Download Unit 4 - Chemical Equilibrium

1
UNIT 4 CHEMICAL EQUILIBRIUM
Day 1
A.
Recognizing Equilibrium (pg 424)
Gaseous Equilibrium : Consider the reaction : H2 (g) + I2 (g) 
i.
2 HI (g)
6.00 mole of H2 and 4.00 mole of I2 are placed in a 1.00 L vessel ; some of the H2 and I2 molecules
collide and react forming HI molecules ; as the HI molecules accumulate they begin reacting with each
other forming H2 and I2. At first the rate of the forward reaction is
than the rate
of the reverse reaction but as time proceeds the forward reaction
in rate and the
reverse
in rate. At some point in time the rate of the forward and reverse
reaction are equal and there appears to be no reaction going on. The above reaction is easy to follow
since I2 is violet in colour and HI is colourless. At first the colour of the system is
and the colour will
as the reaction proceeds.
When there is no further change in colour we know the system is in
Referred to
as
equilibrium since reactions are still going on but in a balanced state.
H2 (g)
[initial][i]
[change][c]
[euilibrium][e]
_____
_____
_____
+
I2 (g) 
_____
_____
_____
2 HI (g)
_____
_____
_____
Percent reaction
-
Similar to percent yield
The yield of product measured at equilibrium compared with the maximum possible yield of
product
2
Percent Reaction = Concentration at Equilibrium x 100%
Theoretical yield
% Reaction =
Consider the graphs above and answer the following :
1.
2.
The initial concentration of H2 _______ I2
.
As the reaction proceeded the concentration of H2 decreased to 2.5 mol/L and I2 decreased to
0.50 mol/L : [HI] increased from 0 to 7.0 mol/L . Why did the [HI] increased twice the value
that both hydrogen and Iodine decreased by ? ________________________
Equilibrium Conditions
1.
2.
3.
No input or output (_________________ system )
Presence of both species ( reactants and products ) or phases in a two phase system
No apparent change in macroscopic properties ( pressure, temp., colour etc.)
Consider the flasks below :
H2 (g)
[initial][i]
[change][c]
[euilibrium][e]
_____
_____
_____
+
I2 (g) 
_____
_____
_____
2 HI (g)
_____
_____
_____
% Reaction =
ii.
-
Solubility Equilibrium
consider sodium chloride dissolving in water ; the crystals begin to ____________________ as more
and more ions accumulate they start _______________________ . When these two rates are equal the
solution is said to be ________________________ and equilibrium has been established.
iii.
Phase equilibrium
H2O (s)
 H2O(l)
4
Consider the following :
When the rate of evaporation is equal to the
rate of condensation, no net change is
occurring. The system is in a state of dynamic
equilibrium since both events occur
simultaneously.
1.
2
3.
4.
5.
6.
System closed or open _____________________
Presence of both species or just one
__________________________
Constant properties or properties changing ______________________________
How would you know if this system is at equilibrium or not? ____________________
How does the rate of evaporation compare to the rate of condensation? ____________
How would the rates change if it was heated ?____________________
Complete pg 428 (1-3)
N2O4 - NO2 Equilibrium(pg 430)
-
consider a container is filled with colourless N2O4 (g)
-
Consider the equation : N2O4 (g) + energy 2NO2 (g)(reddish-brown)
N2O4 is just 2 NO2 molecules stuck together
the N2O4 molecules begin to break down forming _____________ and the colour _____________; at
some point the colour remains _________________ and _______________________ is established.
5
Consider the following experimental data
Time(s)
0
20
40
60
80
100
[N2O4]
0.10
0.07
0.05
0.04
0.04
0.04
[NO2]
0.0
0.06
0.10
0.12
0.12
0.12
Questions :
1.
At what time was equilibrium established ? ___________________
2.
At equilibrium which one has a higher concentration and what colour would exist ?____________
3.
At 20 s is the rate of the forward reaction increasing or decreasing ?
about the reverse reaction _________________________
4.
If the conc. of N2O4 decreases by 0.02 ; by what value does the NO2 increase by and
Why ? __________________________________
5.
Does this data reflect the graph above ? ________________________________________
; what
Example 2
1.
a.
The reversible reaction is : Cl2(g) + PCl3(g)  PCl5(g)
Consider the table below :
Reaction Time (s)
[ Cl2 ]
[ PCl3 ]
[ PCl5 ]
0
1.00
1.00
0.00
20
0.90
0.90
0.10
40
0.80
0.80
0.20
60
0.75
0.75
0.25
80
0.71
0.71
0.29
100
0.71
0.71
0.29
120
0.71
0.71
0.29
When was equilibrium attained ? ________________
5
b.
What would be the conc. of Cl2 after 20 hours ? ____________________
c.
After 40 s which reaction is faster ? ( forward or reverse ) ___________________
d.
At 25 s is the forward reaction increasing or decreasing in its rate ? _____________
e.
At 25 s is the reverse reaction increasing or decreasing in its rate ? _____________
f.
At 100 s which reaction is faster, forward or reverse? __________________
g.
At 100 s which reaction is increasing in its rate, the forward or the reverse? ____________
Ex.3
Consider the equilibrium : N2 + 3 H2  2 NH3
1.0 mol of N2 and 1.0 mol of H2 are placed in a 1.0 L flask and allowed to react. Consider the table
below :
a.
b.
c.
d.
e.
f.
Reaction Time
[ N2 ]
[ H2 ]
[ NH3 ]
0.0 min.
1.00
1.00
0.00
1.0
0.90
0.70
0.20
2.0
0.80
0.40
0.40
3.0
0.75
0.25
0.50
4.0
0.72
0.16
0.56
5.0
0.70
_____
_____
6.0
_____
_____
0.60
Fill in the blanks in the table.
When was equilibrium established ? _________________________
What is equal at equilibrium ? ______________________________
Account for the change in concentration for both reactants and products. __________________
If all the N2 had been converted to NH3 . How many moles of ammonia would form ? _________
Calculate the % yield considering the amount of ammonia that is found at equilibrium ?
_____________________________
Complete pg 437( 1,3 )
Day 2 :
-
6
Equilibrium Calculations (pg 431)
consider the equilibrium N2O4 2 NO2
1.00 mole of N2O4 is placed in a 2.0 L container and allowed to come to equilibrium. At equilibrium
there is 0.200 mole of N2O4 calculate the concentration of NO2
[initial][i]
[change][c]
[euilibrium][e]
N2O4 
_____
_____
_____
2 NO2
_______
_______
_______
To calculate the theoretical yield or percent reaction ; assume all the N2O4 is converted to NO2.
The theoretical yield would be ___________________What is the % yield ? ___________________
Complete pg 437 ( 6,7,4 ) pg 438 (7a-c,8,9)
Law Of Chemical Equilibrium (pg 439)
-
consider the general equation : aA + bB  cC + dD
Ke = kf/kr = [C]c[D]d
------------- = Equilibrium constant
[A]a[B]b
[ ] - refers to _______________ in the units of __________________
- Consider the reaction below :
2H2(g) + O2(g)  2H2O (g)
7
Equilibrium constant
-
Ke = [H2O]2
-----------------[H2]2[O2]
Mass action expression
-
Ke is determined by dividing product of the conc. of products by the product of the conc. of
reactants
high Ke values indicate that the _________________ are favored over the __________________
Write the Ke expressions for the following and state which is favored at equ. reactants of products ;
also note that solids and liquids do not appear in equ. expressions only gases and aqueous solutions.
1.
2 HI (g) I2(g) + H2(g)
Ke = 1.84 x 10-2 =
at 423oC
2.
AgCl(s)  Ag+(aq) + Cl-(aq)
Ke = 1.7 x 10-10 =
at 25oC
3.
SO2(g) + 1/2 O2(g)  SO3(g)
4.
2 C(s) + O2(g)  2 CO(g)
Ke = 7.7 =
Ke = 1.7 x 102 =
General Rule :
If Ke is > 1 than conc. of products is _____________ conc of reactants
If Ke is < 1 than conc. of products is _____________ conc. of reactants
Magnitude of Ke :
The higher the Ke the greater extent the reaction has gone in the forward direction
that is toward completion
Consider the equ. Cu(s) + 2 Ag+(aq)  Cu2+(aq) + 2 Ag(s)
Ke = 2.0 x 1015
Write the Ke expression : Ke =
Because of the high value the
are favored over the _______________________
and thus the ______________ reaction is favoured over the ____________ reaction.
-
If the Ke for the following equ. is 50 : H2 + I2  2 HI than what would be Ke for the following :
a.
1/2 H2 + 1/2 I2  HI
b.
2 HI  I2 + H2
8
Ke =
Ke =
Complete pg 442 (1) pg 444 (4-6) pg 447 (7)
Calculating Ke values(pg 443)
1.
Given the equ. : 2 A(g) + B(g)  C(g) + 2 D(g)
The following equ. conc. were found [A] = 2.0 moL/L ; [B] = 3.0 mol/L ; [C]= 4.0 mol/L ;
[D] = 1.0 mol/L
Calculate Ke =
2.
= 0.33
Given the equ. : A(g) + 2B(g)  2C(g)
1.0 mole of A and 2.0 moles of B are placed in a 1.0 L vessel. At equ. there was 0.50 moles of A.
Calculate the equ. conc. of B and C and calculate the Ke value.
[initial] [i]
A(g)
+
_____
2 B(g) 
______
2 C(g)
______
[change in] [c]
_____
______
______
[final] [e]
_____
______
______
Ke =
=
3.
9
Consider the equ. : H2 + I2  2 HI
0.10 moles of H2 and 0.20 moles of I2 are placed in a 2.0 L container. When equ. was established it
was found that 20 % of the H2 had reacted. Calculate the equ. conc. of each and calculate the Ke
value.
H2
[initial] [i]
+
_____
I2

