chemical equilibrium

Reversible Reactions In a chemical reaction, Reactants are transformed into products.

The products formed can react to re-form the original reactants.

A

B + C

D

A

D

C

B +

A B + C D A D + B C

Reactions that can be traversed in both directions are said to reversible.

Theoretically at least, all chemical reactions are reversible.

Reversible Reactions

HgOHgO

HgOHgO

HgO

HgO

HgOHg

O2

HgHg

HgO

Hg

HgO

O2

HgO

2 HgO (s) 2 Hg (l) + O2 (g)

2 Hg (l) + O2 (g) 2HgO (g)

Reversible Reactions

2 HgO (s) 2 Hg (l) + O2 (g)

12

Upon heating, mercury (II) oxide decomposes to mercury (Hg) and

oxygen (O2) [ Equation 1]:

Under the same conditions, mercury (Hg) and oxygen (O2) recombine

form mercury (II) oxide again [Equation 2]:

Mercury and oxygen combine to form mercury oxide just as fast

as mercury oxide decomposes into mercury and oxygen

2 HgO (s) 2 Hg (l) + O2 (g)

2 Hg (l) + O2 (g) 2HgO (s)

Reversible Reactions

2 HgO (s) 2 Hg (l) + O2 (g)

Both reactions continue to occur, but there is no net change in the

composition of the system.

The amounts of mercury (II) oxide, mercury (Hg), and oxygen

(O2) remain constant as long as the reaction conditions remain

constant There is a state of equilibrium between the

two chemical reactions.

Chemical equilibrium is a dynamic state of balance in which the rates of opposing reactions are exactly equal.

Reversible Reactions

Examine the two sets of plots carefully, noting which substances have zero initial concentrations, and are thus "reactants" in each case. Satisfy yourself that these two sets represent the same chemical reaction system, but with the reactions occurring in opposite directions. Most importantly, note how the concentrations of all the components are identical when the system reaches equilibrium.

2 HI H2 + I2 Dissociation of Hydrogen Iodide

H2 + I2 2 HI Synthesis of Hydrogen Iodide

Reversible Reactions

The equilibrium state is independent of the direction from which it is

approached. Whether we start with an equimolar mixture of H2 and I2 (left)

or a pure sample of hydrogen iodide (shown on the right, using twice the

initial concentration of HI to keep the number of atoms the same), the

composition after equilibrium is attained (shaded regions on the right)

will be the same.

H2 + I2 2HI

For more information, Click Here

Factors affecting the rate of chemical reactions

The rate of a chemical reaction is the time needed for a

chemical reaction to be complete.

Factors that affects the rate of the chemical reactions are:

1. The nature of the reactants

C C

H

H

H

H

C C

H

H

H

H

H

H

Double covalent bond (weak)

More reactiveSingle covalent bond (strong)

Less reactive

The stronger the bond between the elements of a certain molecule, the harder it is to break in a chemical reaction, and the slower the reaction.

Factors affecting the rate of chemical reactions2. The Temperature

A mixture of iron and sulfur doesn’t react unless strongly heated.

An elevation in the temperature makes the reaction goes faster by

increasing the frequency of collisions between reacting particles.

In general all reactions, especially endothermic (because they absorb

energy) ones occur much more quickly when heated. This is due to the

fact that heat gives enough energy to break or form bonds between

different atoms.

Heat

Factors affecting the rate of chemical reactions

3. The presence of a catalyst

A catalyst speeds the rate of any reaction without affecting its products.

Hydrogen peroxide is an antiseptic that decomposes spontaneously

into water and oxygen:

2H2O2 (l) 2H2O (l) + O2 (g)

Because the reaction is slow, H2O2 can be

conserved for many months. But when a platinum wire is immersed in H2O2

solution, oxygen gas release is observed on the

platinum surface.

Factors affecting the rate of chemical reactions

4. The surface area

As the surface area of the reaction mixture decreases, the rate of the reaction increases.

Interpretation: Who can give an interpretation???

