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Chemical Equilibrium
I. A State of Dynamic Balance
1N2 (g) + 3H2(g)
The concentrations
of the reactants
decrease at first…
…while the concentration
of the product increases
2NH3(g) ΔG0 = -33.1 kJ
But, then, before the
reactants are used up,
all concentrations
become constant
The reaction is
spontaneous
under standard
conditions
Chemical Equilibrium
I. A State of Dynamic Balance -when a ________
reaction results in the
almost ________
complete conversion of
________
reactants to ________,
products the
1N2 (g) + 3H2(g)
2NH3(g) ________
reaction is said to go to
completion but _____
most _________
reactions
__________,
1N2 (g) + 3H2(g)
2NH3(g) ___
do ____
not go to __________,
completion most
reactions are __________
reversible
1N2 (g) + 3H2(g)
2NH3(g) _________
At first, only the reactants
are present, so only the
forward reaction can occur
1N2 (g) + 3H2(g)
2NH3(g)
Chemical Equilibrium
I. A State of Dynamic Balance -as soon as the ________
forward ________
reaction
begins, the concentrations
____________ of the
_________
reactants go _____,
down and the
reaction _____
rate goes _____
down as the
_________
collisions per unit
number of __________
____
time goes _____
down
As soon as the products
begin forming, the forward
reaction rate slows and the
reverse reaction begins
1N2 (g) + 3H2(g)
2NH3(g)
Chemical Equilibrium
I. A State of Dynamic Balance -as the _________
reaction proceeds, the
____
rate of the ________
forward _________
reaction
continues to ________
decrease and the ____
rate
reverse ________
reaction continues
of the ________
increase until the two _____
rates are
to ________
_____,
equal and the system has reached a
chemical __________
equilbrium
1N2 (g) + 3H2(g)
2NH3(g) state of ________
Chemical Equilibrium
I. A State of Dynamic Balance -at ___________,
equilibrium the concentrations
____________
of the ________
reactants and ________
products are
not _____,
equal but _______,
constant because the
rate of _________
formation of the ________
products
____
equal to the ____
rate of _________
formation
is _____
of the ________
reactants
The Golden Gate Bridge If all other roads leading
connects San Francisco in and out of the two cities
to Sausalito.
were closed…
…and the number of vehicles
crossing the bridge per hour
in one direction equaled the
number of vehicles crossing
the bridge in the opposite
direction…
What is true of the number
of vehicles in each city throughout
the day?
Are there the same number of
vehicles in each city?
Chemical Equilibrium
II. Equilibrium Expressions -while _____
some chemical systems have
and Constants
little tendency to _____,
react and _____
some
chemical systems _____
react readily and
go to __________,
completion _____
most chemical
___
state of __________,
equilibrium
systems reach a _____
leaving varying amounts of ________
reactant
____________
unconsumed
Waage
-in 1864, Norwegian chemists ______
Guldberg proposed the _______
Law of
and _________
___________________,
Chemical Equilibrium which states,
at a given ___________,
temperature a chemical
state in which a
system may reach a _____
ratio of _______
reactant and
particular _____
product ____________
concentrations has a constant
_______
_______
value
Cato Maximilian Guldberg
1836-1902
Peter Waage
1833-1900
Chemical Equilibrium
II. Equilibrium Expressions -the _______
general ________
equation for a _______
reaction
and Constants
at __________
equilibrium can be written
______________________________,
aA + bB
cC + dD
A and __
B are ________,
reactants __
C and
where __
D are ________,
products __,
a __,
b __,
c and __
d are
__
the ___________
coefficients in the ________
balanced
________,
equation and the __________
equilibrium
_______
constant __________
expression is
[C]c [D]d
Keq =
[A]a [B]b
equilibrium ________
mixtures with ___
Keq
-___________
> __
1 contain more ________
products
values __
reactants at ___________,
equilibrium while
than ________
__________
equilibrium ________
mixtures with ___
Keq values
1 contain more ________
__
< __
reactants than
products at __________
equilibrium
________
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the equilibrium constant expression for the homogeneous
equilibrium for the synthesis of ammonia from nitrogen and hydrogen.
1N2 (g) + 3H2(g)
Keq =
Keq
2NH3(g)
[C]c [D]d
The equilibrium is homogeneous
because all the reactants and
products are in the same physical
state (gas)
[A]a [B]b
c
[C]
=
[A]a [B]b
=
[NH3]c
[N2]a[H2]b
=
[NH3]2
[N2]1[H2]3
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the equilibrium constant expression for the equilibrium for the
synthesis of Hydrogen iodide from iodine and hydrogen.
1H2 (g) + 1I2(g)
Keq =
Keq
2HI(g)
[C]c [D]d
The equilibrium is homogeneous
because all the reactants and
products are in the same physical
state (gas)
[A]a [B]b
c
[C]
=
[A]a [B]b
=
[HI]c
[H2]a[I2]b
=
[HI]2
[H2]1 [I2]1
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the equilibrium constant expression for the equilibrium for the
decomposition of Dinitrogen tetroxide into Nitrogen dioxide.
1N2O4 (g)
Keq =
Keq =
2NO2(g)
[C]c [D]d
The equilibrium is homogeneous
because all the reactants and
products are in the same physical
state (gas)
[A]a [B]b
[C]c
[A]a
=
[NO2]c
[N2O4]a
=
[NO2]2
[N2O4]1
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the equilibrium constant expression for the equilibrium for the
reaction of Carbon monoxide and Hydrogen which produces methane
(Tetrahydrogen monocarbide) and water.
1CO(g) + 3H2 (g)
Keq
[C]c [D]d
=
[A]a [B]b
Keq
c [D]d
[C]
=
[A]a [B]b
=
1CH4(g) +
[CH4]c[H2O]d =
[CO]a [H2]b
1H2O (g)
[CH4]1[H2O]1
[CO]1 [H2]3
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the equilibrium constant expression for the equilibrium for the
decomposition of Dihydrogen monosulfide into diatomic hydrogen and
diatomic sulfur.