2 HI
______
______
[change in] [c]
_____
______
______
[equ.] [e]
_____
______
______
Ke =
4.
=
Consider the equ . 2 SO2(g) + O2(g)  2 SO3
A 10.0 L vessel was filled with 4.0 mol SO2 , 2.2 mol O2 and 5.6 mol of SO3. When equ. is established
there are 2.6 mol of SO2. Calculate equ. conc of each and calculate the Ke value.
2 SO2
+
O2 
2 SO3
[initial] [i]
_____
______
______
[change in][c]
_____
______
______
[final] [e]
_____
______
______
Ke =
=
Complete pg 449 ( 3,4,6,8,9 )
= 48
Day 3 Factors affecting the State of Equlibrium
10
Le Châteliers Principle(pg 450)
-
if a
system at equilibrium is subjected to change or __________
processes tend to occur to counteract the stress
1.
Concentration (pg 450)
-
consider the equilibrium : Fe3+ ( aq ) + SCN- 
FeSCN2+
( yellow )
( colourless ) ( deep red )
-
adding more SCN- ---->
stress too much SCN- to relieve stress equilibrium shifts to the _____________ producing
more
causing the colour to __________________
-
Ex. 2
-
.
adding more Fe3+ has equal affect; See graph below
N2 ( g ) + 3 H2 (g)  2 NH3 ( g )
increase conc. of N2 -----> stress too much
to relieve stress equilibrium
shifts to the
; thus conc. of H2 will
and conc. of NH3 will
____________ Consider the equ. conc to be N2 = 1.0 moL/L; H2 = 3.0 moL/L and NH3 = 2.0 moL/L
If an additional 0.5 moL/L of N2 is added; graphically show how the equ. conc are established; keep in
mind the mole ratios from the equation.
Graph :
H2
NH3
N2
Ex. 3
11
Consider the equ. SO2(g) + 1/2 O2 (g) SO3(g)
What would be the effect on the equilibrium conc. of all substances in the following instances .
Illustrate on the graphs that follow:
a.
More O2 is pumped in :
effect on [ SO2 ]
effect on [ O2 ]
effect on [ SO3 ]
b.
Some
effect
effect
effect
SO2 is removed from the system :
on [ SO2 ]
on [ O2 ]
on [ SO3 ]
[SO 3]
[SO 3 ]
[SO 2]
[SO 2 ]
[O 2]
[O 2 ]
it is important to understand that only substances in the gaseous state or in aqueous form can affect
equilibrium ; that is the addition of solids will not affect the equilibrium position.
Consider the equ. : FeO (s) + CO(g)  Fe (s) + CO2 (g)
What would be the effect on the equ. conc. of CO2 as a result of the following ?Answer each one as follows
the stress is
; to relieve the stress equ.shifts to the
and the equ. conc. of CO2 will
______________
Stress
Increase [CO]
Add more FeO
Decrease [CO2 ]
i.
ii.
iii.
Equilibrium Shift
[CO2 ]
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
Review :
A(g) + B (g)
2A(g) + B (g)
C(g) + D(g)
Some A is added
[A]
[A]
[C]
time
[B]
[B]
[D]
time
Some A is added
2x
time
time
New Equilibrium is established
time
4C(g) + 3D(g)
time
New Equilibrium is established
x
time
4x
[C]
3x
[D]
time
2.
Effect of Temperature (pg 453)
-
consider the equ. N2O4 (g) + 59 kJ  2 NO2 (g)
increase in temp. ; stress is too much energy to relieve the stress equ. shifts to the
using up some of the energy ; reddish brown colour ________________
increase in
conc. and a decrease in
conc.
12
since there
[N 2 O 4 ]
[NO2 ]
Ex.2
consider equ. N2 (g) + 3 H2 (g) 2 NH3 (g) + energy
-
increase in temp. ; stress is
to relieve stress equ. shifts to the
thus conc. of NH3
decrease in temp. ; stress is
to relieve stress equ. shifts to the
; [ NH3] ____________________
-
H2
NH3
N2
Ex.3
Consider equ. N2 (g) + 1/2 O2 (g) + energy  N2O (g)
Complete graph below if the temp. is suddenly increased.
[N 2 ]
[O 2 ]
[N 2 O]
Ex. 4 Consider equ. C2H4 (g) + H2 (g)  C2H6 (g) + energy Complete the graph below if there was a sudden
decrease in temp.
[C 2H 4]
[H 2]
[C 2H 6]
3.
Effect of Pressure (pg 454)
N2O4 (g) + energy  2 NO2 (g)
-
consider a sudden decrease in volume ; partial pressure of gases would
colour of gases
would
; lasts for a long moment then colour
_________________ and becomes constant.
-
stress is too much pressure ; to relieve stress equ. shifts to the
mole(s) exerts less pressure than
mole(s);
# of moles of NO2 will
and # of moles of N2O4 will
will
__________________
since13
but conc. of both
Diagram :
[N 2 O 4 ]
[NO2 ]
Ex.2
-
Consider the equ. N2(g) + 3 H2(g) 
2 NH3(g)
Consider a sudden increase in volume : partial pressure of gases __________; stress is not enough
pressure equ. shifts to the
since
moles exerts more
pressure than
moles ;
# of moles of NH3
and # moles of N2 and H2 both
But the conc. of
each ________________________________
Diagram :
H2
NH3
N2
Pressure affects equilibrium only if :
1.
2.
Ex.2
gases are involved not solids or liquids
# of moles of gas on the left and right of the equation are different.
Consider the equilibrium : SO2(g) + 1/2 O2(g)  SO3(g)
If the volume was decreased ; what effect would have the # of moles of each substance when
equilibrium is re-established ? Also indicate the effect on the conc. of each.
Substance
SO2
O2
SO3
# of moles
Concentration
14
[SO 3 ]
[SO 2 ]
[O 2 ]
Consider the Equ.: CO2 (g)  CO2 (aq) + energy
One brand of soft drink is bottled under a pressure of 101.3 KPa. A competing brand is bottled under
202.6 KPa. How does this affect the conc. of CO2 in the drink ? Stress is
to relieve
stress equ. shifts to the
thus [CO2(aq)]
-
In the bottles of soft drink at constant temp. CO2(aq) is in equ. the CO2(g) present in the space above
the liquid. If P is constant ;
a)
How does the [CO2(aq)] change when the drink is refrigerated ?
Stress is
to relieve stress equ. shifts to the
_________________________ [CO2(aq)] _____________________
How does [ CO2 (aq) ] change if the drink is warmed ?
Stress is
to relieve stress equ. shifts to the
[ CO2(aq) ] _______
b)
Review :
2A(g)+B (g) + heat
4C(g) +
3D(g)
The temperature is increased
4x
[A]
[C]
2x
time
[B]
x
time
IV
3x
[D]
time
Effect of a Catalyst (pg 455)
-
V
time
New Equilibrium is established
a catalyst speeds up a reaction since less
energy is needed but
both the forward and reverse or affected by the same amount so that equilibrium is
established _________________ but it has no overall effect on the equilibrium.
Adding Inert Gases
-
adding a noble gas such as He does not affect equilibrium as long as the volume remain
constant ; if the pressure remains constant the volume must increase thus is a decrease in
partial pressures of gases ; equilibrium shifts to the side containing more moles.
Day 4 Consider the Equ. 2 H2O(g) + heat  2 H2(g) + O2(g)
1.
2.
3.
4.
5.
increase [H2] - stress is
; conc. H2
decrease [H2O] -stress is
; [H2]
increase [O2] - stress is
; [O2]
increase temp.- stress is
[H2]
15
to relieve stress equ. shifts to the
[H2O] ______________
to relieve stress equ. shifts to the
; [O2]
; [H2O] _______________
; to relieve stress equ. shifts to the
[H2]
; [H2O] ______________________
; to relieve stress equ. shifts to the ___________
; [O2]
; [H2O] ______
, [O2]
decrease volume - stress is
since there are
moles H2
, # of moles O2
[H2]
, [O2]
,[H2O]
; to relieve stress equ. shifts to the
moles of gas on that side of the equation; #
, # of moles of H2O _____
;
.
6.
add an inert gas ( helium) so that the volume is constant.
- with the addition of helium the total pressure will
; the partial pressure of the
gases H2, O2 and H2O will _____________; thus stress on the system is _______________
and
equ.__________________________
7.
add an inert gas so that the pressure remains constant
- for pressure to remain constant the volume must ______________ ; this will __________
the partial pressure of the reactants and products ; stress is ____________ to relieve stress equ.
shifts to the ______________ since there are ________ moles on that side.
# of moles H2 _______; O2 ________ and # of moles of H2O __________ ; [H2] _________
[O2] __________ [H2O] _______________________________
8.
A catalyst is added _______________________________________
Complete pg 457 ( 1-6) pg 459 (2,3,5,6,8,12)
16
missing words
added
any
arrow
both
catalysts
closed
concentration
counteract
directions
dynamic
endothermic
equally
equilibrium
escape
exothermic
extra
gaseous
investigated
lowers
original
position
pressure
principle
quantities
raise
rate
reaction
relative
reverse
reversible
scientist
sealed
shift
speeds
temperature
time
volume
⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌⇌
⇌⇌⇌⇌
N2
2.
N2O4
(g)
3.
H2
+ CO2
4.
4NH3
(g)
+ 5O2
5.
2SO3
(g)
⇌ 2SO2
6.
+ 3H2
⇌ 2NH3
1.
(g)
(g)
2NO
(g)
7.
3N2O4
8.
3NO2
9.
CO
10.
PCl5
(g)
⇌ 2NO2
+ O2
(g)
(g)
(g)
(g)
+ H2O
(g)
⇌ PCl3
(g)
+ H2O
⇌ 4NO
+ O2
(g)
(g)
(l)
(l)
⇌ CH3OH
(g)
+ Cl2
(g)
-115
(aq)
(aq)
(g)
+ 2NO
+ NO
(g)
(g)
-103
-117
-92
+124
Reading Assignment : Haber Process (pg 461) Complete pg 462 ( 3-5 )
neither
absent
catalyst
neither
low
pressure
high
neither
-909
(g)
+196
⇌ 4HNO3
⇌ 2HNO3
+40
(g)
+ 6H2O
(g)
⇌ 2NO2
(g)
low
+58
⇌ CO
(g)
temperature
-92
(g)
(g)
+ 2H2O
+ 2H2
(g)
(g)
ΔH for the
forward
reaction
(kJmol-1)
high
According to Le Chatelier's principle, which conditions
give the best yield of products for these chemical
reactions. (Tick neither if changing the condition
makes no difference to the equilibrium position.)
presen
t
Dynamic Equilibrium
When the _ _ _ _ _ in a chemical reaction looks like this ⇌ , it shows that the reaction is
_ _ _ _ _ _ _ _ _ _. This means the products can react together and turn back into the
_ _ _ _ _ _ _ _ reactants. In other words, the reaction can go _ _ _ _ ways.
When a reversible reaction is set up in a _ _ _ _ _ _ container, the forward reaction happens
much faster than the reverse reaction at first because there's a high _ _ _ _ _ _ _ _ _ _ _ _ _
of reactants and hardly
_ _ _ products. As _ _ _ _ goes on, however, the concentration of reactants falls and the
concentration of products rises. The forward reaction gets gradually slower and the
_ _ _ _ _ _ _ reaction gets gradually faster. It follows that there must come a time when the
reactions are happening in both _ _ _ _ _ _ _ _ _ _ at the same _ _ _ _ and the _ _ _ _ _ _ _
_ _ _ of reactants and products stop changing. This balanced situation is called _ _ _ _ _ _ _
equilibrium. It can only be achieved in a _ _ _ _ _ _ system where no materials can _ _ _ _ _
_ or be _ _ _ _ _.
A _ _ _ _ _ _ _ _ _ called Le Chatelier _ _ _ _ _ _ _ _ _ _ _ _ what happens when the
conditions of reactions in _ _ _ _ _ _ _ _ _ _ _ are changed. He developed this _ _ _ _ _ _ _
_ _: 'If a reaction in equilibrium is subjected to a change, then it responds in such a way as to
_ _ _ _ _ _ _ _ _ _ the effect of that change'. In other words; if you change it, it tries to
change back!
Imagine a reversible reaction where the forward reaction is _ _ _ _ _ _ _ _ _ _ and the
reverse reaction is endothermic. If we raise the _ _ _ _ _ _ _ _ _ _ _, Le Chatelier's principle
tells us that more _ _ _ _ _ _ _ _ _ _ _ reverse reaction will happen to remove the _ _ _ _ _
heat we added. The quantity of reactants will increase _ _ _ _ _ _ _ _ to the quantity of
products and we say that the equilibrium position has shifted in favour of the reactants.
Now imagine a reversible reaction where the _ _ _ _ _ _ of gaseous reactants is greater than
the volume of _ _ _ _ _ _ _ products. In this situation the forward reaction decreases gas
volume and, therefore, _ _ _ _ _ _ the _ _ _ _ _ _ _ _. If we _ _ _ _ _ the pressure, Le
Chatelier's principle tells us that more forward _ _ _ _ _ _ _ _ will happen to remove the extra
pressure we added. The equilibrium _ _ _ _ _ _ _ _ will _ _ _ _ _ in favour of the products.
Le Chatelier found that adding _ _ _ _ _ _ _ _ _ to reversible reactions in closed systems does
not change the quantities of reactants and products at equilibrium. He observed, however,
that adding catalysts _ _ _ _ _ _ up the forward and reverse reactions _ _ _ _ _ _ _ so that
dynamic equilibrium is reached more quickly.
17
Complete pg 462 ( 3-5 )
Case Study: The Haber Process: Ammonia for Food and Bombs
- N2 (g) + 3 H2 (g)  2 NH3 (g) + heat
- to increase amount of product made, Le Chatelier’s principle indicates:
- increase [ reactants] – continuous addition of N2 and H2
- decrease [products] – cool bottom of reaction container – ammonia condenses at – 33 oC, cooled
nitrogen and hydrogen gases are recycled back to the top of the reaction container so are not wasted
- increase pressure – happens at about 50 Mpa ( 50 atmospheres ) pressure
- decrease temp – problem – at low temp, rate forward reaction so slow that it is uneconomical
- so temp kept high , at 500 oC to speed up reaction and the other 3 factors compensate
- use a catalyst to speed up reaction further – discovery of an effective catalyst got Haber the Nobel
prize
footnote: Haber was a good German citizen and helped invent mustard gas for Germany to use as a
weapon during WW1 – one of the first uses of chemical weapons and devastating to any who were
exposed to it. - many people objected to him being awarded the Nobel prize because of this other part of
his work
18
The Haber Process & Optimum Conditions ⇌⇌⇌
The Haber Process is an important _ _ _ _ _ _ _ _ _ _ process because it's used to
make the gas _ _ _ _ _ _ _ (NH3). Ammonia is a _ _ _ material for the manufacture
of _ _ _ _ _ _ _ _ _ _ _ and these are vital when it comes to growing food for the
world's enlarging _ _ _ _ _ _ _ _ _ _. Ammonium nitrate is a particularly important
fertilizer. It's made from ammonia using this _ _ _ _ _ _ _ _ _ _ _ _ _ _ reaction:
ammonia + nitric acid ⇌ ammonium nitrate
NH3 (aq) + HNO3 (aq) ⇌ NH4NO3 (aq)
During the Haber Process, _ _ _ _ _ _ _ _ gas from the air and hydrogen _ _ _ from
water or _ _ _ _ _ _ _ gas are fed into a reaction _ _ _ _ _ _ _. Here they combine
to form ammonia:
N2 (g) + 3H2 (g) ⇌ 2NH3 (g)
(ΔH = -92 kjmol-1)
We can see from the _ _ _ _ _ that the Haber Process reaction is
_ _ _ _ _ _ _ _ _ _. As such the reaction will reach equilibrium after a time and the
reaction chamber will contain a _ _ _ _ _ _ _ mixture of nitrogen, hydrogen and
ammonia. This _ _ _ _ _ _ _ is piped from the reaction chamber into a
_ _ _ _ _ _ _ _ _ where the _ _ _ _ _ _ _ _ _ _ _ is lowered to a level just below the
boiling point of ammonia. The ammonia gas then condenses into _ _ _ _ _ _
ammonia so it can be _ _ _ _ _ _ off at the condenser _ _ _ _. Since the _ _ _ _ _ _
_ points of nitrogen and _ _ _ _ _ _ _ _ are _ _ _ _ _ than ammonia, these two _ _
_ _ _ _ _ _ _ remain as gases and can be _ _ _ _ _ _ _ _. They pass through _ _ _
_ _ back into the reaction chamber.
The _ _ _ _ _ _ _ _ _ _ in the Haber Process reaction chamber have been carefully
_ _ _ _ _ _ by industrial _ _ _ _ _ _ _ _ _ _ who must consider two main
_ _ _ _ _ _ _ _ _ _; they want conditions that produce _ _ _ _ of ammonia as
_ _ _ _ _ _ _ as possible. In other words, they must _ _ _ _ _ _ _ _ both _ _ _ _ _
and reaction _ _ _ _.
missing words
added
iron
ammonia
liquid
arrow
lots
atmospheres low
base
lower
better
mixture
boiling
moderate
catalysts
natural
chamber
neutralization
chemicals
nitrogen
chosen
pipes
compromise
population
condenser
position
conditions
pressure
consider
principle
equilibrium
priorities
even
quickly
exothermic
rate
expensive
raw
factors
reasonable
faster
recycled
fertilizers
reverse
forward
reversible
four
running
gas
scientists
gaseous
slow
hazardous
speed
high
tapped
hydrogen
temperature
improve
two
industrial
yield
Since the forward reaction of the Haber Process is _ _ _ _ _ _ _ _ _ _, Le Chatelier's _ _ _ _ _ _ _ _ _ tells us that a _ _
_ temperature will shift the equilibrium _ _ _ _ _ _ _ _ in favour of the products to give a good yield of ammonia.
Unfortunately, a low temperature also makes the reaction very _ _ _ _ and so scientists have had to
_ _ _ _ _ _ _ _ _ _. They have opted for a _ _ _ _ _ _ _ _ temperature (450oC) which gives a reasonable yield at a _ _
_ _ _ _ _ _ _ _ rate.
The _ _ _ _ _ _ _ reaction in the Haber Process lowers the _ _ _ _ _ _ _ _ because _ _ _ _ volumes of nitrogen and
hydrogen react to produce only _ _ _ volumes of ammonia. In this situation Le Chatelier's principle tells us that a
_ _ _ _ pressure is best for producing a high yield of ammonia and high pressure conditions have the _ _ _ _ _ benefit
of making chemical reactions happen _ _ _ _ _ _. Scientists have not had to compromise with pressure and have opted
for a high pressure of 200 _ _ _ _ _ _ _ _ _ _ _. An even higher pressure might seem like a good idea because it would
produce a _ _ _ _ _ _ yield _ _ _ _ more quickly, however, there are other _ _ _ _ _ _ _ to consider. Setting up and _ _
_ _ _ _ _ high pressure factories is _ _ _ _ _ _ _ _ _ and potentially _ _ _ _ _ _ _ _ _.
An _ _ _ _ catalyst is used in the Haber Process. _ _ _ _ _ _ _ _ _ do not affect the position of equilibrium for reversible
reactions; they simply make both forward and _ _ _ _ _ _ _ reactions happen faster so that
_ _ _ _ _ _ _ _ _ _ _ is reached more quickly. Scientists have opted to use iron for _ _ _ _ _; it does not
_ _ _ _ _ _ _ the yield of ammonia at all.
Day 5 : Lab pg 514 “Le Chatelier’s Principle
Independent Research :
19
Analyse the optimal conditions for a specific chemical process related to the principles of equilibrium that
takes place in nature or is used in industry (e.g., the production of sulphuric acid, electrolyte balance in the
human body,sedimentation in water systems) Sample issue: The principle of dynamic equilibrium is used
in industrial processes to maximize the concentration of products and minimize leftover reactants.
Industrial chemists determine ideal pressure and temperature conditions, and proper catalysts, so that
fewer materials and less energy are used.
Questions:
Why are low temperature conditions not used with exothermic reactions?
________________________________________________________________________________________
________________________________________________________________________________________
________________________________________________________________________________________
________________________________________________________________________________________
________________________________________________________________________________________
________________________________________________________________________________________
________________________________________________________________________________________
How do chemicals dissolved in human blood helpmaintain a blood pH level between 7.2 and 7.4?
________________________________________________________________________________________
________________________________________________________________________________________
________________________________________________________________________________________
________________________________________________________________________________________
________________________________________________________________________________________
________________________________________________________________________________________
________________________________________________________________________________________
20
Review Exercises
In the problems below, for each change given in the first column of the table, use Le Chatelier's principle to
predict
• the direction of shift of the equilibrium.
• the effect on the quantity in the third column.
1. For the following reaction
5 CO(g) + I2O5(s)  I2(g) + 5 CO2(g)
Ho = -1175 kJ
for each change listed, predict the equilibrium shift and the effect on the indicated quantity.
Direction
of Shift
( ; 
no
change)
Change
(a)
(b)
(c)
(d)
(e)
decrease in volume
raise temperature
addition of I2O5(s)
addition of CO2(g)
removal of I2(g)
Effect on
Quantity
Effect
(increase, decrease,
or no change)
Ke
amount of CO(g)
amount of CO(g)
amount of I2O5(s)
amount of CO2(g)
2. Consider the following equilibrium system in a closed container:
Ni(s) + 4 CO(g)  Ni(CO)4(g)
Ho = - 161 kJ
In which direction will the equilibrium shift in response to each change, and what will be the effect on the
indicated quantity?
Direction
Effect on
Effect
Change
of Shift
Quantity
(increase, decrease,
( 
no
or no change)
change)
(a) add Ni(s)
Ni(CO)4(g)
(b) raise temperature
Ke
(c) add CO(g)
amount of Ni(s)
(d) remove Ni(CO)4(g)
CO(g)
(e) decrease in volume
Ni(CO)4(g)
(f) lower temperature
CO(g)
(g) remove CO(g)
Ke
Le Chatelier's Principle Exercises
For each of the following indicate the direction the equilibrium would shift and what would happen to the
concentrations of each substance in the equilibrium
1. The following equilibrium reaction may be established with carbon dioxide and steam
CO (g) + H2O (g)