As the surface area decreases, the reacting molecules become

closer the frequency of collision between molecules

increases New molecules are formed

Factors affecting the rate of chemical reactions5. Change in concentration

a) Increasing the concentration of a reactant shifts an equilibrium to the products

(or right hand) side because the rate of the forwards reaction is increased.

b) Increasing the concentration of a product shifts an equilibrium to the reactant

(or left hand) side because the rate of the reverse reaction is speeded up.

c) Decreasing the concentration of a reactant (by removal or by compounding it

with something else or by precipitation) shifts an equilibrium to the reactants

(or left hand) side because the forwards reaction is slowed down.

The reverse reaction will 'overtake' the forwards reaction.

d) Decreasing the concentration of a product shifts an equilibrium to the products (or right hand) side because the reverse reaction is slowed and the forwards reaction 'overtakes'.

The equilibrium constantThe adjacent graph shows the changes in the reaction rates of the forward and backward reactions:

A + B C + D

Initially (t = 0), [A] and [B] were

maximum, while [C] and [D] were zero.

The rate of the forward reaction decreases as A and B are used up.

The rate of the reverse reaction increases as C and D are formed.

Equilibrium is attained when the two rates become equal [A], [B],

[C], and [D] remain constant at equilibrium.

The equilibrium constant

At equilibrium, the ratio of the product [C] x [D] to the product [A] x [B] has a

definite value at a given temperature. It is known as the equilibrium constant of

the reaction and is designated by the letter K. Thus,

[C] x [D]

K = -------------

[A] x [B]

K is independent of the initial concentrations.

K is dependent on the fixed temperature of the system.

The equilibrium constant K shows the extent to which the reactants are converted to the

products of the reaction.

If K = 1, the products of the concentrations of the products and the reactants have the same value.

If the value of K is very small, the forward reaction occurs only very

slightly before equilibrium is established, and the reactants are favored.

A very large value of K indicates an equilibrium in which the original

reactants are largely converted to products.

[C] x [D]

K = -------------

[A] x [B]

The numerical value of K for a particular equilibrium system is obtained experimentally.

The equilibrium constant

Consider the following general balanced equation:

a A + b B c C + d D

[C]c x [D]d

K = ------------------

[A]a x [B]b

The equilibrium constant K is the ratio of the product of the concentration of the substances formed at equilibrium to the product of the concentrations of the reacting substances, each concentration being raised to the power that is the coefficient of that substance in the chemical equation.

The equilibrium constant

Example: Give the expression of the equilibrium constant

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

Given [N2]=0.1M,[H2]=0.125M, [NH3]=0.11M

The equilibrium constant is given by the expression:

[NH3]2

K = ------------------

[N2] [H2]3

The equilibrium constant

Important Notes

Pure solids don’t appear in the K’s expression.

Pure liquids don’t appear in the K’s expression.

Water, as a liquid or a reactant, doesn’t appear

in the expression.

For example: 2 HgO (s) 2 Hg (l) + O2 (g)

K = [O2]

The equilibrium constant

Exercise 1: An equilibrium mixture of H2, I2, and HI gases at 425 ºC is

determined to consist of 4.5647 x 10-3 mole/liter of H2,

0.7378 x 10-3 mole/liter of I2, and also 13.544 x 10-3 mole/liter

of HI. What is the equilibrium constant for the

system at this temperature given that:

H2 (g) + I2 (g) 2 HI (g)

[HI]2 [13.544 x 10-3]-2

K = ------------ = ---------------------------------------- = 54.47

[H2] [I2] [4.5647 x 10-3] [0.7378 x 10-3]

The equilibrium constant

Exercise 2: Find the relationship between K1 and K2, the equilibrium

constants of these two reactions:

2A + 2B 2C K1

C A + B K2

[C]2 [A] [B]

K1 = ------------- ; K2 = --------------- [A]2 [B]2 [C]

By comparing K1 and K2: K1 = 1 /K22

The equilibrium constant

Exercise 3: The following reaction takes place at 460ºC, where the

equilibrium constant K has a value of 85.

SO2(g) + NO2(g) NO (g) + SO3 (g)

At a certain moment, the concentrations of the reactant and

products were measured to be:

[SO2] = 0.04, [NO2] = 0.5M, [NO] = 0.3M, [SO3] = 0.02M

Is this system at equilibrium?

If not, in which direction must the reaction go to reach

equilibrium?

