Chemical Equilibrium 2

7/28/2019 Chemical Equilibrium 2

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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 ofequilibriumbetween the

two chemical reactions.

Chemical equil ibr iumis a dynamic state of balance in which the

rates of opposing reactions are exactly equal.


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Examine the two sets of plots carefully, noting which substances have zero initialconcentrations, 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 equi l ibrium.

2 HI H2 + I2 Dissociation of Hydrogen Iodide

H2 + I2 2 HI Synthesis of Hydrogen Iodide


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The equilibrium state is independent of the dir ectionfrom 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 equi l ibri um is attained(shaded regions on the right)

wil l be the same.

H2 + I2 2HI

For more information, Click Here
http://localhost/var/www/apps/conversion/tmp/My%20Documents/Downloads/links/link%201/Equilibrium%20introduction.htmhttp://localhost/var/www/apps/conversion/tmp/My%20Documents/Downloads/links/link%201/Equilibrium%20introduction.htm


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Factors affecting the rate of chemical reactions

The rateof 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 reactive

Single covalent bond (strong)

Less reactive

The stronger the bond between the elements of a certain molecule,

the harder it is to breakin a chemical reaction, and the slower the

reaction.


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2. The Temperature

A mixture of iron and sulfur doesnt 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 breakor form bonds between

different atoms.

Heat


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3. The presence of a catalyst

A catalystspeeds the rate of any reaction without affecting its products.

Hydrogen peroxideis an antiseptic that decomposes spontaneously

into water and oxygen:

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

Because the reaction is slow, H2O2 can beconserved for many months.

But when a platinum wire is immersed in H2O2

solution, oxygen gas release is observed on the

platinum surface.


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4. The sur face area

As the sur face area of the reaction mixturedecreases, the rate of

the reactionincreases.

Interpretation: Who can give an interpretation???

As the surface area decreases, the reacting molecules become

closer the frequency ofcollision between molecules

increases New molecules are formed


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5. Change in concentr ation

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'.


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The equilibrium constant

The 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

decreasesas A and B are used up.

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

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

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


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At equilibri um, theratioof the product[C] x [D]to the product[A] x [B]has a

defini te valueat agiven temperature. It is known as the equil ibr ium constantof

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.


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The equilibrium constantKshows theextentto which thereactants are converted to the

productsof the reaction.

IfK= 1, the products of the concentrations

of the products and the reactants have the

same value.

If the value ofKis very small, the forward reaction occurs only very

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

A very large value ofKindicates an equilibrium in which the original

reactants are largely converted to products.

[C] x [D]

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

[A] x [B]

The numeri cal value of K for a particular equil ibr ium system is obtained exper imentally.


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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 equil ibrium constant K is the ratio of the product of the concentr ation of

the substances formed at equi l ibrium to the product of the concentrations of

the reacting substances, each concentration being raised to the power that is

the coeff icient of that substance in the chemical equation.


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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


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Important Notes

Pure solidsdont appear in the Ks expression.

Pure liquidsdont appear in the Ks expression.Water, as a liquid or a reactant, doesnt appear

in the expression.

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

K = [O2]


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Exercise 1: An equilibrium mixture of H2, I2, and HI gases at 425 C 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]


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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


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Exercise 3: The following reaction takes place at 460C, 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?


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Solution 3:

[NO] [SO3] 0.3 x 0.02

K = ------------------ = -------------- = 0.3

[SO2] [NO2] 0.04 x 0.5

K = 0.3

Kequilibrium = 85} K < Kequilibrium The 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.


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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

or iginal equil ibrium.

If the original state of equilibrium is disturbed, the system

seeks a new equi l ibrium state.

Equilibrium is shifted in the direction that releases stress

from the system.


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Le Chatel ier ' s principleprovides a means of

predicting the influence of disturbing factors

on equilibrium systems.

Le Chataliers principle states:

I f a system at equi l ibrium is subjected to a stress,

the equil ibr ium is shi fted in the direction that

rel ieves 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?


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1. Effect of temperature

Changes in the temperature of the system affect the position

of the equilibrium by changing the magnitude of the equilibriumconstant 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 chateliers principle to chemical equilibrium, three stresses will beconsidered:


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2. Ef fect 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 shiftedin the

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

toward the products because this reduces the number of particles inthe gas, thereby decreasing the total pressure on the system.


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3. 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.

Ifone substance is removed from the system, the reaction will move in

a direction to produce more of the substance being removed.


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4. 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 effecton the position of

equilibrium or on the concentrations of the reacting substances.


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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


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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 theequilibrium 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


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At equilibrium (tequi): # of moles of IBr = 0.062n

# 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. : (0.062n) mole n mole n mole

Note: The coeff icient of n is always the same as the coeff icient of the

substance.