<
CO2 (g) + H2 (g) + heat
What would be the effect of each of the following on the equilibrium:
(a) The addition of more H20?
(b) The removal of some H2?
(c) Raising the temperature ?
21
(d)
Increasing the pressure?
(e) Addition of a catalyst?
2. What would be the effect on each of the following on the equilibrium involving the synthesis of
methanol?
CO (g) + 2 H2 (g) < CH3OH (g)
(a) The removal of CH3OH .
(b) an increase in pressure
(c) lowering the concentration of H2.
(d) the addition of a catalyst
3. A small percentage of nitrogen gas and oxygen gas in the air combine at the high temperatures found
in automobile engines to produce NO (g) which is air pollutant.
N2 (g) +
O2 (g)
+ Heat

<
2 NO (g)
(a) Higher engine temperatures are used to minimize carbon monoxide production. What effect do these
higher engine temperatures have on the production of NO (g)? Why?
(b) What effect would high pressures have on the production of NO? Why?
4. What would be the effect of each of the following on the equilibrium involving the reaction of coke, C
(s), with steam to give CO and H2?
C (s) + H2O (g)

<
CO (g) + H2 (g)
(a) The addition of steam?
(b) An increase in pressure?
(c) The removal of H2 as it is produced?
22
(d) The addition of a catalyst?
5. The binding of oxygen by hemoglobin (abbreviated Hb), giving oxyhemoglobin (HbO2) is partially
regulated by the concentration of H+ and CO2 in the blood. Although the equilibrium is rather complicated
it can be summarized as
HbO2 + H+ + CO2 <
CO2HbH+ + O2 .
According to Le Chatelier's principle what would be the effect of each of the following on the equilibrium
(a) The production of lactic acid (contains H+) and CO2 in a muscle during exercise.
(b) Inhaling fresh oxygen enriched air
Day 6 : Equilibrium Problems (pg 463) :
A.
Reaction Quotient (Q) Which way will the equilibrium shift ? If Q is < Ke equ. will shift to the
________________; If Q > Ke equ. will shift to the __________________
Consider the equ. N2(g) + 3 H2(g) 
2 NH3(g) Ke = 0.064 . The concentrations are as follows
[ N2 ] = 0.20 mol/L
[H2] = 0.085
[NH3] = 0.80 mol/L
i.
Calculate trial Ke or Q =
ii.
Compare Q to Ke _______________________ ; equilibrium should shift to the ___________
Try pg 465 (1,2)
B.
Given Ke determine equ. conc.
1.
Consider the equ. 2 A(s) + 3 B(g) 2 C (g) + D (s)
Ke = 1.0 x 10-2. If at equ. the conc. of B was found to be 0.20 mol/L; What is the equ. conc. of C?
Let equilibrium concentration of c = x
2 A(s)
[equ.]
+ 3 B(g)  2 C (g) + D (s)
_____
_______
Ke =
[C] =
Practice pg 466 (3,4)
C.
Given initial conc. And Ke find equ conc. ( pg 465 )
23
Consider the equ. A(g) + B(g)  2 C (g)
Ke = 25.0 If 0.50 mol of A and 0.50 mol of B are placed in a 5.0 L container; calculate the equ. conc.
of each substance.
[I]
A(g) + B(g) 
_____
_____
[C]
_____
______
[E]
_____
_____
2 C (g)
_______
______
_______
Since there is no C equ. shifts to the right . Let x = [A]reacting
To simplify take the square root of each side
Ke =
x=
[A] =
3.
[B] =
[C] =
Consider the equ.
A(g)
+ B(g) 
2 C (g)
Ke = 27
3.0 moles of A, B and C are placed in a 2.00 L container. Calculate the equ conc. of each.
A(g)
+ B(g) 
2 C (g)
[I]
_____
_____
______
[C]
_____
_____
______
[E]
_____
_____
i.
______
Calculate Q to see which way the equ. Will towards
ii.
iii.
Let x =[change in concentration of substance reacting]. Fill in chart above.
Ke =
iv.
x=
[A] = ____ [B] = _____ [C] = ______
Complete pg 472 (5,6)
d. Given initial conc. and Ke find equ conc. ( imperfect square ; simplifying assumption
24
required.

