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AP* Chemistry CHEMICAL EQUILIBRIA: GENERAL CONCEPTS *AP is a registered trademark of the College Board, which was not involved in the production of this product.© 2013 by René McCormick. All rights reserved. THE NATURE OF THE EQUILIBRIUM STATE: Equilibrium is the state where the rate of the forward reaction is equal to the rate of the reverse reaction. At these conditions, concentrations of all reactants and products remain constant with time once equilibrium has been established at constant temperature. In stoichiometry, we assumed chemical systems “went to completion”. A chemist would be quick to point out that the vast majority of chemical reactions do not go to completion; they just have very large K values! Never, ever forget that equilibrium is TEMPERATURE DEPENDENT. Reactions are reversible. This is indicated by double arrows. R x Dynamic equilibrium--R indicates that the reaction is proceeding in the forward and in the reverse direction simultaneously and once equilibrium is established, the rate of each direction is equal. This also keeps the concentration of reactants and products constant (which is not to be confused with “equal”). x The nature and properties of the equilibrium state are the same, no matter what the direction of approach. x Examples: Look at the following plot of the reaction between steam and carbon monoxide in a closed vessel at a high temperature where the reaction takes place rapidly. THE EQUILIBRIUM POSITION Whether the reaction lies far to the right (favors products) or to the left (favors reactants) depends on three main factors: x Initial concentrations (more collisions--faster reaction) x Relative energies of reactants and products (nature goes to minimum energy) x Degree of organization of reactants and products (nature goes to maximum disorder) The significance of K: K > 1 means that the reaction favors the products at equilibrium K < 1 means that the reaction favors the reactants at equilibrium THE EQUILIBRIUM EXPRESSION: A general description of the equilibrium condition proposed by Gudberg and Waage in 1864 is known as the Law of Mass Action. Equilibrium is temperature dependent, however, it does not change with concentration or pressure. x equilibrium constant expression--for the general reaction aA + bB R cC + dD Equilibrium constant: K = [C] c [D] d * Note* K, K c , K eq may all be used here! [A] a [B] b
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Page 1: AP* Chemistry - Lundquist Labs

AP* Chemistry CHEMICAL EQUILIBRIA: GENERAL CONCEPTS

*AP is a registered trademark of the College Board, which was not involved in the production of this product.© 2013 by René McCormick. All rights

reserved.

THE NATURE OF THE EQUILIBRIUM STATE: Equilibrium is the state where the rate of the forward

reaction is equal to the rate of the reverse reaction. At these conditions, concentrations of all reactants and

products remain constant with time once equilibrium has been established at constant temperature.

In stoichiometry, we assumed chemical systems “went to completion”. A chemist would be quick to point out

that the vast majority of chemical reactions do not go to completion; they just have very large K values! Never,

ever forget that equilibrium is TEMPERATURE DEPENDENT.

Reactions are reversible. This is indicated by double arrows.

Dynamic equilibrium-- indicates that the reaction is proceeding in the forward and in the reverse

direction simultaneously and once equilibrium is established, the rate of each direction is equal. This

also keeps the concentration of reactants and products constant (which is not to be confused with

“equal”).

The nature and properties of the equilibrium state are the same, no matter what the direction of

approach.

Examples: Look at the following plot of the reaction between steam and carbon monoxide in a closed

vessel at a high temperature where the reaction takes place rapidly.

THE EQUILIBRIUM POSITION

Whether the reaction lies far to the right (favors products) or

to the left (favors reactants) depends on three main factors:

Initial concentrations (more collisions--faster

reaction)

Relative energies of reactants and products (nature

goes to minimum energy)

Degree of organization of reactants and products

(nature goes to maximum disorder)

The significance of K:

K > 1 means that the reaction favors the products at equilibrium

K < 1 means that the reaction favors the reactants at equilibrium

THE EQUILIBRIUM EXPRESSION: A general description of the equilibrium condition proposed by

Gudberg and Waage in 1864 is known as the Law of Mass Action. Equilibrium is temperature dependent,

however, it does not change with concentration or pressure.

equilibrium constant expression--for the general reaction

aA + bB cC + dD

Equilibrium constant: K = [C]c[D]

d * Note* K, Kc, Keq may all be used here!