2H2S (g)
Keq
[C]c [D]d
=
[A]a [B]b
Keq
c [D]d
[C]
=
[A]a
=
2H2 (g) +
[H2]c [S2]d
[H2S]a
=
1S2 (g)
[H2]2 [S2]1
[H2S]2
Chemical Equilibrium
II. Equilibrium Expressions -_________
equilibria in which all ________
reactants and
and Constants
________
products are in the same ________
physical
_____
state are ____________,
homogeneous but
reactions with _________
reactants and
________
products in _____
more than ___
one ________
physical
________
_____
state result in _____________
heterogeneous
gaseous
_________
equilibria
ethanol
1C2H5OH (l)
Keq
1C2H5OH (g)
[C]c [D]d
=
[A]a [B]b
Keq =
[C2H5OH (g)]1
[C2H5OH (l)]1
Keq = [C2H5OH (g)]1
liquid
ethanol
Chemical Equilibrium
II. Equilibrium Expressions -since ______
liquid and _____
solid ________
reactants and
and Constants
________
products don’t change ___________,
concentration
(which is really their ______),
density if the
temperature remains ________,
constant
___________
equilibrium _______
constant
then in the ___________
__________
expression for a ____________
heterogeneous
___________,
equilibrium the ___________
equilibrium
________
constant only depends on the
concentrations of the ________
reactants and
______________
products in the gaseous
________
_______ state of
matter
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the equilibrium constant expression for the heterogeneous
equilibrium for the decomposition of Sodium Hydrogen carbonate into
Sodium carbonate, Carbon dioxide, and water.
2NaHCO3 (s)
Keq =
1Na2CO3 (s) + 1CO2 (g) + 1H2O (g)
[C]c [D]d [E]e
[A]a
Keq = [D]d [E]e = [CO2]d [H2O]e =
The equilibrium is heterogeneous
because the reactants and
products are in different physical
states (gas and solid)
[CO2]1[H2O]1
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the equilibrium constant expression for the heterogeneous
equilibrium for the decomposition of Calcium carbonate into Calcium
oxide and Carbon dioxide.
1CaCO3 (s)
Keq =
1CaO (s) + 1CO2 (g)
The equilibrium is heterogeneous
because the reactants and
products are in different physical
states (gas and solid)
[C]c [D]d
[A]a
Keq = [D]d
= [CO2]d
=
[CO2]1
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the complete, balanced thermochemical equation and equilibrium
constant expression for the heterogeneous equilibrium for the reaction of
monatomic Sulfur and fluorine gas, which produces Sulfur tetrafluoride
gas and Sulfur hexafluoride gas.
2S (s)
Keq =
+
5F2 (g)
[SF4]1 [SF6]1
[F2]5
1SF4 (g) +
1SF6 (g)
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the complete, balanced thermochemical equation and equilibrium
constant expression for the homogeneous equilibrium for the reaction of
hydrazine (Tetrahydrogen dinitride) and Nitrogen dioxide, which
produces nitrogen and water.
2N2H4 (g) +
Keq
2NO2 (g)
3[H O]4
[N
]
2
2
=
[N2H4]2[NO2]2
3N2 (g)
+
4H2O (g)
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the complete, balanced thermochemical equation and equilibrium
constant expression for the homogeneous equilibrium for the reaction of
Sulfur trioxide and Carbon dioxide, which produces Carbon disulfide
and oxygen.
2SO3 (g) + 1CO2 (g)
Keq =
[CS2]1 [O2]4
[SO3]2 [CO2]1
1CS2 (g) +
4O2 (g)
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the complete, balanced thermochemical equation and equilibrium
constant expression for the heterogeneous equilibrium for the reaction of
monatomic Sulfur and fluorine gas, which produces Sulfur tetrafluoride
gas and Sulfur hexafluoride gas.
2S (s)
Keq =
+
5F2 (g)
[SF4]1 [SF6]1
[F2]5
1SF4 (g) +
1SF6 (g)
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the complete, balanced thermochemical equation and equilibrium
constant expression for the heterogeneous equilibrium for the reaction of
magnatite (Fe3O4) and hydrogen gas, which produces iron and water
vapor.
1Fe3O4 (s) +
Keq =
4H2 (g)
[H2O]4
[H2]4
3Fe (s)
+
4H2O (g)
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the equilibrium constant expression for the homogeneous
equilibrium for the synthesis of ammonia and calculate the value of Keq
when [NH3] = 0.933 M, [N2] = 0.533 M, and [H2] = 1.600 M.
1N2 (g) + 3H2(g)
Keq =
[NH3]2
[N2]1[H2]3
Keq =
[0.933]2
[0.533]1 [1.600]3
Keq =
0.399
2NH3(g)
Chemical Equilibrium
II. Equilibrium Expressions
and Constants
Write the equilibrium constant expression for the homogeneous
equilibrium for the decomposition of Sulfur trioxide into Sulfur dioxide
and oxygen gas, and calculate the value of Keq when [SO3] = 0.0160 M,
[SO2] = 0.00560 M, and [O2] = 0.00210 M.
2SO3 (g)
Keq =
[SO2]2[O2]1
[SO3]2
Keq =
[0.00560]2 [0.00210]1
[0.0160]2
Keq =
2.58 x 10-4
2SO2 (g) + 1O2 (g)
Chemical Equilibrium
III. Le Châtelier’s Principle
1. Hypothesis: What is the effect of temperature on equilibrium?
2. Prediction:
3. Gather Data:
A. Safety: The surfaces of the hot plates and the water will be hot
enough to cause burns. Use caution. Cobalt(II)
chloride hexahydrate is toxic, with an LD50 = 80mg/kg
Avoid ingestion (don’t eat or drink it). Wash hands
thoroughly with soap and water before leaving lab.
Ethanol is extremely flammable. No open flame.
B. Procedure:
1. Pick up a sheet of white construction paper and an artist’s
paintbrush.