The equilibrium constantSolution 3:

[NO] [SO3] 0.3 x 0.02K = ------------------ = -------------- = 0.3

[SO2] [NO2] 0.04 x 0.5

K = 0.3

K equilibrium = 85 } K < K equilibriumThe reaction is not at equilibrium

** K = 0.3 < 1 this means that the reactants NO2 and SO2 are favored. In

order for the system to reach equilibrium, it should move forward,

towards the products side.

Factors that disturb equilibrium

What are the factors that affect the rate of the reaction?

Any change that alters the rate of either reactions disturbs the

original equilibrium.

If the original state of equilibrium is disturbed, the system

seeks a new equilibrium state.

Equilibrium is shifted in the direction that releases stress

from the system.

Factors that disturb equilibrium Le Chatelier's principle provides a means of

predicting the influence of disturbing factors

on equilibrium systems.

Le Chatalier’s principle states:

If a system at equilibrium is subjected to a stress,

the equilibrium is shifted in the direction that

relieves the stress.

If you are stressed, what do you do?

Of course, you will go to a place where you can relax and relieve

the stress. The same concept is applied on the equilibrium of a

chemical reaction?

Factors that disturb equilibrium

1. Effect of temperature Changes in the temperature of the system affect the position

of the equilibrium by changing the magnitude of the equilibrium

constant for the reaction. Increasing the temperature of a reaction that gives off heat is the same

as adding more of one of the products of the reaction. It places a stress on

the reaction, which must be alleviated by converting some of the products

back to reactants. If the temperature of the system in equilibrium is lowered, the

reaction will move in a direction to produce more heat, i.e. the

exothermic reaction is favored.

In applying Le chatelier’s principle to chemical equilibrium, three stresses will be considered:

Factors that disturb equilibrium2. Effect of pressure

This applies to reactions involving gases. If the pressure is increased,

the reaction will move to reduce the pressure by reducing the number

of particles present.

A reaction at equilibrium was subjected to a stress results in an increase

in the total pressure on the system. The reaction then shifted in the

direction that minimized the effect of this stress. The reaction shifted

toward the products because this reduces the number of particles in

the gas, thereby decreasing the total pressure on the system.

Factors that disturb equilibrium3. Effect of concentration

If the concentration of one substance is increased, the reaction will

move in a direction to use up the substance whose concentration was

increased.

If one substance is removed from the system, the reaction will move in

a direction to produce more of the substance being removed.

Factors that disturb equilibrium4. Effect of catalyst

Can you predict the effect of Catalyst on the position of equilibrium?

Both the forward and backward reactions are speeded up in the

same amount; therefore, there is no effect on the position of

equilibrium or on the concentrations of the reacting substances.

Factors that disturb equilibrium

Exercise 4: Based on the following system at equilibrium:

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

How is equilibrium restored in following system in each of

the following cases?

a) A decrease in the concentration of N2

b) An increase in temperature

c) An increase in the total pressure of the system

Factors that disturb equilibrium

Exercise 5: Given the following reaction:

2 IBr (g) I2 (g) + Br2 (g)

If 0.06 moles of IBr are placed in a 0.5 liter container, and the

equilibrium constant K is 8.5x10-3, find the concentrations of IBr, I2,

and Br2 at equilibrium.

Initially (t = 0): # of moles of IBr = 0.06 moles

# of moles of I2 = 0

# of moles of Br2 = 0

At equilibrium (tequi): # of moles of IBr = 0.06 – 2n

# of moles of I2 = n

# of moles of Br2 = n

2 IBr (g) I2 (g) + Br2 (g)

At t = 0: 0.06 moles 0 mole 0 mole

At teq. : (0.06 – 2n) mole n mole n mole

Note: The coefficient of n is always the same as the coefficient of the

substance.

[I2] [Br2]

K = ----------------

[IBr]

# of moles of I2 n[I2] =----------------------- = ----------- Volume 0.5

# of moles of Br2 n[Br2] = ----------------------- = ----------- Volume 0.5

# of moles of IBr (0.06 -2n)[IBr] = ----------------------- = ------------- Volume 0.5

[I2] [Br2] [n / 0.5] [n / 0.5]K = ---------------- 8.5 x 10-3 = -------------------------

[IBr] [(0.06 – 2n) / 0.5]2

n2

8.5 x 10-3 = ---------------- (0.06 – 2n)2

n = 4.67 x 10-3 moles

[I2] = [Br2] = n / 0.5 = (4.67 x 10-3 ) / 0.5 = 9.34 x 10-3 M

[IBr] = (0.06 – 2n)/0.5 = [0.06 – 2(4.67 x 10-3)] / 0.5 = 0.101 M

At equilibrium:

Reactions that run to completion

A reaction may be driven in the preferred direction by applying

Le Chatelier principle.