[A]a[B]

b

Page 2: AP* Chemistry - Lundquist Labs

Chemical Equilibria: General Concepts

2

The product concentrations appear in the numerator and the reactant concentrations in the denominator.

Each concentration is raised to the power of its stoichiometric coefficient in the balanced equation.

- [ ] indicates concentration in Molarity (mol/L)

- Kc--is for concentration (aqueous)

- Kp--is for partial pressure (gases)

- “K” values are often written without units

USING EQUILIBRIUM CONSTANT EXPRESSIONS

Pure solids--do not appear in expression—you’ll see this in Ksp problems soon!

Pure liquids--do not appear in expression—H2O(l) is pure, so leave it out of the calculation

Water--as a pure liquid or reactant, does not appear in the expression. (55.5 M will not change

significantly)

o Weak acid and weak base equations are heterogeneous [multi-states of matter; pure liquid and

aqueous components] equilibria.

o Solubility of salts also fits into this category. The initial solid component has a constant

concentration and is therefore left out of the equilibrium expression.

Exercise 1 Writing Equilibrium Expressions

Write the equilibrium expression for the following reaction:

4 NH3(g) + 7 O2(g) 4 NO2(g) + 6 H2O(g)

K = [NO2]4[H2O]

6

[NH3]4[O2]

7

Exercise 2 Equilibrium Expressions for Heterogeneous Equilibria

Write the expressions for K and Kp for the following processes:

a. The decomposition of solid phosphorus pentachloride to liquid phosphorus trichloride and chlorine gas.

b. Deep blue solid copper(II) sulfate pentahydrate is heated to drive off water vapor to form white solid copper(II) sulfate.

A: K = [Cl2]

Kp = PCl2

B: K = [H2O]5

Kp = PH2O5

Page 3: AP* Chemistry - Lundquist Labs

Chemical Equilibria: General Concepts

3

CHANGING STOICHIOMETRIC COEFFICIENTS

when the stoichiometric coefficients of a balanced equation are multiplied by some factor, the K is

raised to the power of the multiplication factor (Kn). Thus, 2x is K squared; 3x is K cubed; etc.

REVERSING EQUATIONS

take the reciprocal of K ( 1/K)

ADDING EQUATIONS

multiply respective Ks (K1 × K2 × K3 …)

Exercise 3 Calculating the Values of K

The following equilibrium concentrations were observed for the Haber process at 127°C:

[NH3] = 3.1 × 10 2 mol/L

[N2] = 8.5 × 10 1 mol/L

[H 2] = 3.1 × 10 3 mol/L

a. Calculate the value of K at 127°C for this reaction.

b. Calculate the value of the equilibrium constant at 127°C for the reaction:

2 NH3(g) N2(g) + 3 H2(g)

c. Calculate the value of the equilibrium constant at 127°C for the reaction given by the equation:

1

2 N2(g) + 3

2 H2(g) NH3(g)

A: K = 3.8 × 104

B: K’ = 2.6 × 10-5

C: K” = 1.9 × 10

2

Page 4: AP* Chemistry - Lundquist Labs

Chemical Equilibria: General Concepts

4

Exercise 4 Equilibrium Positions

The following results were collected for two experiments involving the reaction at 600°C between gaseous sulfur dioxide and oxygen

to form gaseous sulfur trioxide:

Show that the equilibrium constant is the same in both cases.