Chemical Equilibrium
III. Le Châtelier’s Principle
3. Gather Data:
B. Procedure:
2. With a partner, using a top-loading electronic balance,
mass 0.3 grams of CoCl2·6H2O, crush it into a fine powder
using a mortar and pestle, and place it in a test tube.
3. Add 10 mL of ethanol to the test tube, cap, and shake
vigorously until CoCl2·6H2O dissolves. If the solution is
not light pink, add water dropwise until it turns light pink.
4. Use the solution to paint a winter scene on your white
construction paper, including a pink-colored field of snow.
5. To simulate the coming of spring, warm your painting
over the hotplate in the fume hood. Record your
observations.
Chemical Equilibrium
III. Le Châtelier’s Principle
4. Analyze Data:
+ heat
1Co(H2O)62+ (aq) +
Hexahydrate Co2+ ion (pink)
1Co(H2O)62+ (aq) +
Hexahydrate Co2+ ion (pink)
5. Draw Conclusions:
4Cl- (aq)
chloride ion
4Cl- (aq)
chloride ion
1CoCl42-(aq) + 6H2O (l)
Tetrachlorocobaltate ion (blue)
1CoCl42-(aq) + 6H2O (l)
Tetrachlorocobaltate ion (blue)
Chemical Equilibrium
III. Le Châtelier’s Principle -________
reactions that reach __________
equilibrium
instead of going to __________
completion do not
________
produce as much
-in 1888, ________________________
Henry-Louis Le Châtelier
discovered that there are ways to
_______
control _________
equilibria in order to make
reactions more __________
productive
_________
-____________________
Le Châtelier’s Principle states that if a
stress (like a ______
change in
______
temperature is applied to a system
__________)
equilibrium the system _____
shifts in
at __________,
the ________
direction that _______
relieves the
_____
stress
Henry-Louis Le Châtelier
1850-1936
Chemical Equilibrium
III. Le Châtelier’s Principle -stressors that cause a shift in
equilibrium
A. Changes in Concentration
Write the equilibrium constant expression for the equilibrium for the
reaction of Carbon monoxide and Hydrogen to produce methane and
water. Then, calculate the Keq value when [CO] = 0.30000 M, [H2] =
0.10000 M, and [CH4] = 0.05900 M, and [H2O] = 0.02000 M.
1CO(g) + 3H2 (g)
Keq =
Keq
1CH4(g) +
1H2O (g)
[CH4]1[H2O]1
[CO]1 [H2]3
1 [0.02000]1
[0.05900]
=
[0.30000]1[0.10000]3
=
3.933
= first equilibrium
position
Chemical Equilibrium
III. Le Châtelier’s Principle
A. Changes in Concentration -_________
increasing the ____________
concentration of
CO _________
increases the _______
number of
___
_________
___ and
collisions between CO
___,
H2 _________
increasing the _____
rate of the
forward _______
reaction
_______
1CO(g) + 3H2 (g)
1CH4(g) + 1H2O (g)
-the system responds to the
______
stress of the addition of
reactant by forming more
_______
product to bring the system
_______
back into equilbrium
Chemical Equilibrium
III. Le Châtelier’s Principle
A. Changes in Concentration
1CO(g) + 3H2 (g)
0.99254 M
Keq =
Keq
0.07762 M
1CH4(g) +
1H2O (g)
0.06648 M
0.02746 M
[CH4]1[H2O]1
[CO]1 [H2]3
1 [0.02746]1
[0.06648]
=
[0.99254]1[0.07762]3
=
3.933 =
second equilibrium
position
Chemical Equilibrium
III. Le Châtelier’s Principle
A. Changes in Concentration -_________
increasing the ____________
concentration of
reactant causes __________
equilbrium
a ________
to _____
_______
shift to the ____
right to increase
the ____
______
rate of formation of product
1CO(g) + 3H2 (g)
1CH4(g) +
1H2O (g)
1CO(g) + 3H2 (g)
1CH4(g) +
1H2O (g)
decreasing the ____________
concentration of
-_________
product causes __________
equilbrium
a ________
to _____
_______
shift to the ____
right to increase
the ____
______
rate of formation of product
Chemical Equilibrium
III. Le Châtelier’s Principle
A. Changes in Concentration
Predict what should happen to the following equilibrium if hydrogen
bonding due to the addition of acetone binds water and effectively
removes it from the products.
1Co(H2O)62+ (aq) +
Hexahydrate Co2+ ion (pink)
4Cl- (aq)
chloride ion
1CoCl42-(aq) + 6H2O (l)
Tetrachlorocobaltate ion (blue)
Chemical Equilibrium
III. Le Châtelier’s Principle -stressors that cause a shift in
equilibrium
A. Changes in Volume -_________
decreasing the volume
______ of the _______
reaction
container, according to ______,
Boyle
increases the ________,
pressure which in turn
________
1CO(g) + 3H2 (g)
increases the _____
rate of _________
collision
________
particles of the
between the ________
________,
reactants _________
increasing the _____
rate of the
________
forward _______
reaction
1CH4(g) + 1H2O (g)
shift in the equilibrium
-the _____
_________ causes the
stress on the system to be _______
relieved as
_____
for every __
4 _____
moles of _______
gaseous _______
reactant
_________,
_____ of _______
consumed only __
2 moles
gaseous
product are _________,
produced which,
_______
½ the
according to Avogadro
________, occupies __
volume
decreases the pressure
______, which _________
________
Chemical Equilibrium
III. Le Châtelier’s Principle -stressors that cause a shift in
equilibrium
Use Le Châtelier’s Principle to predict how each of these changes would
affect the ammonia equilibrium system.
1N2 (g) + 3H2(g)
2NH3(g)
a. removing hydrogen from the system __________________________
equilibrium shifts to the left
1N2 (g) + 3H2(g)
2NH3(g)
b. adding ammonia to the system _______________________________
equilibrium shifts to the left
1N2 (g) + 3H2(g)
2NH3(g)
Chemical Equilibrium
III. Le Châtelier’s Principle -stressors that cause a shift in
equilibrium
Use Le Châtelier’s Principle to predict how each of these changes would
affect the ammonia equilibrium system.