Consider the equilibrium : 2H2S(g)
2H2(g)
+
S2(g) Ke = 4.20x10-6
0.200 mol of H2S is placed in a 1.00 L vessel. When equilibrium is established what is the concentration of each ?
2H2S(g)

2H2(g)
+
S2(g)
[i]
[c]
[e]
______
______
______
______
______
______
_______
_______
_______
Let change in concentration of H2S be ____________. Fill in the rest of the table,
If Ke is very small assume 0.200 – 2x is  0.20 . No quadratic is needed. You must do the
following check 2x/.200 x 100% must be less than 5%
Ke =
X = ___________. Do check.
[H2S] _________
[H2 ] = __________ [S2]
Practice pg 476 # 8
Day 7
e.
Same as above but Quadratic is needed (pg 476)
1.
Given the equ. N2O4(g)  2 NO2(g) Ke = 6.7 x 10-3
0.060 mole of NO2 and 0.85 mole of N2O4 are placed in a 1.00 L vessel. When equ. Is established;
What are the equ. conc. of NO2 and N2O4?
Step 1
Calculate a trial Ke(Q) to see which way the equ. will shift
Step 2
N2O4  2 NO2
Let x =
[initial]
[change]
____
____
Quadratic is needed
[final]
____
_____
_____
_____
Ke =
X= _________________
Complete page 480 (10) and pg 481(1,2,3,6,8)
[N2O4] = __________
[NO2] = ___________
25
26
Equilibrium Review
1.
A 1.0 L reaction vessel contained 0.750 mol of CO (g) and 0.275 mol of H 2O (g). After one hour,
equilibrium was reached according to the equation: CO (g) + H 2O (g)  CO2 (g) + H2 (g). Analysis
showed that 0.25 mol of CO2 was present. Calculate Kc. ( 5 )
2.
Consider the equilibrium: 3 I2 (g) + 6 F2 (g)  2 IF5 (g) + I4F2 (g). At a certain temperature, 6.0
mol of IF5 and 8.0 mol of I4F2 are introduced into a 5.0 L container. At equilibrium, 6.0 mol of I4F2 are
left. Calculate Kc. ( 5.8 x 10-4 )
3.
At a certain temperature, Kc = 4 for the equilibrium: 2 HF (g)  H2 (g) + F2 (g). Predict the
direction (if at all) in which the equilibrium will shift when the following amounts of gases are
introduced into a 5.0 L vessel.
a)
b)
c)
3.0 mol HF, 2.0 mol H2, 4.0 mol F2 ( shift right )
0.2 mol HF, 0.5 mol H2, 0.6 mol F2 ( shift left )
0.3 mol HF, 1.8 mol H2, 0.2 mol F2 ( no shift )
4.
For the equilibrium: Br2 (g) + Cl2 (g)  2 BrCl (g), Kc = 7.0 at 400 K. If 0.080 mol of Br2(g) and
0.060 mol of Cl2 (g) are introduced into a 2.0 L vessel at 400 K, determine the equilibrium
concentrations of all three gases. ( Br2=0.02M, Cl2=0.01M, BrCl=0.04M )
QUADRATIC
EQUATION
5.
At 425 oC, Kc = 1.82 x 10-2 for the equilibrium: 2 HI (g)  H2 (g) + I2 (g). If equilibrium is reached
by adding only HI (g) to a 1.0 L reaction vessel, determine:
a)
b)
c)
6.
Consider the equilibrium: CO (g) + H2O (g)  CO2 (g) + H2 (g). A mixture containing 1.00 mol of
CO (g) and 1.00 mol of H2O (g) is placed in a 10.00 L container at 800 K. At equilibrium, 0.665 mol of
CO2 and 0.665 mol of H2 are present. Determine:
a)
b)
7.
8.
the concentrations of H2 (g) and I2 (g) in equilibrium with 0.0100 mol/L HI (g);
(H2=I2=1.35x10-3)
the initial concentration of HI (g) placed in the vessel; ( 0.013 mol/L )
the percent of the original HI (g) that dissociated to reach equilibrium. ( 21%)
the equilibrium concentrations of all four gases; ( CO=H2O= 0.0335M, CO2=H2=0.0665M
)
the value of Kc at 800 K. ( 3.9 )
At 585 K, NOCl (g) is 56.4 % dissociated to reach equilibrium according to the system:
2 NOCl (g)  2 NO (g) + Cl2 (g). If 1.00 mol of NOCl (g) was placed in a 1.00 L vessel, determine:
a)
the equilibrium concentrations of all three gases;
( .436mol/L,NO=0.564mol/L,Cl2=0.282mol/L)
b)
the value of Kc at 585 K. ( 0.47 )
The reaction: A (g) + B (g)  C (g) is exothermic as written. Assume you have an equilibrium mixture of
A, B, and C. How would the equilibrium concentration of C (g) change with:
a)
b)
c)
d)
e)
an increase in temperature; ( C decreases )
an increase in pressure due to a decrease in volume; ( C increases )
the addition of A (g); ( C increases )
the addition of a catalyst; ( C no effect )
the removal of B (g). ( C decreases )
9.
11.
27
For each of the following equilibrium systems:
i)
ii)
write the expression for Kc;
state the direction in which each would be shifted upon the application of the stress listed after each
equation.
a)
b)
c)
2 SO2 (g) + O2 (g)  2 SO3 (g) + energy
C (s) + CO2 (g) + energy  2 CO (g)
N2O4 (g)  2 NO2 (g)
d)
CO (g) + H2O (g)  CO2 (g) + H2(g)
e)
2 NOBr (g)  2 NO (g) + Br2 (g)
f)
g)
h)
i)
3 Fe (s) + 4 H2O (g)  Fe3O4 (s) + 4 H2 (g)
2 SO2 (g) + O2 (g)  2 SO3 (g)
CaCO3 (s)  CaO (s) + CO2 (g)
N2 (g) + 3 H2 (g)  2 NH3 (g)
decrease temperature
increase temperature
increase pressure by
decreasing volume
decrease pressure by
increasing volume
decrease pressure by
increasing volume
add Fe (s)
add a catalyst
remove CO2 (g)
add H2 (g)
( shift R )
( shift R )
( shift L)
( no shift)
( shift R )
( no shift)
( no shift)
( shift R )
( shift R )
Consider the following equilibria which occur simultaneously in the same solution in the presence of
undissolved CA (s):
1.
2.
A1- (aq) + H1+ (aq)  B1- (aq) + H2O (l)
CA (s)  A1- (aq) + C1+ (aq)
Use Le Châtelier's Principle to predict each of the following changes in:
a)
b)
c)
d)
the amount of undissolved CA (s) if the concentration of H1+ (aq) is increased; (decrease)
the concentration of B1- (aq) if the concentration of C1+ (aq) is decreased; (increase)
the concentration of H1+ (aq) if the concentration of C1+ (aq) is increased; (increase)
the concentration of B1- (aq) if more solid CA (s) is added. ( no effect )
28
Day 8 Solubility Product Constant (pg 482)
Ksp is defined as the product of molar concentrations of ions in a saturated solution, each raised to the
appropriate power.
The following table is the Solubility Product Constant for various salts at 250C:
Salt
Ksp
Salt
Ksp
Salt
Ksp
AgCH3COO
1.9 x 10-3
AlPO4
9.8 x 10-21
NiCO3
6.6 x 10-9
Ag2CO3
8.4 x 10-12
BaCO3
2.6 x 10-9
PbSO4
1.8 x 10-8
AgCl
1.8 x 10-10
BaSO4
Zn3(PO4)2
9.1 x 10-33
AgI
8.5 x 10-17
Fe(OH)2
4.9 x 10-17
Ag2SO4
1.2 x 10-5
Al(OH)3
1.9 x 10-33
FeS
1.6 x 10-19
1.1 x 10-10
In general, the lower the Ksp value, the more insoluble the salt will be (i.e attraction between oppositely
charged ions is very strong).
e.g Ag2CO3 is much more ___________ than Zn3(PO4)2. It is appropriate to compare Ksp values only if
the charge ratio of the cations and anions are the same.
Write Ksp expressions for the following salts:
1. PbCl2

Ksp = [Pb2+][Cl-]2
2. CaCO3

Ksp =
3. Ca3(PO4)2

Ksp =
4. Al2(CO3)3

Ksp =
5. Fe(OH)3

Ksp =
6. CuS

Ksp =
7. BaSO4

Ksp =
8. Al2S3

Ksp =
29
Problems : Calculating solubility in moles/L and g/L (pg 484)
Example 1:
Ksp = 8.30 x 10-9 for PbI2. What is the molar solubility of PbI2. (i.e )[Pb2+] in a saturated solution of PbI2?
PbI2(s)

Pb2+
0
+x
x
[initial]
[]
[final]
+
2I0
+2x
2x
Step 1. Let molar solubility = x = [Pb2+]
Step 2. Setup Ksp expression:
ksp = [Pb2+][I-]2
x =
Example 2:
Ksp for Mg(OH)2 is 8.9 x 10-12. What is the molar solubility of Mg(OH)2 in moles/L and g/L ?
Let molar solubility of Mg(OH)2 = x = [Mg2+]
Mg(OH)2