K1 = 4.36

K2 = 4.32

Kc & Kp--NOT INTERCHANGEABLE! Kp = Kc(RT)n

where n is the change in the number of moles of gas going from reactants to products:

n = total moles gas produced total moles gas reacting

o R = universal gas law constant 0.0821 L atm/ mol K

o T = temperature in Kelvin

Kc = Kp if the number of moles of gaseous product = number of moles of gaseous

reactant since (RT)n = (RT)

0 = 1

Kp = Kc(RT)n is often referred to as the “politically correct” (pc) equation to help you remember the

order of the Ks in the equation!

Page 5: AP* Chemistry - Lundquist Labs

Chemical Equilibria: General Concepts

5

Exercise 5 Calculating Values of Kp

The reaction for the formation of nitrosyl chloride

2 NO(g) + Cl2(g) 2 NOCl(g)

was studied at 25°C. The pressures at equilibrium were found to be

PNOCl = 1.2 atm

PNO = 5.0 × 10-2 atm

PCl2 = 3.0×10-1 atm

Calculate the value of Kp for this reaction at 25°C.

= 1.9 × 103

Exercise 6 Calculating K from Kp

Using the value of Kp obtained in Sample Exercise 13.4, calculate the value of K at 25° C for the reaction:

2 NO(g) + Cl2(g) 2 NOCl(g)

= 4.6 × 104

Page 6: AP* Chemistry - Lundquist Labs

Chemical Equilibria: General Concepts

6

MAGNITUDE OF K--what does it mean anyway? When greater than one, formation of products is

favored. When less than one, formation of reactants is favored.

Can you...

1. …write an equilibrium constant expression?

2. …tell how K is changed if the stoichiometric coefficients are changed on an equation?

3. … tell how to find K for a summary equation?

4. …tell how K depends on the way equilibrium concentrations are expressed and how to

convert K in terms of Kc vs. Kp?

5. …explain what K is telling you about a reaction?

THE REACTION QUOTIENT

For use when the system is NOT at equilibrium.

For the general reaction

aA + bB cC + dD

Reaction quotient = Qc = [C]c[D]

d

[A]a[B]

b

Qc has the appearance of K but the concentrations are not necessarily at equilibrium.

1. If Q < K, the system is not at equilibrium: Reactants products to make Q = K at equil.

2. If Q = K, the system is at equilibrium.

3. If Q > K, the system is not at equilibrium: Reactants products to make Q = K at equil.

Quite useful for predicting what will happen under special conditions.

Exercise 7 Using the Reaction Quotient

For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0 × 10-2. Predict the direction in which the system

will shift to reach equilibrium in each of the following cases:

a. [NH3]0 = 1.0 × 10-3 M; [N2]0 = 1.0 × 10-5 M; [H2]0 = 2.0 × 10-3 M

b. [NH3]0 = 2.00 × 10-4 M; [N2]0 = 1.50 × 10-5 M; [H2]0 = 3.54 × 10-1 M

c. [NH3]0 = 1.0 × 10-4 M; [N2]0 = 5.0 M; [H2]0 = 1.0 × 10-2 M

A: shift left

B: no shift

C: shift right

Page 7: AP* Chemistry - Lundquist Labs

Chemical Equilibria: General Concepts

7

SOME CALCULATIONS WITH THE EQUILIBRIUM CONSTANT

General steps for solving equilibrium problems.

1. Set up “RICE” TABLE—write a balanced reaction, place initial concentrations into the

table, determine the change in initial concentrations in terms of x, calculate the

equilibrium concentration expressions (0.25 M – x).

R = write a balanced reaction for the predominant reacting species

I = fill in the initial concentrations

C = determine the change that is taking in place in terms of x

E = express the equilibrium concentrations in terms of x

2. Set up the equilibrium expression and set it equal to its value, if given.

3. Celebrate if you are given equilibrium concentrations! Just skip down to the “E” line and

fill them in. You may be asked to work backwards to determine the “change” in

equilibrium.

4. If you are given a K value, then use it to solve for x and use x to calculate the equilibrium

concentrations.

Hints:

Look for very small K values (where K < 105), "x" may be negligible.