1N2 (g) + 3H2(g)
2NH3(g)
c. adding hydrogen to the system _______________________________
equilibrium shifts to the right
1N2 (g) + 3H2(g)
2NH3(g)
Chemical Equilibrium
III. Le Châtelier’s Principle -stressors that cause a shift in
equilibrium
How would decreasing the volume of the reaction container affect each
of these equilibria?
a. 2SO2 (g) + 1O2(g)
2SO3(g) _________________________
equilibrium shifts to the right
b. 1H2 (g) + 1Cl2(g)
2HCl(g) _____________________________
stress has no effect on equilibrium
c. 2NOBr(g)
2NO(g) + 1Br2(g) _________________________
equilibrium shifts to the left
Chemical Equilibrium
III. Le Châtelier’s Principle -stressors that cause a shift in
equilibrium
A. Changes in Temperature -while _______
changes in _____________
concentration
and ________
changes in _______
volume cause
______
shifts in _________,
equilibria they ___
do
not _______
change the __________
equilibrium
___
constant but a ______
change in
_______,
___________
temperature causes ______
change in
both the __________
equilibrium ________
position
equilibrium _______
constant
and the __________
Chemical Equilibrium
III. Le Châtelier’s Principle -stressors that cause a shift in
equilibrium
A. Changes in Temperature -since the _______
reaction for making
_______
methane has a _______
negative ____,
ΔH0 the
________
_______
is _________,
forward
reaction
exothermic
1CO(g) + 3H2 (g)
reverse _______
reaction is
and the _______
endothermic
heat can be
__________, so ____
thought of as a _______
product in the
________
forward _______
reaction and a _______
reactant
reverse reaction
in the _______
_______
1CH4(g) + 1H2O (g) + heat
ΔH0 = -206.5 kJ
Chemical Equilibrium
III. Le Châtelier’s Principle -stressors that cause a shift in
equilibrium
A. Changes in Temperature -_________
increasing the temperature
__________ is
like _______
adding more _______
reactant to the
_______
reaction in which _____
heat acts as a
reactant and is _____
used ___,
up in this
_______
endothermic _______
reverse
case, the __________
_______
reaction
1CO(g) + 3H2 (g)
1CH4(g) +
1H2O (g) + heat
equilibrium shifts to the _____,
left
-__________
_________
decreasing the ___________
concentration of
_______
methane because _______
methane is a
reactant
reverse _______
reaction
_______ in the _______
Chemical Equilibrium
III. Le Châtelier’s Principle -stressors that cause a shift in
equilibrium
In the following equilibrium, would you raise or lower the temperature to
get the following results?
1C2H2 (g) + 1H2O(g)
1CH3CHO(g) ΔH0 = -151 kJ
a. increase the amount of CH3CHO______________________________
lower the temperature
1C2H2 (g) + 1H2O(g)
1CH3CHO(g) + heat
b. decrease the amount of C2H2 ________________________________
lower the temperature
1C2H2 (g) + 1H2O(g)
1CH3CHO(g) + heat
Chemical Equilibrium
III. Le Châtelier’s Principle -stressors that cause a shift in
equilibrium
In the following equilibrium, would you raise or lower the temperature to
get the following results?
1C2H2 (g) + 1H2O(g)
1CH3CHO(g) ΔH0 = -151 kJ
c. increase the amount of H2O _________________________________
raise the temperature
1C2H2 (g) + 1H2O(g)
1CH3CHO(g) + heat
Chemical Equilibrium
III. Le Châtelier’s Principle -stressors that cause a shift in
equilibrium
In the following equilibrium, what effect does decreasing the volume of
the reaction vessel have?
1CO(g) + 1Fe3O4(s)
1CO2 (g) + 3FeO(s)
__________________________________________________________
None. The solids do not change their concentrations, and the number
__________________________________________________________
of moles of gaseous reactant equals the number of moles of gaseous
__________________________________________________________
product
In the following equilibrium, what effect does simultaneously increasing
the temperature and the pressure have?
1CO (g) + 1Cl2(g)
1COCl2(g) + heat
ΔH0 = -151 kJ
__________________________________________________________
Cannot predict. An increase in temperature causes a shift in the
__________________________________________________________
equilibrium to the left, while an increase in pressure causes a shift in
__________________________________________________________
equilibrium to the right.
Chemical Equilibrium
III. Le Châtelier’s Principle
1. Hypothesis: What is the effect of a change in concentration of
reactants and a change in temperature on equilibrium?
2. Prediction:
3. Gather Data:
A. Safety: The surfaces of the hot plate will be hot enough to
cause burns. Use caution. Cobalt(II) chloride
hexahydrate is toxic, with an LD50 = 80mg/kg Avoid
ingestion (don’t eat or drink it). Wash hands
thoroughly with soap and water before leaving lab.
Concentrated Hydrochloric acid is extremely
corrosive. Avoid contact with eyes, skin, and
clothing. Goggles, aprons, and gloves mandatory.
Chemical Equilibrium
III. Le Châtelier’s Principle
3. Gather Data:
B. Procedure:
1. With a partner, measure out 2 mL of 0.1 M CoCl2
solution into a test tube. Record initial color. __________
2. Add 3 mL (60 drops) of concentrated HCl to the test tube.
Record color. _____________
3. Add water dropwise to the test tube until a color change
occurs. Record color. ______________
4. Add 2 mL of 0.1 M CoCl2 solution to another test tube.
Add concentrated HCl dropwise until the solution turns
purple. If the solution turns blue, add water until it turns
purple.