[initial]
[]
[final]
Mg2+
____
____
____
+
2OH_____
_____
_____
Complete :
Example 3:
Ksp for AgCl is 1.1 x 10-10. Calculate solubility in moles /L and grams/Litre.
Step 1. Setup Ksp expression.
Step 2. Calculate molar solubility (mol/L).
Step 3. Calculate solubility in g/L.
Complete :
Determine the solubility of AgI, Ag2CrO4, and Ca3(PO4)2 in
a) moles/L and b) g/L at 25oC. You will have to refer to the Ksp tables in your textbook.
Day 9
Problems : Calculating Ksp given solubility
30
Example 1:
At 200C, it is found that 1.94 x 10-4 g of AgCl saturates 100 ml of water. If the molar mass of AgCl is
143.4 g/mol, calculate Ksp.
Review:
concentration
=
# moles of solution/ Volume of solution
C= ______ = __________
Step 1. Find molar solubility in mol/L (i.e concentration)
c= n =
m
=
1.94 x 10-4 g
= 1.35 x 10-5 mol/L
----- ----------------------------------v
MxV
143.4 g/mol x 0.10 L
Step 2. Write dissociation equation:
AgCl(s)  Ag+(aq) + Cl-(aq)
H2O
Step 3. Calculate [Ag+] and [Cl-]
[Ag+] = 1 mol Ag+ x 1.35 x 10-5 mol/L AgCl = 1.35 x 10-5 mol/L = [Cl-]
-----------1 mol AgCl
Step 4. Set up Ksp expression
Ksp = [Ag+][Cl-] = (1.35 x 10-5)(1.35 x 10-5) = 1.82 x 10-10
Alternate solution : when you calculate the molar solubility that is your x value; setup the Ksp expression and
substitute in for x.
X =1.35 x 10-5 mol/L Ksp = (x)(x) = (1.35 x 10-5)(1.35 x 10-5) = 1.82 x 10-10
Example 2:
The solubility of CaF2 at 250C is 1.70 x 10-3 g
----100 ml H2O
. Calculate Ksp.
Step 1. [CaF2] =
Step 2. CaF2  _____________ +
H2O
Step 3. [Ca+2] =
[F-]
Step 4. ksp =
=
=
________________
31
Example 3:
A saturated solution of Ag2CO3 can be made by dissolving 1.27 x 10-4 moles of solid Ag2CO3 in 1.00 L
of H2O. What is the Ksp for Ag2CO3 ? (Ans: 8.19 x 10-12)
Ag2CO3(s)  2Ag+ + CO3-2
1.
[Ag2CO3] =
2.
[Ag+]
=
[CO3-2] =
3.
Ksp =
Complete pg 486 (1,3,4b)
Applications : Sparingly Soluble Solutes in Animals:
- proteins and amino acids digested – nitrogenous wastes
-
1. ammonia – water-dwellers – very soluble in water so diffuses into water, very toxic but [ ammonia ] is low
-
2. urea NH2 CONH2 - terrestrial animals with no shells – soluble, not very toxic, removed from blood by
kidneys – active transport!
-
3. uric acid – terrestrial animals with egg-laying reproduction – almost insoluble and solution is more toxic than
urea – ppt out so can’t harm the organism, even the developing embryo
32
ASSIGNMENT: Ksp PROBLEMS
1.
The solubility of PbSO4 in water at 25C is 0.035 g/L. Calculate Ksp for PbSO4. ( 1.3 x 10 -8 )
2.
Mg(OH)2 has a solubility of 7.05 x 10-3 g/L. Calculate Ksp of Mg(OH)2. ( 7 x 10 -12 )
3.
The solubility of Ag2CrO4 is 1.32 x 10
4.
Ksp of NiCO3 is 1.4 x 10-7. Calculate the molar solubility of NiCO3. ( 3.7 x 10 -4 mol/L )
5.
Given the Ksp for Ag2CrO4 is 1.9 x 10-12, what is the concentration in mol/L of Ag+ ions in a saturated
solution of Ag2CrO4? ( 1.6 x 10 -4 mol/L )
6.
The solubility of BaF2 is 1.30 g/L. Calculate its Ksp. ( 1.62 x 10 -6 )
7.
The formula of a sparingly soluble salt A3B2. The molar mass of A is 40.0 g/mol and B is 30.0 g/mol. If
2.00 g of A3B2 saturates 250 mL of solution, calculate the Ksp for A3B2.
( 1.9 x 10 -5 )
8.
The Ksp of Ca3(PO4)2 is 2.0 x 10
( 4.4 x 10 -4 g )
Read pages 482 -483
-4
mol/L. Calculate Ksp for Ag2CrO4. ( 9.0 x 10 -12 )
. How many grams of Ca3(PO4)2 is required to saturate 2.0 L of solution?
-29
-
Day 10 PROBLEMS: Predicting Precipitate Formation ( pg. 487 )
consider the equilibrium: CaCO3(s) 
Ca2+ + CO3-2
for a saturated solution the ion product: [Ca2+] x [CO3-2] is exactly equal to ksp.
precipitate will form if ion product > Ksp. (supersaturated solution)
no precipiate will form if:
ion product = Ksp. (Saturated solution)
ion product < Ksp. (Unsaturated solution)
33
Example 1: Predicting Precipitate Formation
When the product of the concentration of the ions exceed the value of Ksp they cannot exist in equilibrium
anymore. They will form a precipitate in order to reduce the concentrations of the ions in solution back to
the equilibrium value.
To determine whether the Ksp is exceeded, just substitute the ion concentrations into an expression similar
to the Ksp expression and determine trial Ksp(Q)
If the trial Ksp. exceeds the Ksp actual then a precipitate forms. If the Ksp trial is larger than the values of K sp
then the ions are above the saturation line and the ions are in a super-saturated situation. The ions will
precipitate until their individual concentrations multiplied together equal the value of K sp.
If the trial ksp. does not exceed the value of Ksp then the ions are in an unsaturated situation, which means
there aren't enough ions dissolved to form a precipitate.
If the trial ksp equals the value of Ksp then the solution is saturated. The solution can hold no more
dissolved ions without a precipitate forming.
Problem:
Does a ppt. of AgCl form when 1 mL of 0.1 mol/L AgNO3 is added to a beaker containing
1 L of tap water with a Cl- ion concentration of 1.0 x 10-5 mol/L?
[Cl-]f =______________
[AgNO3] f = ________________
[Ag+] =
Ksp trial = Q = [Ag+][Cl-]
=
The Ksp for AgCl is 1.8 x 10-10 therefore Ksp trial ____________ Ksp actual so_______ ppt. forms.
Ex .2 Does a ppt of AgI form when 10 mL of 0.1 mol/L AgNO3 gets added to 90 mL of a solution
containing 1.0 x 10-10 mol/L of KI? The Ksp for AgI is 8.5x 10-17
[AgNO3] =
[Ag+] =
[KI] =
[I-] =
Q=
= ___________
________________________________________________________________________________
Example 3:
1.0 ml of 0.02 mol/L of Pb(NO3)2 and 1.0 ml of 0.0040 mol/L of AlCl3 are mixed. Will a
precipitate of PbCl2form? [ksp(actual) for PbCl2 is 1.2 x 10-5]
34
Example 4:
If 50.0 ml of 0.0010 mol/L CaCl2 solution was added to 50.0 ml of 0.010 mol/L Al2(SO4)3, would a
precipitate of CaSO4 form? (ksp CaSO4 = 2.4 x 10-5)
Step 1.
i) [CaCl2]final =
ii) [Ca2+]
=
iii) [Al2(SO4)3] =
iv) [SO42-]
=
Step 2. Q =
Step 3. Ksp(trial) is
Complete pg 489 (6)
pg 493 (5,6,7,8,10,12)
Ksp(actual) ___________________
35
Review Problems
1.
The Ksp for CdS is 6.0 x 10-27. Calculate the molar solubility in mol/L
[ 7.8 x 10-14 mol/L ]
2.
The Ksp for Mg(OH)2 is 1.0 x 10-11. Calculate the molar solubility. [ 1.4 x 10-4 mol/L ]
3.
The Ksp for PbCO3 is 1.0 x 10-13. Calculate the solubility in g/L. [ 8.4 x 10-5 g/L ]
4.
The molar solubility of Ag2CO3 is 1.27 x 10-4 mol/L. Calculate the Ksp. [ 8.19 x 10-12 ]
5.
1.49g of AgBrO3 dissolves in 1L of water. Calculate the Ksp. [ 3.99 x 10-5 ]
6.
The salt A3B2 has a solubility of 2.5 x 10-8 mol/L. Calculate the Ksp. [ 1.1 x 10-36 ]
7.
Does a precipitate form when 20.0 ml of 1.0 x 10-4 mol/L Ba(OH)2 is added to 60.0 ml of 2.4 x 10-5
mol/L Zn(NO3)2. The Ksp for Zn(OH)2 = 1.8 x 10-14 [ Ksp(expt) = 4.5 x 10-14, ppt forms ]
8.
Does a precipitate form when 50.0 ml of 0.0020 mol/L Ca(NO3)2 is added to 100.0 ml of 0.050 mol/L
Na3PO4. Ksp for Ca3(PO4)2 = 2.1 x 10-33. [ Ksp(expt) = 3.33 x 10-13, ppt. forms ]
36
Chemical Energy and Equilibrium (pg 494)
Define :Entropy (S) –
Entropy increases :
1.
2.
3.
moles of products in gas state is ______________ than moles of reactants
temperature ________________________
change of state from ______________ to ________________ to ____________________
Define :Entropy Change (S) –
-
chemical reactions have two natural drives one toward _______________ enthalpy and one towards
_______________ entropy. If both drives are met in one direction the reaction will be ______________i.e.
tends to occur on its own.
e.g.
Mg(s) + 2 HCl (aq) --> MgCl2 (aq)
-
the forward reaction proceeds toward ____________ entropy and _________ enthalpy thus this is a
_________________________ reaction.
-
if one drive can be met in one direction and other drive in the other direction the reaction will be in
_____________________________
e.g.
-
N2O4 (g)
+ H2 (g) + energy
 2 NO2(g)
the following system is at equilibrium since the drives are met in different directions. Max entropy favors the
_________________ side and max entropy favors the ______________ side. Thus equilibrium represents a
compromise between the drive towards max entropy and the drive toward min enthalpy.
Complete pg 498 (1)
Define : 2nd Law of Thermodymanics - nature is always proceeding to a state of higher entropy.
Determine whether the following reactions show an increase or decrease in entropy.
1.) 2 KClO3 (s) --> 2 KCl (s) + 3 O2 (g)
2.) H2O (l) --> H2O (s)
3.) N2 (g) + 3 H2 (g) --> 2 NH3 (g)
4.) NaCl (s) --> Na+1 (aq) + Cl-1 (aq)
5.) KCl (s) --> KCl (l)
6.) CO2 (s) --> CO2 (g)
7.) H+1 (aq) + C2H3O2-1 (aq) --> HC2H3O2 (l)
8.) C (s) + O2 (g) --> CO2 (g)
9.) H2 (g) + Cl2 (g) --> 2 HCl (g)
37
10.) Ag+1 (aq) + Cl-1 (aq) --> AgCl (s)
11.) 2 N2O5 (g) --> 4 NO2 (g) + O2 (g)
12.) 2 Al (s) + 3 I2 (s) --> 2 AlI3 (s)
13.) H+1 (aq) + OH-1 (aq) --> H2O (l)
14.) 2 NO (g) --> N2 (g) + O2 (g)
15.) H2O (g) --> H2O (l)
Enthalpy & Entropy
For each of these processes, predict if Entropy increases or decreases.
⇄
2H2O(g)
⇄
2SO2(g)
+
+
Cl-(aq)
1.
2H2(g) +
O2(g)
2.
2SO3(g)
3.
Ag+(aq)
4.
Cl2(g)
⇄
2Cl(g)
5.
H2O(l) ⇄
H2O(g)
6.
CaCO3(s)
+
7.
I2(s)
608 kJ ⇄
8.
4Fe(s)
+
+
O2(g)
⇄ AgCl(s)
180 kJ ⇄ CaO(s)
3O2(g) ⇄
+
CO2(g)
I2(aq)
2Fe2O3(s)
+
1570 kJ
Consider both Enthalpy and Entropy and determine if each reaction will
a) go to completion
b) not occur or
c) go to equilibrium
9.
H2O(l) ⇄
10.
CaCO3(s)
11.
I2(s)
H2O(g)
+
180 kJ ⇄ CaO(s)
608 kJ ⇄
+
H = 150 kJ
4Fe(s) + 3O2(g) ⇄ 2Fe2O3(s)
13.
Cl2(g)
14.
Ag+(aq) + Cl-(aq) ⇄ AgCl(s) +
2Cl(g)
CO2(g)
I2(aq)
12.
⇄
+
∆H = -1570 kJ
H = +26.8 kJ
86.2 kJ
Consider both Enthalpy and Entropy and determine if each reaction will
38
a) have a large Keq
b) have a small Keq
c) have a Keq about equal to 1
15.
H2SO4(aq)
+ Zn(s)
NH4NO3(s) ⇄
16.
NH4+(aq)
N2(g) + 3H2(g) ⇄
17.
H2O(l) + 150 kJ ⇄
18.
19.
Ca(s)
⇄ ZnSO4(aq)
+
+ H2(g)
H = -207 kJ
H = +30 kJ
NO3-(aq)
2NH3(g) + 92 kJ
H2O(g)
+ 2 H2O(l) ⇄ Ca(OH)2(aq)
+
H2(g)
H =
-210 kJ
Day 11 Experiment Ksp of Calcium oxalate (pg 517)
Independent research :
-
Assess the impact of chemical equilibrium processes on various biological, and technological systems ( e.g
remediation in areas of heavy metal contamination, development of gallstones, use of buffering in medications,
use of barium sulphate in medical diagnosis. Heavy metals such as copper, lead and zinc can accumulate to toxic
levels in the human body. A process called chelation which causes a chemical reaction involving an equilibrium
shift, removes the metals from the body before permanent organ damage occurs. Sample questions :
1.
Why are headache tablets buffered?
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
2.
Why is barium sulphate safe to use for X-rays of the digestive system even though barium ions are
poisonous.
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
3.
How do kidney stones form?
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
4.
Read article no sweat pg 457. summarize what you learned
39
__________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
-
Complete pg 522 ( 1-18) ( omit 7,8 )pg 523 (1,2,3, 5,6,10,11,12,13,14,15a,16,17,18,19,20,24,25)
40
Day 12 ACID AND BASES (ionic equilibrium)(pg.528)
Arrhenius Concept :
Acids : produce ___________ ions in solution
Bases : produce ___________ ions in solution
Bronsted - Lowry Theory (S.H. Bronsted and T.M. Lowry - 1923)
ACID - a substance that can donate a _____________ [Hydrogen ion (H+)] to a base.
BASE - a substance that can accept a _______________ from an acid.
e.g. HCl
+
NH3

