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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