Chemical Equilibrium
III. Le Châtelier’s Principle
3. Gather Data:
B. Procedure:
5. Place the test tube in an ice bath. Record color. ________
6. Place the test tube in a hot water bath. Record color.
___________
4. Analyze Data:
A. The equation for the reversible reaction in this experiment is:
+ heat
1Co(H2O)62+ (aq) +
Hexahydrate Co2+ ion (pink)
4Cl- (aq)
chloride ion
1CoCl42-(aq) + 6H2O (l)
Tetrachlorocobaltate ion (blue)
Chemical Equilibrium
III. Le Châtelier’s Principle
4. Analyze Data:
A. Use the equation to explain the colors of the solution in steps 1,
2, and 3
In Step 1, the solution is initially a pink color, because the reaction
arrives at an equilibrium in which the concentration of the pink-colored
1Co(H2O)62+(aq) is at a higher concentration than the blue-colored
1CoCl42-(aq). In Step 2, the addition of HCl increases the concentration of
Cl-, shifting the equilibrium to the right to favor the formation of the blue
1CoCl42-(aq), so the solution turns blue. In Step 3, the increase in
concentration of water shifts the equilibrium left, re-establishing a new
equilbrium where the concentration of 1CoCl42-(aq) is higher than it was
orginally, so the purple color shows more of a balance of pink and blue.
Chemical Equilibrium
III. Le Châtelier’s Principle
4. Analyze Data:
B. Explain how the equilibrium shifts when heat energy is added
or removed.
In Step 5, since heat acts like a product in the exothermic reverse
reaction, removing heat by lowering the temperature causes the
equilibrium to shift to the left, increasing the rate of the reverse reaction
and causing the solution to turn pink. In Step 6, since heat acts like a
reactant in the endothermic forward reaction, adding heat by increasing
the temperature causes the equilibrium to shift to the right, increasing the
rate of the forward reaction and causing the solution to turn blue.
5. Draw Conclusions:
Chemical Equilibrium
IV. Using Equilibrium Constants -when a ________
reaction has a _____
large
Keq the __________
___,
equilibrium _______
mixture
contains _____
more ________
products than
reactants at __________
equilibrium
________
-when a ________
reaction has a _____
small
Keq the __________
___,
equilibrium _______
mixture
contains _____
more ________
reactants than
products at __________
equilibrium
________
A. Calculating Equilibrium Concentrations -__________
equilibrium
________
constants can also
be used to
________
calculate the
equilibrium
__________
____________
concentration of
any ________
substance
in the _______
reaction
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations
At 1200 K, the Keq for the following reaction equals 3.933. What is the
concentration of the methane produced, if [CO] = 0.850 M, [H2] = 1.333
M, and [H2O] = 0.286 M?
1CO(g) + 3H2 (g)
0.850 M
Keq =
3.933 =
1.333 M
[CH4]1[H2O]1
[CO]1 [H2]3
[CH4]1[0.286]1
[0.850]1[1.333]3
[CH4] =
27.7 M
1CH4(g) +
?M
1H2O (g)
0.286 M
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations
At 1405 K, the Keq for the following reaction equals 2.27 x 10-3. What is
the concentration of the Hydrogen gas produced, if [S2] = 0.0540 M, and
[H2S] = 0.184 M?
2H2S (g)
0.184 M
Keq =
2.27 x
10-3
[H2]
[H2]2 [S2]1
[H2S]2
[H2]2 [0.0540]1
[0.184]2
=
=
0.0377 M
2H2(g) +
?M
1S2 (g)
0.0540 M
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations
If Keq for the following reaction equals 10.5, what is the equilibrium
concentration of Carbon monoxide, if [H2] = 0.933 M, and [CH3OH] =
1.32 M?
1CO(g) + 2H2 (g)
?M
Keq =
10.5
=
0.933 M
[CH3OH]1
[CO]1 [H2]2
[1.32]1
[CO]1 [0.933]2
[CO] = 0.144 M
1CH3OH (g)
1.32 M
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
If the Keq for the following reaction equals 64.0, what are the equilibrium
concentrations of I2, H2, and HI, if [I2]0 = 0.200 M, [H2]0 = 0.200 M and
[HI] = 0.000 M?
1I2(g) + 1H2 (g)
?M
Initial
Change
2HI(g)
?M
?M
[H2]
[I2]
[HI]
0.200
0.200
0.000
-1x
-1x
+2x
0.200 - 1x
2x
Equilibrium 0.200 - 1x
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
1I2(g) + 1H2 (g)
Initial
Change
[H2]
[I2]
[HI]
0.200
0.200
0.000
-1x
-1x
+2x
0.200 - 1x
2x
Equilibrium 0.200 - 1x
Keq =
[HI]2
[I2]1 [H2]1
2HI(g)
8.00
=
2
[2x]
64.0 =
[0.200 – 1x]1 [0.200 – 1x]1
[2x]
[0.200 – 1x]1
x
[H2]
[I2]
[HI]
=
=
=
=
0.160
0.040 M
0.040 M
0.320 M
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
If the Keq for the following reaction equals 16.0, what are the equilibrium
concentrations of PCl3, Cl2, and PCl5, if [PCl5]0 = 1.00 M?
1PCl5(g)
1PCl3(g) + 1Cl2 (g)
?M
?M
?M
[PCl3]
[Cl2]
[PCl5]
Initial
0.00
0.00
1.00
Change
+1x
+1x
-1x
0.00 + 1x
1.00 – 1x
Equilibrium 0.00 + 1x
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
1PCl5(g)
1PCl3(g) + 1Cl2 (g)
[PCl3]
[Cl2]
[PCl5]
Initial
0.00
0.00
1.00
Change
+1x
+1x
-1x
0.00 + 1x
1.00 – 1x
Equilibrium 0.00 + 1x
1 [Cl ]1
[PCl
]
3
2
Keq =
x2 = 16.0 – 16.0x
x = -b ± √b2 – 4ac
[PCl5]1 x2 + 16.0x - 16.0 = 0
2a
1 [x]1
[x]
ax2 + bx + c = 0
16.0 =
[1.00 - x]1
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
x2 + 16.0x - 16.0 = 0
ax2 + bx + c = 0
x = -b ± √b2 – 4ac
2a
x = -16.0 ± √(-16.0)2 – 4(1.00)(-16.0)
2(1.00)
x = 0.950 (but not -17.0)
[PCl3] = 0.950 M
[Cl2] = 0.950 M
[PCl5] = 0.05 M
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
If the Keq for the following reaction equals 0.680, what are the
equilibrium concentrations of COCl2, CO, and Cl2, if [CO]0 = 0.500 M
and [Cl2]0 = 1.00 M?