If "x" is large enough to impact the equilibrium values, then you must subtract it

from the initial concentration. Your math will be simplified if the problem is a

perfect square. If not, you must use the quadratic formula. You are allowed to use

your calculator with a solver or program for solving the quadratic.

If none of the initial concentrations are zero, then Q must be calculated first to

determine the direction of the shift before following the above general steps.

Exercise 8 Calculating Equilibrium Pressures I

Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander for the NASA Apollo missions.

In the gas phase it decomposes to gaseous nitrogen dioxide:

N2O4(g) 2 NO2(g)

Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a temperature

where Kp = 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of

NO2(g).

= 0.600 atm

Page 8: AP* Chemistry - Lundquist Labs

Chemical Equilibria: General Concepts

8

Exercise 9 Calculating Equilibrium Pressures II

At a certain temperature a 1.00-L flask initially contained 0.298 mol PCl3(g) and 8.70 × 10-3 mol PCl5(g). After the

system had reached equilibrium, 2.00 × 10-3 mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes according to

the reaction

PCl5(g) PCl3(g) + Cl2(g)

Calculate the equilibrium concentrations of all species and the value of K.

[Cl2] = 2.00 × 10-3

M

[PCl3] = 0.300 M

[PCl5] = 6.70 × 10-3

M

K = 8.96 × 10-2

Exercise 10 Calculating Equilibrium Concentrations I

Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is 5.10.

Calculate the equilibrium concentrations of all species if 1.000 mol of each component is mixed in a 1.000-L flask.

[CO] = [H2O] = 0.613 M

[CO2] = [H2] = 1.387 M

Page 9: AP* Chemistry - Lundquist Labs

Chemical Equilibria: General Concepts

9

Exercise 11 Calculating Equilibrium Concentrations II

Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium

constant of 1.15 × 102 at a certain temperature. In a particular experiment, 3.000 mol of each component was added to

a 1.500-L flask. Calculate the equilibrium concentrations of all species.

[H2] = [F2] = 0.472 M

[HF] = 5.056 M

Exercise 12 Calculating Equilibrium Pressures

Assume that gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where the

equilibrium constant is 1.00 × 102. Suppose HI at 5.000 × 10-1 atm, H2 at 1.000×10-2 atm, and I2 at 5.000 × 10-3 atm are

mixed in a 5.000-L flask. Calculate the equilibrium pressures of all species.

PHI = 4.29 × 10-1

atm

PH2 = 4.55 × 10-2

atm

PI2 = 4.05 × 10-2

atm

Page 10: AP* Chemistry - Lundquist Labs

Chemical Equilibria: General Concepts

10

EXTERNAL FACTORS AFFECTING EQUILIBRIA

Le Chatelier’s Principle: If a stress is applied to a system at equilibrium, the position of the

equilibrium will shift in the direction which reduces the stress.

Shifts occur to reestablish equilibrium positions. Think about K! Generally, products

reactants

Adding or removing a reagent causes the equilibrium to shift to reestablish K.

As long as you milk a “mamma” cow, she keeps making more milk!

Increasing pressure causes the equilibrium to shift to the side containing the fewest number of

moles of gas. The converse is also true.

Adding a catalyst to a reaction has causes no shift and has NO EFFECT on K. It just causes

equilibrium to be established faster!

Changing the temperature is a lot like adding or removing a “reactant or product”. Well, as

long as you think of heat energy as a “reactant” or “product”.