A reaction reaches a state of equilibrium unless one of the products

escapes or is removed.

Some reactions appear to go to completion in the forward direction:

1. Burning a paper ( complete reaction).

2. Decomposition of potassium chlorate to oxygen and potassium chloride.

Reactions that run to completion

1. Formation of a gas

The reaction between sodium hydrogen carbonate (Baking soda) and

hydrochloric acid releases carbon dioxide gas as illustrated in the

given figures:

Reactions that run to completionIllustration

NaHCO3 + HCl NaCl + H2CO3

The ionic reaction

Na+ + HCO3- + H3O+ + Cl- Na+ + Cl- + H2CO3

The net ionic equation: HCO3- + H3O+ H2O + H2CO3

CO2 (g) + H2O

The net ionic equation: HCO3- + H3O+ 2H2O + CO2 (g)

Weak acid

Carbonic acid

Reactions that run to completion2. Formation of precipitate

When solutions of sodium chloride and silver nitrate are

mixed, a white precipitate of silver chloride immediately

forms.

The reaction effectively runs to completion because an “insoluble”

product is formed.

Reactions that run to completion2. Formation of precipitate (illustration)

AgNO3 + NaCl AgCl (s) + NaNO3

The ionic equation

Ag+ + NO3- + Na+ + Cl- AgCl (s) + Na+ + NO3

-

The net ionic equation

Ag + + Cl- AgCl (s)

White precipitate

Reactions that run to completion3. Formation of a slightly ionized product

Water is a typical compound that ionizes slightly into H3O+ and OH-.

Water can be formed as a product in the neutralization reaction.

The reaction effectively runs to completion because the product

(H2O) is only slightly ionized.

Reactions that run to completion3. Formation of a slightly ionized product (illustartion)

NaOH + HCl NaCl + H2O

The ionic equation

Na+ + OH- + H3O+ + Cl- Na+ + Cl- + 2H2O (l)

The net ionic equation

H3O+ + OH- 2 H2O (l)

The common ion effect is an application of Le Chatelier's Principle.

The Common Ion Effect

If we mix a soluble salt containing an ion common to a slightly soluble salt, we will affect the position of the equilibrium of the slightly soluble salt system.

Adding the common ion to the salt solution by mixing the soluble salt will add to the concentration of the common ion.

According to Le Chatelier's Principle, that will place a stress upon the slightly soluble salt equilibria (added concentration).

The equilibrium will respond so as to undo the stress of added common ion.

The Common Ion EffectExample Na+Cl- (aq) Na+ (aq) + Cl- (aq)

Bubble hydrogen chloride gas (HCl) in a saturated solution of

sodium chloride (NaCl).

As sodium chloride dissolves, NaCl separates as a precipitate.

Interpretation

The concentration of the common ion (Cl-) increases on the right side of

the reaction, while that of sodium ions decreases.

The equilibria will shift so that the common ion will be reduced which means a shift to the left, thus REDUCING the solubility of the slightly soluble salt system (NaCl).

The Common Ion EffectExample

A 0.1 M acetic acid solution (CH3COOH) has a pH of 2.9. When sodium acetate (CH3COO-Na+) is dissolved in the given solution, the pH increases to 4.9.

Interpretation

CH3COOH + H2O CH3COO- + H3O+

When sodium acetate is dissolved in the acetic acid solution, the concentration of the acetate ion (CH3COO-) on the right side of the equation will increase. The equilibrium will shift to the right (backward) so as to decrease the concentration of the added ion More CH3COOH is formed the concentration of H3O+ in solution decreases pH increases.

Common ion

Equilibrium Constant of Weak Acids

Weak acids ionize to a slight extend, producing small number of the

acidic H3O+ ions, according to the following general reaction:

HA + H2O H3O+ + A-

The extend to which a weak acid ionizes into ions is referred to as

ionization percentage (α).