1CO (g)
?M
+
1Cl2 (g)
1COCl2 (g)
?M
?M
[COCl2]
[CO]
[Cl2]
Initial
0.00
0.500
1.00
Change
+1x
-1x
-1x
0.500 - 1x
1.00 – 1x
Equilibrium 0.00 + 1x
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
1CO (g)
+
1Cl2 (g)
[COCl2]
[CO]
[Cl2]
Initial
0.00
0.500
1.00
Change
+1x
-1x
-1x
0.500 - 1x
1.00 – 1x
Equilibrium 0.00 + 1x
Keq =
0.680 =
1COCl2 (g)
[COCl2]1
[CO]1 [Cl2]1
x = 0.340 - 0.340x – 0.680x + x2
0.680x2 - 2.02x + 0.340 = 0
[x]1
[0.500 - x]1 [1.00 - x]1
ax2 + bx + c = 0
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
0.680x2 -
2.02x
+ 0.340 =
0
ax2 + bx + c = 0
x =
x = -b ± √b2 – 4ac
2a
2.02 ± √(-2.02)2 – 4(0.680)(0.340)
2(0.680)
x = 2.79 or 0.176
[COCl2] = 0.176 M
[CO] = 0.324 M
[Cl2] = 0.82 M
x = 0.176
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
If the Keq for the following reaction equals 36.0, what are the equilibrium
concentrations of H2, Br2, and HBr, if [H2]0 = 0.250 M and
[Br2]0 = 0.250 M?
1H2(g)
?M
+ 1Br2 (g)
2HBr(g)
?M
?M
[H2]
[Br2]
[HBr]
Initial
0.250
0.250
0.000
Change
-1x
-1x
+2x
0.250 - 1x
0.000 + 2x
Equilibrium 0.250 - 1x
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
1H2 (g)
+ 1Br2 (g)
[H2]
[Br2]
[HBr]
Initial
0.250
0.250
0.000
Change
-1x
-1x
+2x
0.250 - 1x
0.000 + 2x
Equilibrium 0.250 - 1x
Keq =
36.0 =
2HBr (g)
[HBr]2
[H2]1 [Br2]1
[2x]2
[0.250 - x]1 [0.250 - x]1
6.00 =
2x
(0.250 – x)
x = 0.188
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
32x2 - 18.0x + 2.25 = 0
ax2 + bx + c = 0
x = -b ± √b2 – 4ac
2a
x = 18.0 ± √(-18.0)2 – 4(32)(2.25)
2(32)
x = 0.375 or 0.188
[H2] = 0.062 M
[Br2] = 0.062 M
[HBr] = 0.376 M
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
If the Keq for the following reaction equals 20.0, what are the equilibrium
concentrations of H2, Cl2, and HCl, if [H2]0 = 1.00 M and
[Cl2]0 = 2.00 M?
1H2(g)
?M
+ 1Cl2 (g)
2HCl(g)
?M
?M
[H2]
[Cl2]
[HCl]
Initial
1.00
2.00
0.00
Change
-1x
-1x
+2x
Equilibrium
1.00 - 1x
2.00 - 1x
0.00 + 2x
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
1H2 (g)
20.0 =
2HCl (g)
[H2]
[Cl2]
[HCl]
Initial
1.00
2.00
0.00
Change
-1x
-1x
+2x
Equilibrium
Keq =
+ 1Cl2 (g)
[HCl]2
[H2]1 [Cl2]1
1.00 - 1x
2.00 - 1x
0.00 + 2x
4.00x2 = 40.0 – 60.0x + 20.0x2
16.0x2 - 60.0x + 40.0 = 0
[2x]2
[1.00 - x]1 [2.00 - x]1
ax2 + bx + c = 0
Chemical Equilibrium
IV. Using Equilibrium Constants
A. Calculating Equilibrium Concentrations from Initial
Concentrations Using ICE (Initial, Change, Equilibrium)
16.0x2 - 60.0x + 40.0 = 0
ax2 + bx + c = 0
x = -b ± √b2 – 4ac
2a
x = 60.0 ± √(-60.0)2 – 4(16.0)(40.0)
2(16.0)
x = 2.88 or 0.869
[H2] = 0.13 M
[Cl2] = 1.13 M
[HCl] = 1.74 M
Chemical Equilibrium
V. Solubility Equilibria -like a few _________
chemical _________
reactions that go
to completion
_________, upon __________,
dissolving some
______
ionic __________
compounds _________
dissociate
ions
completely into _____
1Na+ (aq) +
1NaCl (s)
1Cl- (aq)
-some _____
ionic __________,
compounds however, are
sparingly _______,
soluble and quickly reach
only ________
solubility __________
equilibrium
a ________
1BaSO4 (s)
Keq =
1Ba2+ (aq) +
[Ba2+ ]1 [SO42-]1
[BaSO4]1
1SO42- (aq)
Chemical Equilibrium
V. Solubility Equilibria -in the __________
equilibrium constant
_______ __________,
expression
______
Barium ______
sulfate is a _____,
solid so the _______
[BaSO4]
is _______,
constant and can be combined with the
Keq value to form the ________
solubility _______
product
___
constant
_______
Keq x
[BaSO4]1 = [Ba2+ ]1 [SO42-]1
Ksp = [Ba2+ ]1 [SO42-]1
Write the solubility constant expression for the following solubility
equilibrium:
1Mg(OH)2 (s)
1Mg2+ (aq) + 2OH- (aq)
Ksp = [Mg2+ ]1 [OH-]2
Chemical Equilibrium
V. Solubility Equilibria
A. Calculating Solubilities from
Solubility Product Constants
What is the solubility, in M, of Silver iodide at 298 K?