Think of it this way:

ENDOTHERMIC: A + B C + D + H value (heat is added into the system heat is a reactant)

Rewrite: A + B + heat C + D

EXOTHERMIC: A + B C + D H value (heat is lost from the system heat is a product)

Rewrite: A + B C + D + heat

Exercise 13 Using Le Chatelier’s Principle I

Arsenic can be extracted from its ores by first reacting the ore with oxygen (called roasting) to form solid As4O6, which is

then reduced using carbon:

As4O6(s) + 6 C(s) As4(g) + 6 CO(g)

Predict the direction of the shift of the equilibrium position in response to each of the following changes in conditions.

a. Addition of carbon monoxide

b. Addition or removal of carbon or tetraarsenic hexoxide (As4O6)

c. Removal of gaseous arsenic (As4)

A: shift left

B: no shift

C: shift right

Page 11: AP* Chemistry - Lundquist Labs

Chemical Equilibria: General Concepts

11

Exercise 14 Using Le Chatelier’s Principle II

Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced.

a. The preparation of liquid phosphorus trichloride by the reaction:

P4(s) + 6 Cl2(g) 4 PCl3(l)

b. The preparation of gaseous phosphorus pentachloride according to the equation:

PCl3(g) + Cl2(g) PCl5(g)

c. The reaction of phosphorus trichloride with ammonia:

PCl3(g) + 3 NH3(g) P(NH2)3(g) + 3 HCl(g)

A: shift right

B: shift right

C: no shift

Exercise 15 Using Le Chatelier’s Principle III

For each of the following reactions, predict how the value of K changes as the temperature is increased.

a. N2(g) + O2(g) 2 NO(g) H° = 181 kJ

b. 2 SO2(g) + O2(g) 2 SO3(g) H° = 198 kJ

A: increases

B: decreases

Page 12: AP* Chemistry - Lundquist Labs

AP* Chemistry

Solubility Equilibrium

*AP is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.© 2008 by René

McCormick. All rights reserved

SOLUBILITY EQUILIBRIA (The Solubility-Product Constant, Ksp)

We’ve got good news and we’ve got bad news… Which do you want first?

The good news: Solubility equilibrium is really simple.

The bad news: You know all those solubility rules that state a substance is insoluble? They are actually a little

bit soluble after all. Only the future attorneys among you read the fine print. Soluble is often defined as

“greater than 3 grams dissolving in 100 mL of water”. So, there is a lot of wiggle room for solubility up to 3

grams! This type of equilibria deals with that wiggle room. Apologies for “lying” to you.

If you can actually see that a salt is insoluble, usually evidenced by solid “stuff” sitting on the bottom of the

container, then the solution is actually saturated. Saturated solutions of salts present due to a chemical reaction

taking place present yet another type of chemical equilibria.

Slightly soluble salts establish a dynamic equilibrium with the hydrated

cations and anions in solution. Examine the formation of AgCl (s) as a

solution of AgNO3(aq) is squirted into a solution of NaCl(aq).

When the solid is first added to water, no ions are initially present.

As dissolution proceeds, the concentration of ions increases until

equilibrium is established. This occurs when the solution is

saturated.

The equilibrium constant, the Ksp, is no more than the product of the

ions in solution. (Remember, solids do not appear in equilibrium

expressions.)

For a saturated solution of AgCl, the equation would be:

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

The solubility product expression for the AgCl(s) precipitate would be:

Ksp = [Ag+][Cl ]

The AgCl(s) does not appear in the equilibrium expression since solids are left out. Why?

Because, the concentration of the solid remains relatively constant.

A table of Ksp values follows on the next page.

Page 13: AP* Chemistry - Lundquist Labs

Solubility Equilibria 2

Write the Ksp expression for each of the following reactions and find its value in the table above.

CaF2(s) Ca+2

+ 2 F Ksp =

Ag2CrO4(s) 2 Ag+ + CrO4

2 Ksp =

CaC2O4(s) Ca+2

+ C2O42

Ksp =

DETERMINING Ksp FROM EXPERIMENTAL MEASUREMENTS

In practice, Ksp values are determined by careful laboratory measurements using various spectroscopic

methods.

Remember STOICHIOMETRY? Surely, you’ve made peace with the concept by now…

Page 14: AP* Chemistry - Lundquist Labs

Solubility Equilibria 3

Example: Lead(II) chloride dissolves to a slight extent in water according to the equation below.

PbCl2(s) Pb+2

+ 2Cl

Calculate the Ksp if the lead ion concentration has been found to be 1.62 × 102M.