NH4+
+
ACID
BASE
Conjugate Acid
(proton donor) (proton acceptor)
ClConjugate Base
Conjugate Base - remove a _______from the acid and add a _________ charge.
Conjugate Acid - add a ___________ to the base and add a ___________ charge.
Amphoteric substances :
-
can act as both acids or bases depending on the conditions :
HCl (g)
+
H2O(l) --------> ____________________ + _______________________
_____________
NH3(g)
+
H2O(l)  ____________________ + _________________________
______________
HS-(g)
+ H2O(l) 
___________
S-2
+
+
H2O(l) 
HS______________
S-2
+
H3O+
OH-
Exercise 1
Complete the following acid - base equilibriums : identify acids, bases and conjugate pairs.
a.
b.
c.
d.
HNO2(g)
____________
HF(aq)
___________
+
NH3(l)
________
+
HSO4- (aq)
_____________
NH3(aq)
_____________
Cl- (aq)
_______
 ___________________ + ___________________
____________________ ___________________
 ___________________ + ____________________
__________________
____________________
+
CN- (aq)

__________
+
HF(aq)

__________
HCN
____________
+
______________ +
______________
_____________________
____________________
____________________
____________________
Neutralization Reactions :
Acid
+ Base
--------> Salt
+
Water
HCl(aq)
+ NaOH(aq) --------> _______________ + H2O(l)
H3O+ + _____ + ______ + ______ ------> ________ + _______ + 2 H2O(l)
Net reaction :
_____________
+ ______________ ------> __________________
In pure water : [ H3O+ ] = [ OH- ]
Let x = [ H3O+ ] = [ OH- ]
Ke = Kw = [ H3O+ ] x [ OH- ] = 1.0 x 10-14
41
Kw = (x)(x) = 1.0 x 10-14
x = ______________
-
H2O is a weak electrolyte -----> poor conductor of electricity due to low concentration of ions in solution
-
Consider the equ. : 2 H2O + energy  H3O+ + OH-
-
As the temperature increases equilibrium shifts to the _______________ and the [ H + ] and [ OH- ]
________; thus Kw ____________ e.g. at 50o C Kw = 5.5 x 10-14 [ H+ ] and [OH-] both equal
__________________; the solution will be ________________ ( acidic, basic or neutral )
or H2O + energy 
H+ + OH-
Complete pg 532 (1)
The pH Scale (pg.540)
Every aqueous solution is either acidic, basic or neutral. There is a quantitative relationship between the
concentration of hydronium and hydroxide ions in the solution.
Neutral solution [H3O+] = [OH-]
Acid solution [H3O+] > [OH-]
Basic solution [H3O+] < [OH-]
The brackets as usual denote molar concentrations.
The pH scale is a numerical scale which, for most applications extends from 0 through to 14. The numbers
on the scale represent the relative acidity of solutions and can be converted into actual hydronium ion
concentrations.
The pH scale is based on the self-ionization of pure water.
Two water molecules will sometimes combine into hydronium and hydroxide ions.
H2O + H2O <=====> H3O+ + OH-
Pure water is considered to neutral and the hydronium ion concentration is 1.0 x 10-7 mol/L which is equal
to the hydroxide ion concentration.
ie.
[H3O+] = [OH-] = 1.0 x 10-7mol/L
The scale reaches a maximum at 14. Please note again that the hydronium and hydroxide concentrations
multiply out to 10-14 M. The pH scale was derived around this relationship:
ie.
[H3O+] = 10-pH mol/L. So the pH is the -log of the [hydronium ion]. See page 541 for calculator tips.
Strong acids and strong Bases
-
strong acids are completely ionized in water e.g HCl
Ex.1
What is the pH of 0.15 mol/L HCl
What is [OH-] ?
[H+] =.15
[i]
[c]
[e[
HCl ----- H+
+ Cl0.15
0
0
-.15
+.15
+.15
0
+.15
+.15
pH = -log10[H+] = -log100.15 = ____________
[OH-] = Kw/[H+] = ____________ = ______________
Ex. 2
200.0 mL of 0.10 mol/L H2SO4 is diluted to 1.0 L. Calculate pH and [OH-]
Ex. 3
The pH of a 100.0 mL solution of HNO3 is 2.2. Calculate the [HNO3]
42
[H+] = 10-pH = 10-2.2 = ____________
[HNO3] = __________
Strong Bases - completely dissociated in solution
H2O
e.g.
NaOH
------ Na+
+
OHThe pOH scale is the corollary of the pH scale. ie. pH + pOH = 14
Thus a solution that has a pH = 7 must also have a pOH = 7.
a.
What is the pOH of a 0.010 mol/L NaOH solution?
[OH-] = _____________
b.
Ex. 2
pOH = -log10[OH-] = ______________ = ___________________
What is it's pH? _________________
25.0 g of Ba(OH)2 in 500.0 mL of water. Calculate pOH, pH and [OH-]
i.
[Ba(OH)2] = _______________ = _____________________
ii.
[OH-] = ________________
iii.
pOH = ________________ = _____________________
iv.
pH = __________________ = ____________________
v.
[H+] = ______________ = _________________________
Complete pg 540 ( 8,10) pg 546 (12c,13,14) pg 549(18,19)
43
Day 13 Weak Monoprotic Acids(pg 551)
-
partially ionized in solution
The general equation for a weak acid is:
HX + H2O  H3O+ + XApplying the equilibrium law to this equation we get:
Ex. 1
Ka = [H3O+ ][ X- ]
-------------- =
[ HX ]
[H+][X-]
--------------[HX]
(a)
Calculate the [H+], (b) the pH and (c) the % ionization for a 0.100 mol/L solution of acetic acid at
25oC, Ka for CH3COOH is 1.8 x 10-5.
The equation for the dissociation is:
CH3COOH  H+(aq) + CH3 COO-(aq)
Let 'x' be the concentration of acetic acid that dissociates:
CH3COOH 
H+(aq) + CH3 COO-(aq)
[i]
_______
______
_______
[c]
_______
______
_______
[e]
_______
______
_______
Ka =
[H+] = __________________
x= ______________
pH = ______________
The % ionization is __________ x100% = ____________%
0.100 mol/L
Note x can be neglected if the % ionization is less than 5%
Ex. 2
Given the equation : HF 
H+ +
F- . Calculate the Ka and % ionization if 0.15mol/L HF has a pH
of 2.00
HF 
H+ +
F[i]
_____
_____
______
[c]
[e]
_____
_____
_____
_____
______
______
Step 1 Calculate [H+]
Step 2 Complete chart
Step 3 Set up Ka expression and solve for Ka
Step 4 Calculate % ionization
Complete page 554 (1) pg 556(3) pg 568 (7) pg 570(10)
44
Weak Bases ( pg 557 )
Similar calculations are used for weak bases.
eg.
NH3(g) + H2O(l) <---------> NH4+(aq) + OH-(aq)
Kb = _____________________
The dissociation constant for any of the conjugate bases can be obtained from the value of K w divided by
the appropriate value of Ka. (Ka can also be obtained by dividing Kw by Kb).
eg. NH3 Kw = 1.0 x 10-14
Ka for NH4+ = 5.6 x 10-10
Kb = 1.0 x 10-14 = 1.8 x 10-5
5.6 x 10-10
1.
What is the [OH-] of a 0.10 mol/L solution of NaCN? Calculate pH. Ka HCN = 6.2 x 10-10
When NaCN dissolves in water it dissolves completely because it is a sodium salt.
NaCN
--------> Na+ + CN0.1mol/L
0.1mol/L 0.1mol/L
However the CN- ions that are produced then react with water and set up an equilibrium.
CN-(aq) + H2O <--------> HCN(aq) + OH_______
_______
______
[i]
[c]
_______
_______
______
[e]
_______
_______
______
Kb = _____________________
X = ____________________ = [OH]
Ex. 2
Calculate the pH of a 0.10 mol/L NaCH3COO solution? (Ka CH3COOH = 1.8 x 10-5)
NaCH3COO
CH3COO-
-------> Na+
+
H2O

+ CH3COOCH3COOH
+
Complete pg 574(13) pg 579(3,5,6,13,16,17,18)
OH-
45
Assignment Ka and Kb
1.
The [ H + ] of a 0.10 M solution of a weak acid HF is found to be 8.2 x 10-3 ml/L.
HF + H2O 
H3O+ + F-
a)
What is the [ F- ]?
( 8.2 x 10-3 mol/L )
b)
Write the Ka expression.
c)
Calculate the Ka and % ionization.
[Ka = [ H3O+ ] [ F- ] ]
-------------[ HF ]
( 6.7 x 10-4 , 8.2% )
2.
The initial concentration of an acid HA is 0.0010 mol/L. The pH is 4.0. Calculate the Ka.
( 1 x 10-5 )
3.
0.010 moles of an acid HA is dissolved in a volume of water giving 1000.0 mL of solution. The pH of the
solution was 4.0. Determine the Ka ( 1.0 x 10-6 )
4.
What is the [ H+ ] in a 0.10 mol/L CH3COOH. The Ka = 1.8 x 10-5
5.
A 2.4 g sample of acetic acid was dissolved in 2.00 L of water. The [ H+ ] = 6.0 x 10-4 M. Calculate Ka and
pH of CH3COOH.
CH3COOH  CH3COO- + H+ ( Ka = 1.8 x 10-5, pH = 3.2 )
6.
Kb for NH3 is 1.8 x 10-5. Calculate [ H+ ], [ OH- ], pH, pOH and the percent ionization of a 0.020mol/L NH3
solution.
NH3 + H2O  NH4+ + OH-
( 1.34 x 10-3 mol/L )
( [ H+ ] = 1.58 x 10-11, [ OH- ] = 6 x 10-4, pH = 10.8, pOH = 3.2, 3% )
7.
Kb = 1.7 x 10-6 for N2H4. If a solution of N2H4 has a pH of 10.5 what is [ N2H4 ] in the solution?
N2H4
8.
+ H2O