For example, acetic acid (CH3COOH) has an ionization percentage of 1.4%. This means that if there are 100 moles of acetic acid, then only 1.4 moles will dissociate into the corresponding ions:

CH3COOH + H2O CH3COO- + H3O+

100 moles 1.4 moles 1.4 moles

Equilibrium Constant of Weak Acids

HA + H2O H3O+ + A-

t = 0 c 0 0

teq c – x x x

The portion of HA that dissociates into the ions depends on the acid

ionization percentage and on its concentration:

x = c α

HA + H2O H3O+ + A-

t = 0 c 0 0

t eq c – cα cα cα

c is the initial concentration

[A-][H3O+]K= ----------------

[HA][H2O]

[A-][H3O+]

K [H2O] = ---------------- [HA]

K and [H2O] are constants Ka = K [H2O], where Ka is the acid-ionization constant.

[A-][H3O+]Ka = ---------------- [HA]

The weaker the acid is, the smaller the value of Ka and α, due to the fewer number of ionized species in the numerator.

Equilibrium Constant of Weak Acids

Equilibrium Constant of Weak Acids

[A-][H3O+] (cα) (cα) Ka = ---------------- = ---------------- [HA] c – cα

c2α2 cα2

Ka = -------------- = --------- c (1- α) 1- α

Since α is very small for weak acids α <<< 1 1 - α ~ 1

Ka = cα2 / 1 Ka = c α2

Equilibrium Constant of Weak Acids

Another relation

[A-][H3O+]Ka = ---------------- [HA]

But [A-] = [H3O+];

[H3O+]2

Ka = ---------------- [HA]

[H3O+]2

Ka = ---------------- c

[H3O+]2 = Ka . c

[H3O+] = √ Ka . c

Equilibrium Constant of Weak AcidsIn summary

[H3O+] = √ Ka . c

Ka = c α2

[A-][H3O+]Ka = ---------------- [HA]

Equilibrium Constant of Weak AcidsExercise 6: a) Find the acid-ionization constant of a solution of 0.1M

acetic acid that has an ionization percentage of 1.4%.

b) Find the pH of the solution.

a) c = 0.1 M ; α = 1.4 x 10-2 ; Ka = ????

Ka = c α2 = (0.1) (1.4 x 10-2)2 Ka = 1.96 x 10-5

b) pH = -log [H3O+]

[H3O+] = √ Ka . c = √ (1.96 x 10-5) (0.1) [H3O+] = 2.744 x 10-3 M

pH = -log [H3O+] = - log (2.744 x 10-3) pH = 2.56

Ionization constant of water

The equation of the self-ionization of water:

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

The equilibrium constant: [H3O+] [OH-]K = -------------------- [H2O]2

But [H2O] is constant K [H2O]2 = [H3O+][OH-]

Kw = [H3O+] [OH-] = 1 x 10-14

The equilibrium constant for water is nothing but the ionization constant of water, Kw.

Equilibrium Constant of Weak Bases

B + H2O BH+ + OH-

t = 0 c 0 0

teq c – x x x

The portion of B that dissociates into the ions depends on the base

ionization percentage and on its concentration:

x = c α

B + H2O BH+ + OH-

t = 0 c 0 0

t eq c – cα cα cα

c is the initial concentration

[OH-][BH+]K= ----------------

[B][H2O]

[OH-][BH+]

K [H2O] = ---------------- [B]

K and [H2O] are constants Kb = K [H2O], where Kb is the base-

ionization constant. [OH-][BH+]

Kb = ----------------

[B]

The weaker the base is, the smaller the value of Kb and α, due to the

fewer number of ionized species in the numerator.

Equilibrium Constant of Weak bases

Equilibrium Constant of Weak bases

Using the same approach as with weak acids, we can derive a

similar set of formulas:

Kb = c α2

[H3O+] = √ Kb . c

Equilibrium Constant of Weak bases

Exercise 7: Prove that for any acid-conjugate base pair, Ka of

the weak acid and Kb of its conjugate base are

related through the following formula:

Ka. Kb = Kw

Consider any weak acid, HA, in water:

HA + H2O H3O+ + A-

[A-][H3O+]Ka = ---------------- [HA]

Equilibrium Constant of Weak basesIts conjugate base, A- , would undergo the following reaction:

A- + H2O HA + OH-

[HA] [OH-]Kb = ---------------- [A-]

Ka x Kb = [A-][H3O+] ---------------- x [HA]

[HA] [OH-]---------------- [A-]

Ka x Kb = [H3O+] [OH-] = Kw

•This formula is always true, and can be directly applied.