If lead’s concentration is x, then chloride’s concentration is 2x. So. . . .

Ksp = (1.62 × 102)(3.24 × 10

2)

2 = 1.70 × 10

5

Exercise 1 Calculating Ksp from Solubility I

Copper(I) bromide has a measured solubility of 2.0 × 104 mol/L at 25°C. Calculate its

Ksp value.

Ksp = 4.0 × 108

Exercise 2 Calculating Ksp from Solubility II

Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of

1.0 × 1015

mol/L at 25°C.

Ksp = 1.1 × 1073

Page 15: AP* Chemistry - Lundquist Labs

Solubility Equilibria 4

ESTIMATING SALT SOLUBILITY FROM Ksp

Relative solubilities can be deduced by comparing values of Ksp BUT, BE CAREFUL!

These comparisons can only be made for salts having the same ION:ION ratio.

Please don’t forget solubility changes with temperature! Some substances become less soluble in cold water

while other increase in solubility! Aragonite is an example.

Example: The Ksp for CaCO3 is 3.8 × 109 @ 25 C. Calculate the solubility of calcium carbonate in pure

water in (a) moles per liter & (b) grams per liter:

Exercise 3 Calculating Solubility from Ksp

The Ksp value for copper(II) iodate, Cu(IO3)2, is 1.4 × 107 at 25°C. Calculate its solubility at 25°C.

= 3.3 × 103 mol/L

Exercise 4 Solubility and Common Ions

Calculate the solubility of solid CaF2 (Ksp = 4.0 × 1011

) in a 0.025 M NaF solution.

= 6.4 × 10-8

mol/L

Page 16: AP* Chemistry - Lundquist Labs

Solubility Equilibria 5

Ksp AND THE REACTION QUOTIENT, Q

With some knowledge of the reaction quotient, we can decide

whether a precipitate (ppt) will form AND

what concentrations of ions are required to begin the precipitation of an insoluble salt.

1. Q < Ksp, the system is not at equil. (unsaturated)

2. Q = Ksp, the system is at equil. (saturated)

3. Q > Ksp, the system is not at equil. (supersaturated)

Precipitates form when the solution is supersaturated!!!

Precipitation of insoluble salts

Metal-bearing ores often contain the metal in the form of an insoluble salt, and, to complicate

matters, the ores often contain several such metal salts.

Dissolve the metal salts to obtain the metal ion, concentrate in some manner, and ppt. selectively

only one type of metal ion as an insoluble salt.

Exercise 5 Determining Precipitation Conditions

A solution is prepared by adding 750.0 mL of 4.00 × 103 M Ce(NO3)3 to 300.0 mL of 2.00 × 10

2 M KIO3.

Will Ce(IO3)3 (Ksp = 1.9 × 1010

) precipitate from this solution?

yes

Page 17: AP* Chemistry - Lundquist Labs

Solubility Equilibria 6

Exercise 6 Precipitation

A solution is prepared by mixing 150.0 mL of 1.00 × 10-2

M Mg(NO3)2 and

250.0 mL of 1.00 × 101 M NaF. Calculate the concentrations of Mg

2+ and F

at equilibrium with solid MgF2 (Ksp = 6.4 × 109).

[Mg2+

] = 2.1 × 10-6

M

[F-] = 5.50 × 10

-2 M

SOLUBILITY AND THE COMMON ION EFFECT

Experiments show that the solubility of any salt is always less in the

presence of a “common ion”.

Why? LeChâtelier’s Principle, that’s why! Be reasonable and use

approximations when you can.

The pH can also affect solubility. Evaluate the equation to see which

species reacts with the addition of acid or base.

Would magnesium hydroxide be more soluble in an acid or a base? Why?

Mg(OH)2(s) Mg2+

(aq) + 2 OH (aq) (milk of magnesia)

Page 18: AP* Chemistry - Lundquist Labs

Solubility Equilibria 7

SOLUBILITY, ION SEPARATIONS, AND QUALITATIVE ANALYSIS

This section will introduce you to the basic chemistry of various ions.