N2H5+
+ OH-
( 5.8 x 10-2 mol/L )
What is the pH of a 0.01 M solution of NaNO2?. The Ka of NaNO2 is 5.1 x 10-4
NaNO2 ------------->
NO2- + H2O 
Read pg 591 Applications
+
+
__________________
( pH = 7.7 )
Day 14 Volumetric Analysis (pg 595)
46
In many acid-base reactions, the equilibrium is displaced almost completely toward the product side. These
reactions may be considered quantitative and can be used as the basis for the analysis of the amount of acid or
base in a given sample. The process is termed volumetric analysis.
The requirements are:
1)
2)
3)
Only a single, specific reaction must take place between the unknown substance and the known substance
used for the analysis.
The unknown substance must react completely and rapidly with the added standard reagent. ie; it must be
a quantitative reaction.
An indicator or method must be available to signal when all the unknown substance has reacted with the
added standard reagent.
The usual objective is to determine the mass or percentage of a qualitatively identified component in a sample
whose quantitative amount is unknown. If the sample is a solution, the objective may be to determine its molar
concentration.
A 0.660 mol/L NaOH solution is used to determine the molar concentration of H2SO4 solution. What is the molar
concentration of the acid, 20.0 mL of which is just neutralized by 36.0 mL of the standard base?
This is a standard problem which can be solved using:
NaCaVa =Nb CbVb where Na and N b are the # of moles of H+ and # moles of OH - as taken from their formulas;
C is the molar concentration, V is the volume, a is 'of the acid', b is 'of the base'.
Therefore : 2 mol x Ca x 20.0 mL = 1 mol x 0.660 mol/L x 36.0 mL
Ca = ____________________________
Complete :
1.
What is the molar concentration of a hydrochloric acid solution, 30.0 mL of which is just neutralized by
48.0 mL of 0.100 mol/L NaOH?
2.
How many mL of 0.100 mol/L HCl are required to neutralize 25.0 mL of 0.100 mol/L Ba(OH)2?
Standard Solutions
To make up a standard solution you usually add a liquid solution of unknown concentration to a solid that
has been accurately weighted.
eg.
An HCl solution is standardized using pure Na2CO3(mm=106.00) as a primary standard. What
is the molar concentration of the acid if 30.00 mL of the acid solution is required to react
completely with a 0.500 gram sample of Na2CO3?
Na2CO3(s) + 2 HCl(aq) ---------> 2 NaCl(aq) + H2O(l) + CO2(g)
moles of Na2CO3 = 0.500 grams = 4.72 x 10-3 moles
---------------106.00 g/mol
Reaction works on a 1:2 basis 1 Na2CO3 = 2 HCl
------------ ---------4.72 x 10-3
x
47
x = 9.44 x 10-3 moles of HCl
C = ____________________ = ___________________
Alternate Solution
NaCaVa =Nbmb
_____
Mb
Example 2
A sodium hydroxide solution is standardized by reaction with benzoic acid, HC7H5O2,
(M=122.00 grams/mole). A 2.00 gram sample of benzoic acid required 35.00 mL of NaOH to
reach the endpoint. What is the molar concentration of the base? Benzoic acid is monoprotic)
Acid-Base Titration (pg 595)
An acid-base titration is just a method by which we can perform a volumetric analysis. The concentration of an
acid-base solution may be determined by measuring the volume of the base of known concentration needed to
react completely with a specific volume of an acid solution.
Definitions
Standard solution:
Titration:
Buret:
Equivalence Point:
Chemical Indicators:
Endpoint:
A solution of ________________ concentration.
the process of adding the standard solution(titant) from a graduated tube in controlled
amounts.
a graduated tube with a dispensor control at the bottom.
the point at which equal molar quantities of reactants are present.
a chemical that will change colour at or very near the ph of the stoichiometric point.
The place in the titration when the indicator changes colour.
Titration Curves
A. Strong Acid and a Strong Base
Calculate pH :
a. no base added [HCl] = 0.300 mol/L --- [H+] = __________
pH = __________
48
b. 5.00 mL of 0.300mol/L NaOH is added
[OH-] = _____________
[NaOH]f = c2 = c1v1
------v2
[HCl]f = c2 = c1v1
[H+] = _____________
------v2
[H+]excess = _______________________ pH = _____________ = _______________
c.
20.0 mL of 0.300 mol/L NaOH is added
[NaOH]f = ______________ = ______________________
[OH-] = ___________________
[HCl]f = _________________ = ______________________ [H+] = ____________________
the H+ is consumed completely by the OH- forming water. H2O - H+ + OH-.
Kw = (x)(x) = 1x 10-14
x = 1x 10-7 = [H+] = [OH-]
pH = __________ =__________
d. 30.0mL of 0.300 mol/L NaOH is added
[NaOH]f = ______________ = ______________________
[OH-] = ___________________
[HCl]f = _________________ = ______________________ [H+] = ____________________
[OH-]excess = ____________________
pOH = ______________
pH = _______________
Titration of a Weak Acid by a Strong Base
At the equivalence point, the solution contains sodium acetate, which has an anion, but not a cation, that
hydrolyses. The acetate ions however do ionize to give a slightly basic solution. For this reason, at the
equivalence point where all the acid has neutralized all the base, there is still the acetate ions in solution which
ionize the water.
a.
Ka = 1.8 x 10-5
Initial pH ---> 20.0mL of 0.300 mol/L CH3COOH
CH3COOH 
[I]
[C]
[E]
CH3COO- +
H+
Ka = 1.8 x 10-5 = ______________________
[H+] = ___________________
pH = ________________
b.
10.0 mL of 0.300 mol/L NaOH is added
[i]
NaOH(aq)
__________
[c]
__________
____________
______________
[e]
__________
____________
______________
Na+
+ OHCH3COO-
+
+
CH3COOH(aq)
___________

CH3COOH
+ H2O


NaCH3COO(aq)
0
Na+
CH3COOH
+
H2O (l)
+ CH3COO- + H2O
OH-
+
[i]
__________
___________
______________
[c]
__________
____________
______________
[f]
__________
____________
______________
Kb = _________________ = _____________________
[OH-] = _________________
pOH = _______________
pH = ____________________
c. after 20.0 mL of 0.300 mol/L NaOH has been added
NaOH(aq)
+

CH3COOH(aq)
NaCH3COO(aq)
[i]
__________
___________
[c]
__________
____________
______________
[e]
__________
____________
______________
Na+
+ OHCH3COO-
+
+ H2O
CH3COOH


+
H2O (l)
0
Na+
CH3COOH
+ CH3COO- + H2O
+
OH-
[i]
__________
___________
______________
[c]
__________
____________
______________
[e]
__________
____________
______________
Kb = _________________ = _____________________
[OH-] = _________________
pOH = _______________
pH = ____________________
49
50
d) after 30.0 mL of 0.300 mol/L NaOH has been added
Titration of a Weak Base by a Strong Acid
At the equivalence point, when all the acid has neutralized all the base there will still be the cation NH 4+ in
solution. This ion will cause the water to ionize and force the resulting pH of the solution downwards.
a.
Initial pH with no acid added
NH3
[i]
[c]
[e]
+
H2O
_____
_____
_____

NH4+
+
0
_____
_____
OH-
0
_____
_____
Kb = ________________ = ____________________
[OH-] = ______________ pOH = ____________
b.
After 20.0 mL of 0.100 mol/L HCl is added
NH3
+ HCl
 ___________ + _____________
[i] _______
_______
__________
___________
[c] _______
_______
__________
___________
[e] _______
_______
__________
___________
NH4+
[i] _______
[c] _______
[e] _______

NH3
__________
__________
__________
+
H+
___________
___________
___________
Ka = ____________________ = _______________________
[H+] = _________________
pH = ____________________
pH = ___________
c.
After 30.0 mL of 0.100 mol/L HCl is added
51
Try These
1.
Calculate the pH of the resulting solution after 20.00 mL of 0.20 mol/L NaOH has been added to 25.00 mL
of 0.20 mol/L HC2H3O2. (Ka CH3COOH = 1.8 x 10-5 )
2.
Calculate the pH of the resulting solution after 10.00 mL of 0.20 mol/L NaOH has been added to 25.00 mL
of 0.20 mol/L NH4Cl. (Ka NH4+= 5.6 x10-10 )
Complete pg 607 ( 5) pg 614(6,9)
52
Day 15 Titration of a Weak Acid
In this activity, you will standardize a sodium hydroxide solution, then determine the concentration of an unknown
acid by titration with the standardized base. You will then determine the pH of the acid using a probe and discuss
the differences.
Experimental Design
This is a two-part activity. In the first part, you will standardize a sodium hydroxide solution by titrating it with the
primary standard, potassium hydrogen phthalate, KHC8H4O4(aq) . The concentration of the unknown acid solution is
then determined by titrating it with the standardized sodium hydroxide solution.
Materials
lab apron
eye protection
sheet of blank white paper
electronic balance
pH meter
distilled water
wash bottle with distilled water
125-mL Erlenmeyer flask
1000-mL glass or plastic bottle
rubber stoppers
NaOH(s)
KHC8H4O4(s)
1% phenolphthalein
vinegar, lemon juice, or other acid solution of unknown
concentration
dropper
stirring rod
buret
buret stand
100-mL graduated cylinder
10-mL graduated cylinder
weighing paper
Procedure
Part I Standardization of NaOH(aq)
1. In a 125-mL Erlenmeyer flask, dissolve approximately 10 g of NaOH(s) in 50 mL of distilled water.
2. Transfer the entire solution to a 1000-mL bottle and dilute with 500 mL distilled water. Stir the solution but do
not shake it. Assume volumes are additive.
3. Weigh approximately 0.4 g KHC8H4O4(s) and record the mass to three significant digits.
4. Place the KHC8H4O4(s) into a clean, dry Erlenmeyer flask. Add 50.0 mL distilled water and two to three drops of
phenolphthalein. Swirl to mix.
5. Allow several millilitres of the NaOH(aq) solution to flow through a buret, making sure that the solution wets all
of the inside surfaces.
6. Fill the wetted buret with the NaOH(aq) solution and record the volume in a suitable chart.
7. Place the Erlenmeyer flask containing KHC8H4O4(aq) over a sheet of white paper and titrate with the NaOH(aq)
solution in the buret until the endpoint is reached.
8. Repeat steps 3 to 7 two more times and calculate the mean of the three volumes of NaOH(aq) used to reach the
endpoint. Rinse out the flask.
Part II Determining the [H+(aq)] in a solution of unknown concentration
9. Refill the buret you used in Part A with standardized NaOH(aq) solution.
10. Place 10.00 mL of an unknown acidic solution into a clean, dry Erlenmeyer flask. Add 2 to 3 drops of
phenolphthalein. Swirl to mix.
11. Place the Erlenmeyer flask containing the acid solution over a sheet of white paper and titrate with
standardized NaOH(aq) solution in the buret until the endpoint is reached.
12. Repeat steps 10 to 11 two more times, and calculate the mean of the three volumes of NaOH(aq) used to reach
the endpoint. Use this average value to calculate
the concentration of H+ (aq) in the acidic solution.
13. Calibrate a pH meter and use it to measure the pH of a sample of the unknown acid solution. Record the value.
14. Discard all solutions in the sink with lots of running water. Return materials and equipment to their proper
location. Wash your hands with soap and water.
Analysis
Part 1
(a) Use the average of your titration volumes to calculate the exact concentration of the NaOH(aq) solution.
(b) Theoretically (through calculation), what should have been the concentration of the acid
(c) What was the pOH and pH of the basic solution?
(d) Why was KHC8H4O4(s) used as the acid in the standardization titration?
53
Part II
(d) Use the evidence from your titration to calculate the concentration of the vinegar solution.
(e)Calculate the pH of the acidic solution (need to find Ka of acetic acid)
(f) Calculate the % error based on the reading of the calibrated pH meter.
(g) After the addition of 10mL of NaOH, what was the approximate pH of the sample solution?
Part III
Evaluation
(g) Comment on the reasons for the difference in the actual and applied results (sources of error) and evaluate the
Procedure and suggest changes that might correct any sources of error.
**For write up of this lab, it is required that you
a.
b.
c.
d.
e.
f.
g.
Develop an appropriate question to be answered.
Develop an appropriate hypothesis, including dependant and independent variables.
Develop the procedure as a flow chart
Present the data tables appropriately
Clearly show your calculations
Communicate effectively the evaluation
Develop an appropriate conclusion
54
CATEGORIES