1Ag+ (aq) + 1I- (aq)
1AgI (s)
Ksp = [Ag+ ]1 [I-]1
8.5 x 10-17
= [Ag+ ]1 [I-]1
8.5 x 10-17
=
s2
√ 8.5 x 10-17
=
√ s2
s = 9.2 x 10-9 M
Chemical Equilibrium
V. Solubility Equilibria
A. Calculating Solubilities from
Solubility Product Constants
What is the solubility, in M, of Copper(II) carbonate at 298 K?
1CuCO3 (s)
1Cu2+ (aq) + 1CO32- (aq)
Ksp = [Cu2+ ]1 [CO32-]1
2.5 x 10-10 = [Cu2+ ]1[CO32-]1
2.5 x 10-10 = s2
√ 2.5 x 10-10 = √ s2
s = 1.6 x 10-5 M
Chemical Equilibrium
V. Solubility Equilibria
B. Calculating Ion Concentration from Ksp
What is [OH-] at 298 K in a saturated solution of Mg(OH)2 at
equilibrium?
1Mg(OH)2 (s)
1Mg2+ (aq) +
2OH- (aq)
Ksp = [Mg2+ ]1 [OH-]2
let x
= [Mg2+ ]
so 2x = [OH-]
5.6 x 10-12 = (x)(2x)2
5.6 x 10-12 = 4x3
1.4 x 10-12 = x3
3
√1.4 x 10-12 = x
1.1 x 10-4 = x
2.2 x 10-4 = 2x = [OH-]
Chemical Equilibrium
V. Solubility Equilibria
B. Calculating Ion Concentration from Ksp
What is [Ag+] at 298 K in a saturated solution of AgBr at equilibrium?
1Ag+ (aq) +
1AgBr (s)
1Br- (aq)
Ksp = [Ag+ ]1 [Br-]1
let x
= [Ag+ ]
so x = [Br-]
5.4 x 10-13 = x2
√5.4 x 10-13 = x
7.3 x 10-7 = x
= [Ag+]
Chemical Equilibrium
V. Solubility Equilibria
B. Calculating Ion Concentration from Ksp
What is [F-] at 298 K in a saturated solution of CaF2 at equilibrium?
1Ca2+ (aq) +
1CaF2 (s)
2F- (aq)
Ksp = [Ca2+ ]1 [F-]2
3.5 x 10-11 = (x)(2x)2
3.5 x 10-11 = 4x3
8.8 x 10-12 = x3
3
√8.8 x 10-12 = x
2.1 x 10-4 = x
4.2 x 10-4 = 2x = [F-]
Chemical Equilibrium
V. Solubility Equilibria
C. Predicting Precipitates -besides being used to calculate the
_________
solubility of an _____
ionic _________
compound
and the ___________
concentration of ____
ions in a
saturated _______,
solution ___
Ksp values can
_________
predict if a _________
precipitate
be used to _______
will form if two
___ _____
ionic __________
compounds
are mixed
Predict whether PbCl2 will form as a precipitate if 100 mL of 0.0100 M
NaCl is added to 100 mL of 0.0200 M Pb(NO3)2:
initial
-the concentrations
____________ of the ______
solutions
________ allow you to calculate
_______ the
____________
concentrations of ____
Pb2+ and ___
Cl- ions
in the _____
mixed _________,
solutions which
multiplied together,
when _________
ion _______,
product or ___
Qsp
determine the ___
Chemical Equilibrium
V. Solubility Equilibria
C. Predicting Precipitates
Predict whether PbCl2 will form as a precipitate if 100 mL of 0.0100 M
NaCl is added to 100 mL of 0.0200 M Pb(NO3)2:
1Pb2+ (aq) +
1PbCl2 (s)
2Cl- (aq)
Qsp = [Pb2+]1 [Cl-]2
[Pb2+] =
[Cl-]
=
0.0200 M
2
0.0100 M
= 0.0100 M
= 0.00500 M
2
Qsp = [0.0100]1 [0.00500]2
Qsp = 2.50 x 10-7 < 1.7 x 10-5 =
Ksp
Chemical Equilibrium
V. Solubility Equilibria
C. Predicting Precipitates -if the ___
Qsp is ___
< the ___,
Ksp the
_______ is __________,
solution
unsaturated and a
_________
precipitate ____
will ___
not ____,
form and if
Qsp is ___
= the ___,
Ksp the _______
solution
the ___
saturated and ___
no change
is _________
______ will
Qsp is ___
Ksp a
occur, but if ___
> the ___,
__________
precipitate will form, reducing the
Qsp ___
Ksp
___
ion ___________
concentration until ___
= ___,
and the system arrives at
equilibrium and the _______
solution
__________
becomes ________
saturated
Qsp =
2.50 x 10-7 < 1.7 x 10-5 =
No precipitate should form
Ksp
Chemical Equilibrium
V. Solubility Equilibria
C. Predicting Precipitates
Predict whether Ag2SO4 will form as a precipitate if 500 mL of 0.010 M
AgNO3 is added to 500 mL of 0.25 M K2SO4:
1Ag2SO4 (s)
2Ag+ (aq) +
1SO42- (aq)
Qsp = [Ag+]2 [SO42-]1
[Ag+]
=
[SO42-] =
Qsp
Qsp
0.010 M
2
0.25 M
= 0.0050 M
= 0.012 M
2
= [0.0050]2 [0.012]1
= 3.0 x 10-7 < 1.2 x 10-5 =
No precipitate should form
Ksp
Chemical Equilibrium
V. Solubility Equilibria
C. Predicting Precipitates
Predict whether a precipitate will form if 200 mL of 0.20 M MgCl2 is
added to 200 mL of 0.0025 M NaOH:
1Mg(OH)2 (s)
1Mg2+ (aq) +
2OH- (aq)
Qsp = [Mg2+]1 [OH-]2
[Mg2+] =
[OH-]
=
0.20 M
2
0.0025 M
= 0.10 M
= 0.0012 M
2
Qsp = [0.10]1 [0.0012]2
Qsp = 1.4 x 10-7 > 5.6 x 10-12 = Ksp
A precipitate of Mg(OH)2 should form
Chemical Equilibrium
V. Solubility Equilibria
D. Common Ion Effect -the ________
solubility of _______
PbCrO4 in _____
water is
________
4.8 x 10-7 mol/L, which means that
you can ________
dissolve ________
4.8 x 10-7 of
PbCrO4 in ____
1.00 L of _____
pure _____,
water
_______
4.8 x 10-7 of _______
PbCrO4 will ____
not
but _________
_______
dissolve in ____
1.00 L of a ______
0.10 M
solution of _______,
K2CrO4 because of the
________
common ___
ion ______
effect
1PbCrO4 (s)
1Pb2+ (aq) +
1CrO42- (aq)
Ksp = [Pb2+]1 [CrO42-]1 = 2.3 x 10-13
product of the concentrations
-since the _______
____________
ions is _____
equal to a constant
of both ____
_______,
(the _________
solubility _______
product _______),
constant if
2+]
[CrO42-] goes up
_______
__, [Pb
_____
must go _____
down
Chemical Equilibrium
V. Solubility Equilibria
D. Common Ion Effect -adding a _______
solution to an __________
equilibrium
that contains a ________
common ___
ion _______
lowers
the ________
solubility of a _________
substance
ion or, according to
containing that ___,
Le Châtelier’s ________,
Principle stresses
_____________
the __________
equilibrium and causes the
_______
system to _____
shift the __________
equilibrium in
the _______
direction that _______
relieves the ______
stress
1PbCrO4 (s)
1Pb2+ (aq) +
1CrO42- (aq)