It also illustrates how principles of chemical equilibria can be applied in the laboratory.

Objective: Separate the following metal ions from an aqueous sample containing ions of silver, lead(II),

cadmium(II) and nickel(II).

From your knowledge of solubility rules, you know that chlorides of lead and silver will form

precipitates while those of nickel and cadmium will not. Adding dilute HCl to sample will ppt. the lead

and silver ions while the nickel and cadmium will stay in solution.

Separate the lead and silver precipitates from the solution by filtration. Heating the solution causes

some of the lead chloride to dissolve. Filtering the HOT sample will separate the lead (in the filtrate)

from the silver (solid remaining in funnel with filter paper).

Separating cadmium and nickel ions require precipitation with sulfur. Use the Ksp values to determine

which ion will precipitate first as an aqueous solution of sulfide ion is added to the portion of the sample

that still contains these ions. Which precipitates first?

Page 19: AP* Chemistry - Lundquist Labs

Solubility Equilibria 8

Exercise 7 Selective Precipitation

A solution contains 1.0 × 104

M Cu+ and 2.0 × 10

3 M Pb

2+. If a source of I

- is added gradually to this

solution, will PbI2 (Ksp = 1.4 × 108) or CuI (Ksp = 5.3 × 10

12) precipitate first?

Specify the concentration of I necessary to begin precipitation of each salt.

CuI will precipitate first

Concentration in excess of 5.3 × 10-8

M required

SOLUBILITY AND COMPLEX IONS

The formation of complex ions can often dissolve otherwise

insoluble salts.

Often as the complex ion forms, the solubility equilibrium

shifts to the right (away from the solid) and causes the

insoluble salt to become more soluble.

Example: If sufficient aqueous ammonia is added to silver chloride, the

latter can be dissolved as [Ag(NH3)2]+

forms, which is soluble.

That is a significant improvement with regard to the solubility of AgCl(s).

The equilibrium constant for dissolving silver chloride in ammonia is not large, but, if the concentration of

ammonia is sufficiently high, the complex ion and chloride ion must also be high, and silver chloride will

dissolve. Of course, this process is quite smelly!

Page 20: AP* Chemistry - Lundquist Labs

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ConcentrationsatEquilibriumEquilibrium calculations – Rice Tables

1. Dinitrogen tetroxide in its liquid state was used as one of the fuels on the lunar lander for the NASA Apollo

missions. In the gas phase it decomposes to gaseous nitrogen dioxide:

N2O4 (g) ⇄ 2 NO2 (g)

Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a

temperature where Kp = 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the

equilibrium pressure of NO2(g).

2. At a certain temperature a 1.00-L flask initially contained 0.298 mol PCl3(g) and 8.70 × 10-3

mol PCl5(g). After

the system had reached equilibrium, 2.00 × 10-3

mol Cl2(g) was found in the flask. Gaseous PCl5 decomposes

according to the reaction

PCl5 (g) ⇄ PCl3(g) + Cl2 (g)

Calculate the equilibrium concentrations of all species and the value of K.

3. Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium

constant is 5.10. Calculate the equilibrium concentrations of all species if 1.000 mol of each component is

mixed in a 1.000-L flask.

4. Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an

equilibrium constant of 1.15 × 102 at a certain temperature. In a particular experiment, 3.000 mol of each

component was added to a 1.500-L flask. Calculate the equilibrium concentrations of all species.

5. Assume that gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature

where the equilibrium constant is 1.00 × 102. Suppose HI at 5.000 × 10

-1 atm, H2 at 1.000×10

-2 atm, and I2 at

5.000 × 10-3

atm are mixed in a 5.000-L flask. Calculate the equilibrium pressures of all species.

Page 23: AP* Chemistry - Lundquist Labs

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Page 39: AP* Chemistry - Lundquist Labs