K sp and Molar Solubility

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AP CHEMISTRY

CHAMBERLAIN

1

CHAPTER 16

The Solubility Product Constant, K

sp

Many important ionic compounds are only slightly soluble in water (we used to call them “insoluble” – Chapter 4)

 An equation can represent the equilibrium between the compound and the ions present in a saturated aqueous solution:

Solubility product constant, K sp

:

 K sp

for the BaSO

4

example above:

 Write the dissociation reaction for each solid: o Al(OH)

3 o Mg

3

(PO

4

)

2

2

3

Try It Out: Write a chemical equation AND the solubility constant expression representing solubility equilibrium for Ag

3

AsO

4

, K sp

=

1.0 x 10 -22 .

4

K

sp

and Molar Solubility

K sp

is an ____________________________________

 Molar solubility is the

Molar solubility is ______________________ to the value of

K sp

, but molar solubility and K sp

are ___________ the same thing.

In fact, ____________________________________ doesn’t always mean ________________________________________

 Solubility depends on both K sp

and the form of the equilibrium constant expression

Example 16.2 At 20 °C, a saturated aqueous solution of silver carbonate contains 32 mg of Ag

2

CO

3

per liter of solution. Calculate

K sp

for Ag

2

CO

3

at 20 °C. The balanced equation is

Ag

2

CO

3

(s) ↔ 2 Ag + (aq) + CO

3

2– (aq) K sp

= ?

5

Example 16.3 From the K sp

value for silver sulfate, calculate its molar solubility at 25 °C.

Ag

2

SO

4

(s) ↔ 2 Ag + (aq) + SO

4

2– (aq) K sp

= 1.4 x 10 –5 at 25 °C

Try It Out: Calculate the K sp

value for cadmium iodate, solubility

0.097 g Cd(IO

3

)

2

/100 mL H

2

O.

6

Common Ion Effect in Solubility Equilibria

The common ion effect affects solubility equilibria as it does other aqueous equilibria.

The solubility of a slightly soluble ionic compound is

______________________ when a second solute that furnishes a common ion is added to the solution.

Try It Out: Calculate the molar solubility of Mg(OH)

2

in 0.25 M

NaOH (aq).

7

Solubilities and Activities

Ions that are ____________ common to the precipitate can also affect solubility o CaF

2

is ____________ soluble in 0.010 M Na

2

SO

4

than it is in water.

Increased solubility occurs because of interionic attractions.

 Each…

The effective concentrations, or _____________________, of

Ca +2 and F -1 are lower than their actual concentrations.

Will Precipitation Occur? Is it complete?

Q sp

is the

Q ip

can then be compared to K sp

. o Precipitation should occur if ______________. o Precipitation cannot occur if ______________. o A solution is just saturated if ______________.

 In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered.

8

Example 16.6 If 1.00 mg of Na

2

CrO

4

is added to 225 mL of

0.00015 M AgNO

3

, will a precipitate form?

Ag

2

CrO

4

(s) ↔ 2 Ag + (aq) + CrO

4

2– (aq) K sp

= 1.1 x 10 –12

To determine whether precipitation is complete:

A slightly soluble solid does not precipitate _______________ from solution, but we generally consider precipitation to be

“complete” if 99.9% of the target ion is precipitated (0.1% or less is left in solution)

Three conditions generally _____________________________

__________________________ of precipitation:

1.

A _____________________ value of K sp

.

2.

A _________________________ of the target ion.

3.

A concentration of common ion that _________________

________________________that of the target ion.

9

Example 16.9 To a solution with [Ca 2+ ] = 0.0050 M, we add sufficient solid ammonium oxalate, (NH

4

)

2

C

2

O

4

(s), to make the initial [C

2

O

4

2– ] = 0.0051 M. Will precipitation of Ca 2+ as CaC

2

O

4

(s) be complete?

CaC

2

O

4

(s) ↔ Ca 2+ (aq) + C

2

O

4

2– (aq) K sp

= 2.7 x 10 –9

10

Effect of pH on Solubility:

• If the anion of a precipitate is that of a weak acid, the precipitate will dissolve somewhat when the pH is lowered:

CaF

2

(s) ↔ Ca 2+ (aq) + 2 F – (aq)

If, however, the anion of the precipitate is that of a strong acid, lowering the pH will have no effect on the precipitate.

AgCl(s) ↔ Ag + (aq) + Cl – (aq)

Will pH effect the solubility of: o CaCO

3 o Mg(OH)

2 o AgCl o KNO

3

Example 16.11 What is the molar solubility of Mg(OH)

2

(s) in a buffer solution having [OH – ] = 1.0 x 10 –5 M, that is, pH = 9.00?

Mg(OH)

2

(s) ↔ Mg 2+ (aq) + 2 OH – (aq) K sp

= 1.8 x 10 –11

11

Complex Ion Formation

A _____________________ consists of a central metal atom or ion, with other groups called _________________bonded to it.