Equilibrium Constant of Weak bases

Exercise 8: a) Find the ionization percentage of a weak base, B, of

concentration 3M, if the base-ionization constant of

7.8 x 10-4?

b) Find the pH of the above solution.

a) Kb = c . α2

Kb 7.8 x 10-4

α2 = ------------- = --------------- c 3

α = 1.6 %

Equilibrium Constant of Weak bases

b) pH = ???

pH = -log [H3O+]

[OH-] = √ Kb . c = √ (7.8 x 10-4) (3) [OH-] = 0.048 M

[H3O+][OH-] = 10-14 [H3O+] [0.048] = 10-14

[H3O+] = 2.06 x 10-13 M

pH = -log [H3O+] = - log (2.06 x 10-13) pH = 12.68

Buffers Buffers are special type of solutions made up of a weak acid and

the salt of its conjugate base, or of a weak base, mixed with the salt

of its conjugate acid.

Examples of Buffers may include:

1. Acetic acid (HC2H3O2) mixed with sodium acetate (NaC2H3O2)

Weak acid Conjugate base

2. Ammonia (NH3) mixed with ammonium chloride (NH4Cl)

Weak base Conjugate base

Buffers Buffers resist changes in pH when an acid or a base is added in

small amounts.

Suppose a diluted acid is added in small amounts to a buffer made

of HC2H3O2 and NaC2H3O2.HC2H3O2 + H2O H3O+ + C2H3O2

-

When the diluted acid is added to the buffer, the concentration of

[H3O+] on the left side of the equation increases According to

Le Chatelier principle, the equilibrium will shift to the right, thus

reducing the concentration of H3O+ ions The pH will remain at

its initial level unchanged.

HC2H3O2 + H2O H3O+ + C2H3O2-

Now, when a base is added to the solution, it will react with H3O+ ions

causing a decrease in their concentration According to Le Chatelier

principle, the equilibrium will shift to the right to increase the

concentration of H3O+ ions pH will return to its original value.

Buffers

Solubility equilibrium

Solubility is defined as the amount of salt (in grams) that can be

dissolved in 100 g of water.

In general, salts are classified into 3 broad categories:

1. Soluble, when more than 1g of the salt can dissolve in a 100 g of water.2. Insoluble, when less than 0.1g of the salt can dissolve in a 100 g of water.

3. Slightly soluble when the mass of salt dissolved in a 100 g of water falls

between 0.1g and 1g.

Solubility equilibrium

A saturated solution is defined as a solution that contains the

maximum amount of salt dissolved in water.

Saturated solutions exhibit the behavior of equilibrium system, since

some of the salt is dissolved in water, while the rest is precipitated in

the bottom of the beaker.

Solubility equilibrium

Consider the case of silver chloride (AgCl) having a solubility of 8.9 x 10-1

g/100 g of water AgCl is considered insoluble in water

AgCl (s) Ag+ + Cl-

The equilibrium constant: [Ag+] [Cl-]K = ---------------- [AgCl]

K [AgCl] = [Ag+] [Cl-]

K and [AgCl] are considered to be constant (since AgCl is a solid,

so its concentration does not affect the equilibrium):

Ksp = [Ag+] [Cl-]Ksp = solubility-product

constant

Solubility equilibrium

The solubility-product constant is then the product of the molarities of

the ions in a saturated solution, each ion being raised to the power of its

coefficient.

Ksp = [Ag+] [Cl-]

Exercise 9 : Find the expression of the Ksp of calcium fluoride, CaF2

CaF2 Ca2+ + 2F-

Ksp = [Ca2+] [F-]2

Solubility equilibrium

The lower the Ksp, the less soluble the salt is.

Least soluble

Most soluble

AgCl (s) Ag+ + Cl-

t = 0 M 0 0

teq M – c c c

M = initial concentration

Ksp = [Ag+] [Cl-] = c . c = c2

c is the molarity in mole/l , while the solubility is in given in

mass of AgCl / 100 g of water.

