[Aa+]b [Bb-]a
= Ksp
The Solubility Product, Ksp
Examples:
Ksp (AgCl) = [Ag+] [Cl-]
Ksp (Fe2S3) = [Fe3+]2 [S2-]3
Ksp (BaSO4) = [Ba2+] [SO42-]
Calculation of Solubility
Ksp (AgCl) = [Ag+] [Cl-] = 1.8 x 10-10 mol2 dm-6
Ag+(aq) + Cl-(aq)
AgCl(s)
Ksp = s x s = 1.8 x 10-10
s 
1.8 x 1 0-10
s = 1.34 x 10-5 mol dm-3
Solubility Products at 25°C
Compound
AgBr
AgCl
AgI
AgIO3
Ag3PO4
Al(OH)3
Ba(OH)2
BaSO4
Bi2S3
CaCO3
CaC2O4
CaSO4
CdS
CoS
CuS
Ksp
5.0 x 10-13
1.8 x 10-10
8.3 x 10-17
3.1 x 10-8
1.3 x 10-20
2.0 x 10-32
5.0 x 10-3
1.1 x 10-10
1.0 x 10-97
4.8 x 10-9
4.0 x 10-9
1.2 x 10-6
8.0 x 10-27
2.0 x 10-25
6.3 x 10-36
Compound
Fe(OH)3
FeS
HgS
Mg(OH)2
MgC2O4
Mn(OH)2
MnS
NiS
PbCl2
PbSO4
PbS
SrSO4
Zn(OH)2
ZnS
Ksp
4.0 x 10-38
6.3 x 10-18
1.6 x 10-52
1.8 x 10-11
8.6 x 10-5
1.9 x 10-13
2.5 x 10-13
1.0 x 10-24
1.6 x 10-5
1.6 x 10-8
8.0 x 10-28
3.2 x 10-7
3.3 x 10-17
1.6 x 10-23
Calculation of Solubility #2
Ksp (PbF2) = [Pb2+] [F-]2 = 3.7 x 10-8 M3
PbF2(s)
Pb2+(aq) + 2F-(aq)
Ksp = s x (2s)2 = 3.7 x 10-8
3
 s  1.85 x 10-8
s = 2.1 x 10-3 M
[Pb2+] = 2.1 x 10-3 M, [F-] = 4.2 x 10-3 M
The Common Ion Effect
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:
[Pb2+] = 2.1 x 10-3 M, [F-] = 4.2 x 10-3 M
What happens if PbF2 is placed in a solution of
0.02 M KF? (F- is the “common ion”)
The Common Ion Effect
What happens if PbF2 is placed in a solution of
0.02 M KF? (F- is the “common ion”)
In this case, [F-] = 2s + 0.02 M
Ksp = s x (2s + 0.02)2 = 3.7 x 10-8
Ksp = 4s3 + 0.08s2 + 0.0004s = 3.7 x 10-8
Solve as a cubic, or if 0.02 >> s
3.7 x 10-8 = s x (0.02)2
s = [Pb2+] = 9.3 x 10-5 (c.f. 2.1 x 10-3 M)
Application of Le Chatelier’s Principle
• Recall that:
if possible, systems react to an imposed change by reducing the
impact of that change
PbF2(s)
Pb2+(aq) + 2F-(aq)
• Increase [F-] (by adding KF - soluble)
• Force equilibrium to left, and therefore reduce
[Pb2+]
Precipitation Reactions
• A solid will form (precipitate) if the concentrations of the
ions exceed the solubility product.
bAa+(aq) + aBb-(aq)
AbBa(s)
First calculate the initial “reaction quotient”, Qo :
Qo = [Aa+]ob [Bb-]oa
Then compare Qo with Ksp
Precipitation Reactions #2
First calculate the initial “reaction quotient”, Qo :
Qo = [Aa+]ob [Bb-]oa
Then compare Qo with Ksp :
If Qo > Ksp Precipitation Occurs
If Qo < Ksp No Precipitation
Precipitation Reactions #3
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?
First calculate the initial “reaction quotient”, Qo :
Qo = [Ag+]o [Cl-]o = 2 x 10-4
1 x 10-5
= 2 x 10-9 M2
x
Then compare Qo with Ksp (1.8 x 10-10 M2):
Is Qo > Ksp Yes! Precipitation Occurs
Precipitation Reactions #4
Example 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?
Cd2+(aq) + 2OH-(aq)
Cd(OH)2(s)
Precipitation Reactions #5
Example 2 cont...
Ksp(Cd(OH)2) = 5.3 x 10-15 M3
Cd2+(aq) + 2OH-(aq)
Cd(OH)2(s)
Ksp = 1 x 10-6 x s2 = 5.3 x 10-15
s 
5.3 x 1 0-9
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 Precipitation
Problem: Ag+(aq) as silver nitrate is added to a
mixture containing 0.001 M Cl- and 0.001 M CrO42-.
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 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
Objectives Covered in Lecture 2
After studying this lecture 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.