Chemical Equilibrium
V. Solubility Equilibria
1. Hypothesis: How Do Solubility Product Constants Compare?
2. Prediction:
3. Gather Data:
A. Safety: Silver nitrate stains skin and clothing and is highly
toxic, with an LD50 = 50mg/kg Avoid ingestion (don’t
eat or drink it). Wash hands thoroughly with soap and
water before leaving lab. Goggles mandatory.
B. Procedure:
1. Using a pipette, place 20 drops of AgNO3 solution into
test well A1 of a 20-well microplate. Place 20 more drops
of the same solution in test well A2.
Chemical Equilibrium
V. Solubility Equilibria
3. Gather Data:
B. Procedure:
2. Add 10 drops of NaCl solution to both test well A1 and
test well A2. Record observations___________________
3. To test well A2 only, add 10 drops of Na2S solution.
Record observations______________________________
4. Compare the contents of test wells A1 and A2. Record
observations_____________________________________
Chemical Equilibrium
V. Solubility Equilibria
4. Analyze Data:
A. Write the complete thermochemical equation for the reaction
that occurred in Step 2.
B. Write the net ionic equation for the reaction in Step 2.
C. Write the equation for the solubility equilibrium that was
established in test wells A1 and A2 during Step 2.
D. Write the solubility constant expression for the equilibrium
established in test wells A1 and A2 during Step 2.
E. Write the equation for the solubility equilibrium that was
established in test well A2 during Step 4.
Chemical Equilibrium
V. Solubility Equilibria
4. Analyze Data:
F. Match the chemical formula of each precipitate with its color.
G. Compare the two Ksp values for the two precipitates. Infer
which is the more soluble.
H. Use Le Châtelier’s Principle to explain how the addition of
Na2S in Step 4 affected the equilibrium in test well A2.
Chemical Equilibrium
V. Solubility Equilibria
4. Analyze Data:
I. Calculate the molar solubilities of both precipitates in the
experiment. Which of the precipitates is more soluble?
Chemical Equilibrium
V. Solubility Equilibria
4. Analyze Data:
A. Write the complete thermochemical equation for the reaction
that occurred in Step 2.
AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
B. Write the net ionic equation for the reaction in Step 2.
Ag+ (aq) + Cl- (aq)
AgCl (s)
C. Write the equation for the solubility equilibrium that was
established in test wells A1 and A2 during Step 2.
1AgCl (s)
1Ag+ (aq) + 1Cl- (aq)
D. Write the solubility constant expression for the equilibrium
established in test wells A1 and A2 during Step 2.
Ksp = [Ag+]1 [Cl-]1 = 1.8 x 10-10
E. Write the equation for the solubility equilibrium that was
established in test well A2 during Step 4.
1Ag2S (s)
2Ag+ (aq) + S2- (aq)
Chemical Equilibrium
V. Solubility Equilibria
4. Analyze Data:
F. Match the chemical formula of each precipitate with its color.
AgCl (s) is white; Ag2S (s) is black
G. Compare the two Ksp values for the two precipitates. Infer
which is the more soluble.
Ksp for AgCl (s) = 1.8 x 10-10
Ksp for Ag2S (s) = 8 x 10-48; AgCl is more soluble
H. Use Le Châtelier’s Principle to explain how the addition of
Na2S in Step 4 affected the equilibrium in test well A2.
1AgCl (s)
1Ag+ (aq) + 1Cl- (aq)
The addition of S2- to the equilibrium removes Ag+ from the equilibrium,
causing the system to shift to the right in favor of the formation of Ag+,
so at the same time the black precipitate Ag2S is forming, the white AgCl
is dissolving to relieve the stress on the equilibrium.
Chemical Equilibrium
V. Solubility Equilibria
4. Analyze Data:
I. Calculate the molar solubilities of both precipitates in the
experiment. Which of the precipitates is more soluble?
1Ag+ (aq) + 1Cl- (aq)
1AgCl (s)
2Ag+ (aq) + 1S2- (aq)
1Ag2S (s)
Ksp = [Ag+ ]1 [Cl-]1
Ksp = [Ag+ ]2 [S2-]1
1.8 x 10-10
= [Ag+ ]1 [Cl-]1
8 x 10-48
= [Ag+ ]2 [S2-]1
1.8 x 10-10
=
s2
8 x 10-48
=
=
√ s2
√ 1.8 x
10-10
3
√ 2x
10-48
s = 1.3 x 10-5 M
AgCl is more soluble
=
4s3
3
√ s3
s = 1 x 10-16 M