To find the concentration, apply this relation:

mass of salt

C = ----------------------------------------------

(Molecular weight of the salt) x 0.1L

Solubility equilibrium

Solubility equilibrium

Exercise 10: Find the Ksp of CaF2 if its solubility is 1.7 x 10-3g/100 g

of water.

Ca = 40 ; F = 19

CaF2 Ca2+ + 2F-

t = 0 M 0 0teq M – c c 2c

mass of CaF2

c = ----------------------------------------------

(Molecular weight of CaF2) x 0.1L

Molecular weight of CaF2= 40 + 2(19) = 78 g/mole

;

Solubility equilibrium

1.7 x 10-3

c = --------------- 78 x 0.1

c = [Ca2+] = 2.18 x 10-4 mole/l

[F-] = 2 c = 2 x (2.18 x 10-4) = 4.36 x 10-4 mole/l

Ksp = [Ca2+] [F-]2 = (2.18 x 1o-4) (4.36 x 10-4)2

Ksp = 4.14 x 10-11

Solubility equilibrium

Exercise 11: Find the solubility of cadmium sulfide, CdS, in g/100g

of water, if its Ksp is 8 x 10-27.

Cd = 112 ; S = 32

CdS Cd2+ + S2-

t = 0 M 0 0teq M – c c c

Ksp = [Cd2+][S2-] = c . c = c2 c = √Ksp = √8 x 10-27

c = 8.944 x 10-14 M

mass of CdS

c = ----------------------------------------------

(Molecular weight of CdS) x 0.1L

Molecular weight of CdS = 112 + 32 = 144 g / mole

mass of CdS8.944 x 10-14 = ------------------- 144 x 0.1

mass of CdS = 1.287 x 10-12 g / 100 g of water

Solubility equilibrium

Solubility equilibrium

Exercise 12: Find the solubility of CdS in mole/l, given its Ksp to be 8 x 10-27

Ksp = [Cd2+][S2-] = c . c = c2

CdS Cd2+ + S2-

t = 0 M 0 0teq M – c c c

c = √Ksp = √8 x 10-27

c = 8.944 x 10-14 M

Solubility equilibriumPrecipitation calculation

A precipitate is the formation of an insoluble salt in solution.

The precipitate may form when mixing two soluble salts.

XY X+ + Y-

For the precipitate XY to form, [X+][Y-] > Ksp

If [X+][Y-] < Ksp no precipitate XY will form.

Solubility equilibrium

Exercise 13: Will a precipitate form when 20 ml of 0.01M BaCl2 is mixed

with 20 ml of 0.005M Na2SO4? Ksp of BaSO4 = 1.1 x 10-10?

BaCl2 (aq) + Na2SO4 (aq) BaSO4 (s) + 2 NaCl (aq)

The equation of the reaction between the two salts:

The dissolution reaction of the precipitate formed is:

BaSO4 (s) Ba2+ + SO42-

Solubility equilibrium

[Ba2+] = 5 x 10-3 M

# moles of BaCl2 = 2 x 10-4 moles

# moles of BaCl2 = # moles of Ba2+ = 2 x 10-4 moles

[Ba2+] = (# moles of Ba2+) / (total volume) = (2 x 10-4) / (40 x 10-3 L)

# moles of BaCl2

[BaCl2] = ----------------------- volume of BaCl2

# moles of BaCl2

0.01 = ------------------------- 20 x 10-3 L

Solubility equilibrium

# moles of Na2SO4 = # moles of SO42- = 1 x 10-4 moles

[SO42-] = (# moles of SO4

2-) / (total volume) = (1 x 10-4) / (40 x 10-3 L)

[SO42-] = 2.5 x 10-3 M

# moles of Na2SO4

[Na2SO4] = ------------------------- volume of Na2SO4

# moles of Na2SO4

0.005 = ------------------------- 20 x 10-3 L

# moles of Na2SO4 = 1 x 10-4 moles

Solubility equilibrium

[Ba2+] [SO42-] = (5 x 10-3) (2.5 x 10-3) = 1.25 x 10-5

[Ba2+] [SO42-] = 1.25 x 10-5

Ksp = 1.1 x 10-10 } [Ba2+] [SO42-] > Ksp

A precipitate of BaSO4 will form in this solution