The __________ ion acts as a Lewis acid (accepts electron pairs).

__________________ act as Lewis bases (donate electron pairs).

The equilibrium involving a complex ion, the metal ion, and the ligands may be described through a ______________________

Examples

• A complex ion contains cobalt (III) and 6 ammonia atoms.

Write the formula for this complex ion.

• A complex ion contains zinc and 4 cyanide ions. Write the formula for this complex ion.

12

13

Example 16.14 If 1.00 g KBr is added to 1.00 L of the solution described in Example 16.13, should any AgBr(s) precipitate from the solution?

AgBr(s) ↔ Ag + (aq) + Br – (aq) K sp

= 5.0 x 10 –13

14

Example 16.15 What is the molar solubility of AgBr(s) in 3.0 M

NH

3

?

AgBr(s) + 2 NH

3

(aq) ↔ [Ag(NH

3

)

2

] + (aq) + Br – (aq)

K c

= 8.0 x 10 –6

Amphoteric Species

 Certain metal hydroxides, insoluble in water, are amphoteric; they will react with both strong acids and strong bases

________________, ___________________, and

________________ are amphoteric.

15

CHAPTER 17

Introduction

________________________________ examines the relationship between heat and work.

Why study thermodynamics? o With a knowledge of thermodynamics and by making a few calculations before embarking on a new venture, scientists and engineers can save themselves a great deal of time, money, and frustration. o Thermodynamics tells us what processes are

_______________.

 Kinetics tells us whether the process is

_____________

2

nd

Law of Thermodynamics

 Says that any spontaneous process…

More order requires a great deal of work and is often impossible to achieve o To put one system in order requires that you put others in __________________.

_______________________ is a mathematical concept describing the distribution of energy within a system.

16

o It is a measure of molecular randomness, or disorder.

 ________________________ means not requiring an outside source of energy to proceed. o It doesn’t mean that the process happens

________________. It just means that it can happen without outside energy _________________________.

Spontaneous Change

o A __________________________ process is one that cannot take place in a system left to itself. o If a process is spontaneous… o Example: o Kinetics determines the ____________________ by which equilibrium is reached. o A high ____________________

______________________ can effectively block a reaction that is thermodynamically favored. o Example: o Early chemists proposed that spontaneous chemical reactions should occur in the direction of

_________________________

17

o It is true that many ____________________________ processes are spontaneous and that many

__________________________ reactions are nonspontaneous o However, enthalpy change is not a

______________________ criterion for predicting spontaneous change

What about spontaneous endothermic reactions?

 The idea that exothermic reactions are spontaneous and that endothermic reactions are not works in many cases, not all cases.

 Example:

We need to look at factors other than enthalpy (H), which is where entropy (S) comes in.

Entropy (S)

The gases in the figure on p. 19 mix because of entropy.

18

The ___________________ of the system remains unchanged in the mixing of the gases …

 … but the number of ____________________for the

______________________of that energy increases.

 The spreading of the energy among states, and increase of entropy, often correspond to a greater physical disorder at the microscopic level (however, entropy is not “disorder”).

 There are two driving forces behind spontaneous processes: the tendency to achieve a lower energy state (enthalpy change) and the tendency for energy to be distributed among states (entropy).

 In many cases, however, the two factors work in opposition.

One may increase and the other decrease or vice versa. In these cases, we must determine which factor predominates.

Assessing Entropy Change

The difference in entropy (S) between two states is the

____________________________.

 The greater the number of configurations of the microscopic particles (atoms, ions, molecules) among the energy levels in a particular state of a system, the ___________________ is the entropy of the system.

19

Entropy generally increases when:

1. Solids melt to form liquids.

2. Solids or liquids vaporize to form gases.

3. Solids or liquids dissolve in a solvent to form nonelectrolyte solutions.

4. A chemical reaction produces an increase in the number of molecules of gases

5. A substance is heated.

Examples: o Melting Ice o Crystallization o Evaporation o Dissolving a Solid o Freezing

Example 17.2 Predict whether each of the following leads to an increase or decrease in the entropy of a system. If in doubt, explain why.

(a) The synthesis of ammonia:

N

2

(g) + 3 H

2

(g)  2 NH

3

(g)

20

(b) Preparation of a sucrose solution:

C

12

H

22

O

11

(s) → C

12

H

22

O

11

(aq)

(c) Evaporation to dryness of a solution of urea, CO(NH

2

)

2

, in water:

CO(NH

2

)

2

(aq)  CO(NH

2

)

2

(s)

Try It Out: Which reactions below would have an INCREASE in entropy?

1.

H

2

O (s)  H

2

O (g)

2.

CO

2

(g)  CO

2

(l)

3.

C

2

H

4

(g) + H

2

(g)  C

2

H

6

(g)

4.