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[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


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[I2] [Br2] [n / 0.5] [n / 0.5]

K = ---------------- 8.5 x 10-3 = -------------------------

[IBr] [(0.062n) / 0.5]2

n2

8.5 x 10-3 = ----------------

(0.062n)2n = 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] = (0.062n)/0.5 = [0.062(4.67 x 10-3)] / 0.5 = 0.101 M

At equilibrium:


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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.


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1. Formation of a gas

The reaction between sodium hydrogen carbonate (Baking soda) and

hydrochloric acid releases carbon dioxide gas as illustrated in thegiven figures:


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Illustration

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


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2. Formation of precipitate

When solutions ofsodium chloride and silver nitrate are

mixed, a whi te precipitateofsi lver chlor ideimmediately

forms.

The reaction effectively runs to completionbecause an insoluble

product is formed.


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2. Formation of precipitate (i l lustration)

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


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3. Formation of a slightly ionized product

Wateris a typical compound that ionizes slightlyinto H3O+ and OH-.

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

The reaction effectively runs to completionbecause the product

(H2O) is only slightly ionized.


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3. Formation of a sli ghtly ionized product (il lustartion)

NaOH + HCl NaCl + H2O

The ionic equation

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

The net ionic equation

H3O+ + OH- 2 H2O (l)


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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 theslightly soluble salt equilibria (added concentration).

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


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Example 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 wi l l be reducedwhich means

a shift to the left, thus REDUCING the solubility of the slightly soluble salt

system (NaCl).


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Example

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, theconcentration 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


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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().

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


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HA + H2O H3O+ + A-

t = 0 c 0 0

teq cx x x

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

ionization percentage and on its concentration:

x = c

HA + H2O H3O+ + A-

t = 0 c 0 0

t eq cc c c

c is the initial

concentration


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[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 , due to the fewer

number of ionized species in the numerator.



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[A-][H3O+] (c) (c)

Ka = ---------------- = ----------------

[HA] cc

c22 c2

Ka = -------------- = ---------

c (1- ) 1-

Since is very small for weak acids


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Another relation

[A-][H3O+]

Ka = ----------------

[HA]

But [A-] = [H3O+];

[H3O+]2

Ka = ----------------

[HA]

[H3O+]2

Ka = ----------------

c

[H3O+]2 = Ka .

c

[H3O+] = Ka . c


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In summary

[H3O+] = Ka . c

Ka = c 2

[A-][H3O+]

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


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Exercise 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 ; = 1.4 x 10-2 ; Ka = ????

Ka = c 2 = (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


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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 equil ibrium constant for water is nothing but the ionization constant

of water, Kw.


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Equilibrium Constant of Weak Bases

B + H2O BH+ + OH-

t = 0 c 0 0

teq cx x x

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

ionization percentage and on its concentration:

x = c

B + H2O BH+ + OH-

t = 0 c 0 0

t eq cc c c

c is the initial

concentration


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[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 smal ler the value of Kband , due to the

fewer number of ionized species in the numerator.

Equilibrium Constant of Weak bases


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

similar set of formulas:

Kb = c 2

[H3O+] = Kb . c


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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]


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Its 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.


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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 . 2Kb 7.8 x 10-4

2 = ------------- = ---------------c 3

= 1.6 %


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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


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Buffers

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

the salt ofits 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


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Buffers

Buffersresist changes in pHwhen 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 toLe 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.


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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


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Solubility equilibrium

Solubilityis def ined as theamount of salt(in grams) that can be

dissolved in 100 g of water.

In general, salts are classified into 3 broad categories:

1. Soluble, whenmore 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 solublewhen the mass of salt dissolved in a 100 g of water falls

between 0.1g and 1g.


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A saturated solutionis def ined 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.


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Consider the case ofsilver 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


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The solubil ity-product constant is then the product of the molari ties 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


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The lower the Ksp, the

less soluble the salt is.

Least soluble

Most soluble


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AgCl (s) Ag+ + Cl-

t = 0 M 0 0

teq Mc 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



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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 0

teq Mc c 2c

mass of CaF2

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

(Molecular weight of CaF2) x 0.1L

Molecular weight of CaF2=

40 + 2(19) = 78 g/mole;


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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


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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 0

teq Mc c c

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

c = 8.944 x 10-14 M


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mass of CdS

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

(Molecular weight of CdS) x 0.1L

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

mass of CdS

8.944 x 10-14 = -------------------

144 x 0.1

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



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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 0

teq Mc c c

c = Ksp = 8 x 10-27

c = 8.944 x 10-14 M


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Precipitation 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.


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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+ + SO4

2-


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[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 BaCl20.01 = -------------------------

20 x 10-3 L


76/77


# 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 Na2SO40.005 = -------------------------

20 x 10-3 L

# moles of Na2SO4 = 1 x 10-4 moles


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[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+] [SO

4

2-] > Ksp

A precipitate of BaSO4 wil l f orm in thi s solution