Question/Hyp
othesis
Question Generating
LAB REPORT RUBRIC Names ____________________________________
LEVEL 1
LEVEL 2
LEVEL 3
(50-59%)
(60-69%)
(70-79%)
Question is unclear.
Question is stated without
using appropriate scientific
terminology.
Unreasonable association
between problem and predicted
results, and variables not
defined correctly.

Hypothesis and
Controls

Procedural Planning
Develops disorganized or
unworkable procedures.
Includes few of the important
steps.

Procedure Format
Procedure included but not
written in past tense,
impersonal and sequentially
numbered.
LEVEL 4
(80-100%)
Question is stated clearly using
appropriate scientific
terminology.
Somewhat reasonable
association between the
problem and the predicted
results, or some variables
improperly defined.
Question is stated
using appropriate
scientific terminology
but is not clear.
Reasonable association
between the problem
and the predicted
results, and most key
variables are defined.
Develops an appropriate
incomplete set of
procedures which may lack
efficiency or clarity.
Includes some of the
important steps.
Some procedure steps
written in past tense,
impersonal and
sequentially numbered.
Develops an
appropriate complete
set of procedures that
are mostly efficient and
clear. Includes most of
the important steps.
Procedure written
mostly in past tense,
impersonal and
sequentially numbered.
Develops an appropriate
complete set of procedures that
are complete, efficient and clear.
Includes all of the important
steps.
Records most relevant
data in an organized
way; adequate detail
and generally accurate.
Makes sufficient
observations to
generate data.
Records all relevant data in an
organized and skillful way;
accurately.
Provides some analysis
of the data. Identifies
most of the obvious
patterns and trends.
Answers most
questions successfully;
a few explanations are
lacking.
Calculations are done
well, with most
information presented
properly with the
proper number of
significant digits.
Provides some
evaluation of
procedure used.
Identifies most of the
sources of error or
limitations.
Draws some valid
conclusions based on
the data; relevant and
clearly stated.
Most requirements for
format/style followed.
Minor errors in the use
of units and
terminology.
Provides insightful analysis of
the data. Identifies both obvious
and subtle patterns and trends.
Insightful association between
the problem and the predicted
results; all variables are clearly
defined.
Procedure
Procedure written completely in
past tense, impersonal and
sequentially numbered.
Observations

Data Recording
Records little data; data is
irrelevant and/or inaccurate.
Records data but
organization is lacking;
some inaccuracies.

Observations
Makes few observations
Makes observations but
may be insufficient to
generate data.
Explains data but provides
limited analysis. Identifies
some obvious patterns or
trends.
Answers some questions
successfully; explanations
are lacking.
Makes insightful observations
but does not draw conclusions.
Analyzing &
Interpreting

Analysis of Data
Provides no analysis of the
data. Identifies few patterns or
trends.

Analysis questions
part I
Answers few questions and/or
explanations are incorrect.

Analysis question
part II
Calculations are done poorly,
with little information
presented properly and
incorrect significant digits.
Calculations are done fairly
well, with most information
presented properly with
the proper number of
significant digits.

Analysis question
part III
Provides no evaluation of
procedure used. Identifies few
sources of error or limitations.
Provides limited evaluation
of procedure. Identifies
some sources of error or
limitations.

Conclusions
Makes illogical or irrelevant
conclusions.

Format

Scientific
Terminology & SI
Units
Few requirements for
format/style followed.
Frequent errors in the use of
units and terminology;
interferes with communication.
Attempts to draw some
conclusions partially based
on the data; not clearly
stated.
Some requirements for
format/style followed.
Some errors in the use of
units and terminology; do
not interfere with
communication.
Answers all questions
successfully; all explanations are
accurate and thorough.
Calculations are done well, with
all information presented in a
neat and organized manner with
the proper number of significant
digits.
Provides a thorough evaluation
of the procedure used. Identifies
all sources of error and
limitations.
Draws clearly stated conclusions
based on the data; relevant and
clearly stated.
All requirements for format/style
followed.
No errors in the use of units and
terminology.
Day 16 Buffer Solutions ( pg 615 )
55
Buffers are solutions with the ability to resist the addition of strong acids or strong bases, within limits.
They play an important role in chemical processes where it is essential that a fairly constant pH is maintained. In
many industrial and physiological processes, specific reactions occur at some optimum pH value. When the pH
varies to any extent from the optimum value, undesirable reactions and effects may occur. For example, the pH of
your blood lies at about 7.35. If this value drops below 7.0 (acidosis) the results are fatal. Also if it rises above 7.7
(alkalosis) the results are as well fatal. Fortunately our blood contains a buffering system which maintains the
acidity at the proper level. If it were not for the protection of the buffering system, we could not eat and adsorb
many of the acidic fruit juices and foods in our diet.
A typical lab buffer is CH3COOH and its salt NaCH3COO. Most buffer solutions are made up using a weak acid and
its sodium salt! When a strong base such as NaOH is added to the buffer, the acetic acid reacts with and consumes
the excess OH- ion. The OH- reacts with the H3O+ ion from the acid in the following reaction:
H2O + CH3COOH <-------> H3O+ + CH3COOH3O+ + OH- <----------> H2O
The OH reduces the H3O+ ion concentration, which causes a shift to the right, forming additional CH3COO- and
H3O+ ions. For practical purposes each mole of OH- added consumes a mole of CH3COOH and produces a mole of
CH3COO-.
OH- + CH3COOH <------------> CH3COO- + H3O+
When a strong acid such as HCl is added to the buffer, the hydronium ions react with the CH3COO- ions of the salt
and form more undissociated CH3COOH.
H3O+ + CH3COO- <-----------> CH3COOH + H2O
As you would expect, there is a limit to the quantity of H+ or OH- that a buffer can absorb without undergoing a
significant change in pH. If a mole of HCl is added to a litre of buffer solution containing 0.5 moles of sodium
acetate/acetic acid buffer the H+ completely consumes the buffer and results in a drastic change in pH.
The blood buffer is made up from the dissolved carbon dioxide in the plasma.
CO2(g) + H2O <-------> H2CO3 <------> HCO3- + H3O+
When a base is added it reacts with the carbonic acid.
OH- + H2CO3 <--------> HCO3- + H2O
When an acid is added it reacts with the bicarbonate ion.
H3O+ + HCO3- <---------> H2CO3 + H2O
Because there are both a base-neutralizer and an acid-neutralizer then we have a buffer.
Buffer Components
A buffer has two components.
HA
NaA ---> Na+ + Aa weak acid &
a soluble salt of the acid
Therefore any extra H3O+ will be neutralized by the A- in the buffer.
H3O+ + A- <-------> HA + H2O
And any extra OH- that is added will be neutralized by the acid.
HA + OH- <------> A- + H2O
Day 17 Reading Assignment pg 621-623 : Case Study ( Acid Deposition )
Complete pg 631( 1,2,5,6,7,10-13,16,17,18,19) pg 632(1,5,8,9,(Calculate final pH:15 b,c,f),25)
Complete pg 636(1-14,17,19,20,21,22,23-27,32-34,36,39)
pg 639(1,2,3,6a,7,9,10,12,16,17,18,19,27,28,30,31,34,44b,f,51,55,56,59,60,62)
56
SCH 4U1 UNIT 4 REVIEW
1. Consider the equilibrium below:
If 1.5 mol of PCl5 was placed in a 1.0 L container and allowed to reach equilibrium, what would the value of Ke be if at
equilibrium [PCl5] = 1.2 mol/L? ( ANS 13 )
PCl3(g) + Cl2(g) <=====> PCl5(g)
2. Consider the equilibrium below:
If 1.6 mol of HI was placed in a 1.0 L container and allowed to reach equilibrium, what would the equilibrium concentrations be
for H2(g), I2(g) and HI(g) if the Ke = 36?
H2(g) + I2(g) <=====>
2] = [I2
3.
-4
. If [A] = 0.24 mol/L, what is the [D]?
mol/L )
-3
Problem
4.
If the solubility of AgBr is 8.8 x 10-7 mol/L, what is its Ksp?(ANS Ksp = (8.8 x 10-7)2 =
5. If the solubility of Mg(OH)2 is 1.3 x 10-4 mol/L, what is its Ksp? ( ANS
-13
-12
-12
6. If 5.6 L of a saturated solution of Ag2CrO4 is found to contain 0.12 g of Ag2CrO4, what is the Ksp of Ag2CrO4.
7.
-5
What is the solubility, in mol/L, of MgCO3 in a 0.65 mol/L solution of MgCl2 if the Ksp of MgCO3
3.9 x 10-5 mol/L
? solubility =
-4
8. If 365 mL of a 0.0054 mol/L solution of Pb(NO3)2
mol/L solution of KI, would a
precipitate form? Calculate the ion product for the potential precipitate. The Ksp of PbI 2 is 7.9 x 10-9. ion product PbI2 is [Pb2+][I1]2 = 3.2 x 10-10 < Ksp, no precipitate forms
9. If 78 mL of a 0.0026 mol/L solution of Zn(NO 3)2 was mixed with 45 mL of a 7.13 x 10-3 mol/L solution of K2CO3, would a
precipitate form? Calculate the ion product for the potential precipitate. The Ksp of ZnCO 3 is 1.0 x 10-10. ion product ZnCO3 is
[Zn2+][CO32-] = 4.3 x 10-6 > Ksp of ZnCO3, yes a precipitate occurs
10. What is the concentration of a monoprotic weak acid if its pH is 5.50 and its K a = 5.7 x 10-10?
-2
mol/L
11. A weak base with a concentration of 1.3 mol/L has a percent ionization of 0.72 %. What is the Kb of this weak base? Ka =
((0.72)(1.3)/100)2 / 1.3 = 6.7 x 10-5
12. What is the percent ionization of a 0.48 mol/L weak acid if its Ka = 1.4 x 10-9?
-3
%
13. What is the pH of a 1.47 mol/L solution of HCN(aq) if its Ka = 3.5 x 10-11? pH = 5.14
14. 25 mL of standardized 0.45 mol/L NaOH is titrated with 21 mL of 0.35 mol/L acetic acid. Calculate the pH of the solution. pH
= 14 – 1.07 = 12.93
57