2 KClO

3

(s)  2 KCl (s) + 3 O

2

(g)

So far, Spontaneity is more likely if o Enthalpy ( ∆ H) is ____________________________ o Entropy ( ∆ S) is __________________________

21

Standard Molar Enthalpies

• According to the _______________________________, the entropy of a pure, perfect crystal can be taken to be zero at 0

K.

• The ________________________________________ is the entropy of one mole of a substance in its standard state.

• Since entropy increases with temperature, standard molar entropies are positive—even for elements.

• Entropy is a state function, so it is calculated in the same way that we calculated enthalpy first semester!

S =

Σ

v p

S°(products) – Σ

v r

S°(reactants)

22

Try It Out: The following reaction may occur in the high-temperature, catalyzed oxidation of NH

3

(g). Use data from Appendix C to calculate the standard molar entropy change for this reaction at 25°C.

4 NH

3

(g) + 3 O

2

(g)  2 N

2

(g) + 6 H

2

O (g)

Entropy & Molecular Structure

In general, the more atoms in its molecules, o Molecules with more atoms have more ways of moving with respect to each other – more vibrational modes.

The more ways that molecules can absorb energy, the more energy levels available for dispersal of this energy, and the greater the entropy.

The Second Law of Thermodynamics (AGAIN)

• Entropy can be used as a sole criterion for spontaneous change, but the entropy change of both the ____________ and its _______________________must be considered.

23

• ____________________________________ establishes that all spontaneous processes increase the entropy of the

______________________(system and surroundings).

• If entropy increases in both the _______________ and the

__________________, the process is spontaneous. o Is it possible for a spontaneous process to exhibit a

decrease in entropy?

Free Energy & Free Energy Change (Gibbs)

• We do not have an effective way to measure the entropy of the universe so we have another thermodynamic function –

• Gibbs Free Energy is the amount of energy in a system that is available to do _____________________________.

• For a process at constant temperature and pressure:

_______________________ = enthalpy-driven o The flow of thermal energy provides most of the free energy for the reaction.

_______________________ = entropy-driven o Reaction disorder provides most of the free energy for the reaction.

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o If __________(negative), a process is _________________. o If ________ (positive), a process is ___________________. o If ___________, neither the forward nor the reverse process is favored; there is no net change, and the process is at _________________________.

Example 17.4 Predict which of the four cases in Table 17.1 you expect to apply to the following reactions.

(a) C

6

H

12

O

6

(s) + 6 O

2

(g)  6 CO

2

(g) + 6 H

2

O(g) ΔH = –2540 kJ

(b) Cl

2

(g)  2 Cl(g)

25

Example 17.4 A Predict which of the four cases in Table 17.1 you expect to apply to the following reactions.

(a) N

2

(g) + 2 F

2

(g)  N

2

F

4

(g) + 6 H

2

O(g) ΔH = –7.1 kJ

(b) COCl

2

(g)  CO(g) + Cl

2

(g) ΔH = +110.4 kJ

Standard Free Energy Change, ∆G°

• The _______________________________, of a reaction is the free energy change when reactants and products are in their standard states.

• The _______________________________ is the free energy change for the formation of 1 mol of a substance in its standard state from the elements in their standard states.

26

27

Example 17.6 A

Use the standard-state form of the Gibbs equation to determine the standard free energy change at 25 ᵒ C for the reaction:

2 NO (g) + O

2

(g)  2 NO

2

(g) ∆ H ᵒ = -114.1 kJ ∆ S ᵒ = -146.2 J/K

Example 17.6 B

Use standard free energies of formation from Appendix C to determine the standard free energy change at 25 ᵒ C for the reaction:

CS

2

(l) + 2 S

2

Cl

2

(g)  CCl

4

(l) + 6 S (s)

28

Relationship of ∆G° to the Equilibrium Constant K

eq

• ______________ is a criterion for equilibrium at any temperature.

• ______________ is a criterion for equilibrium at a single

temperature, that temperature at which the equilibrium state has all reactants and products in their standard states.

• ΔG and ∆ G o are related through the reaction quotient, Q:

• When ∆ G = 0, then Q = K eq

, and the equation becomes:

• The concentrations and partial pressures we have used in K eq are approximations.

• ___________________ (a) are the correct variables for K eq

.

But activities are very difficult to determine. That is why we use approximations. o For pure solid and liquid phases: a = 1. o For gases: Assume ideal gas behavior, and replace the activity by the numerical value of the gas partial pressure (in atm). o For solutes in aqueous solution: Replace solute activity by the numerical value of the solute molarity (M).

29

Calculating Equilibrium Constants

30

The Sign and Magnitude of ∆ G°

Example 17.9

Determine the value of K eq

at 25 ᵒ C for the reaction:

2 NO

2

(g) ↔ N

2

O

4

(g)

31

Example 17.9 A

Use data from Appendix C to determine K eq

at 25 ᵒ C for the reaction:

2 HgO (s) ↔ 2 Hg (l) + O

2

(g)

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