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Chemistry 30 – Unit 4 Part 2
Acid-Base Equilibrium Systems
Chapter 17, Section 17.1
• Acid rain: pH < 5 “normal rain” is still
acidic due to dissolved CO2:
CO2(g) + H2O(l)  H2CO3(aq)
H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3-(aq)
• Acid Deposition: acid rain, acid snow,
acid hail, acid fog, acid dust
• Causes: Mainly SO2(g) & SO3(g) “sox”
and NO(g) and NO2(g) “nox”
Chapter 17, Section 17.1
• “sox” from combustion of coal and
natural gas containing sulfur and from
smelting metal ores to produce pure
metals
See equations page 680: “sox”
ultimately becomes H2SO3(aq) and
H2SO4(aq)
“scrubber” to remove SO2(g) from
smokestacks
• “nox” from automobile exhaust
see equations, page 681
Chapter 17, Section 17.1
• 1st step:
N2(g) + O2(g) + energy  2 NO(g)
Kc very small – barely occurs at all at
room temperature
At the temperature in a car engine Kc
much larger
Lowering temperature would cause
engines to run very inefficiently,
 car engines produce NO(g) which
ultimately gets converted to
HNO2(aq) and HNO3(aq)
Chapter 17, Section 17.1
• Acid deposition serious problem for
lakes
Some organisms can’t tolerate lower
pH’s
• Acidic lakes also leach minerals from
their lakebeds (happens more in
granitic lakebeds in Canadian Shield
than in alkaline soils of Alberta)
• Toxic metals can accumulate in the
bodies of aquatic organisms
Chapter 17, Section 17.1
• Acid shock – spring run-off from acid
snow
Treatment – liming – calcium
carbonate (limestone)
Raises pH; reduces solubility of toxic
metal ions; adds calcium ions (plant
nutrient)
Chapter 17, Section 17.2
• The Brønsted-Lowry Theory
acid: proton (H+) donor
base: proton (H+) acceptor
• allows explanation of non-aqueous
acid-base reactions
• acids and bases defined by their
behaviour in a chemical reaction, not
by their empirical properties
Chapter 17, Section 17.2
acid
base
conjugate
acid
conjugate
base
conjugate pair
conjugate pair
conjugate acid/base pair: two chemical
entities differing by 1 proton (H+)
All Brøntsted-Lowry acid-base reactions
contain 2 conjugate acid/base pairs
In above reaction conjugate acid/base
pairs:
HCl(aq)/Cl‾(aq) and H3O+(aq)/H2O(l)
Chapter 17, Section 17.2
• second example – bottom of page 685
do Practice Problems 2 and 4, page 687
Chapter 17, Section 17.2
• Equations can be written for the
reaction of water with some
substances that will produce H3O+(aq)
and OH‾(aq), each with a familiar
balancing entity
familiar:
e.g.
HCO3‾(aq) + H2O(l)
H3O+(aq) + CO32-(aq)
HCO3‾(aq) + H2O(l)
OH‾(aq) + H2CO3(aq)
Called amphiprotic substances
Chapter 17, Section 17.2
• Examples of amphiprotic substances:
• H2O(l)
• anions containing an H that can ionize
as H+(aq):
• HSO4‾(aq)
• HOOCCOO‾(aq)
• C3H5O(COOH)2COO‾(aq)
• C3H5OCOOH(COO)22‾(aq)
• HSO3‾(aq)
• H2PO4‾(aq)
• HPO42‾(aq) H2PO4‾(aq)
Chapter 17, Section 17.2
• Polyprotic acid: acid that can donate 2 or
more protons
(note that amphiprotic ions are between
polyprotic acids and polyprotic bases)
e.g.
H2CO3(aq) + H2O(l)  H3O+(aq) + HCO3‾(aq)
polyprotic acid
amphiprotic base
HCO3‾(aq) + H2O(l)  H3O+(aq) + CO32‾(aq)
amphiprotic acid
polyprotic base
Polyprotic acid: H2CO3(aq) can donate 2 protons
Polyprotic base: CO32‾(aq) can accept 2 protons
Amphiprotic entity: HCO3‾(aq) can donate or accept a
proton (depending on what it’s reacting with)
Chapter 17, Section 17.2
• Do Practice Problems 5-7, page 688
Chapter 17, Section 17.2
• Predicting equilibrium position for acid-
base reactions: 2 methods; see example
HF(aq) + CH3COO‾(aq)  CH3COOH(aq) + F ‾(aq)
A
B
Using your acid-base chart,
compare the acid and base
on the left to the acid and
base on the right
Stronger pair will transfer
a proton better and
equilibrium will favour the
opposite side
In the example HF(aq) is
stronger acid;
CH3COO‾(aq) is stronger
base – equilibrium favours
products
A
B
Look at position of acid
and base on left side of
equation
If acid
base
If
favours
products
base favours
acid
reactants
In example HF(aq) is above
CH3COO‾(aq)
equilibrium favours
products
Chapter 17, Section 17.2
• Example: Practice Problem 8a, page
690
NH4+(aq) + H2PO4‾(aq)  NH3(aq) + H3PO4(aq)
H3PO4(aq) is stronger acid; NH3(aq) is
stronger base, therefore equilibrium
favours reactants
• It is unnecessary to compare both
acids and bases – the stronger acid
will always be paired with the
stronger base – why?
• the stronger an acid, the weaker its
conjugate base
Chapter 17, Section 17.2
• Do WS 67
Chapter 17, Section 17.3
• When dealing with acid ionizations
(or reactions with water) we use Ka:
For the acid HA(aq)
HA(aq) + H2O(l)  H3O+(aq) + A‾(aq) or
HA(aq)  H+(aq) + A‾(aq)
H3O    A 
Ka 
HA
or
H    A 
HA
Ka is a special case of Kc
Chapter 17, Section 17.3
• Strong acids have large Ka; weak
acids have very small Ka
• Ka’s are present in your acid-base
chart on pages 8-9 of the Data
Booklet
• % ionizations are valid for acids only
over a very narrow concentration
range; Ka’s better but ……….
You may be required to calculate %
ionization for the Diploma Exam
Chapter 17, Section 17.3
• Sample Problem, page 692-3
initial [CH3COOH] = 0.10 mol/L
pH at equilibrium = 2.96
Review:
Find Ka and % ionization
• equilibrium [H3O+(aq)] = 10-2.96 = 0.0011 mol/L
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO‾(aq)
I
0.10
0
0
C -0.0011
+0.0011
+0.0011
E 0.099
0.0011
0.0011
H3O   CH3COO    0.0011 0.0011
Ka 

 1.2  10 5
0.099
CH3COOH 
equilibrium product concentration
0.0011 mol L
% ionization=
 100 
 100  1.1%
maximum possible concentration
0.10 mol L
Chapter 17, Section 17.3
• Try Practice Problems 14 and 17,
page 693
14. equilibrium [H3O+] = 10 2.50  0.0032 mol L
HC4H3N2O3(s) + H2O(l)  H3O+(aq) + C4H3N2O3‾(aq)
I
0.10
0
0
C -0.0032
+0.0032
+0.0032
E
0.097
0.0032
0.0032
0.00322
Ka 
 1.0  10 4
0.097
0.0032 mol L
% ionization 
 100%  3.2%
mol
0.10 L
Chapter 17, Section 17.3
• 17.
nPABA 
4.7 g
 0.034 mol
g
137.15 mol
PABA 
0.034 mol
 0.034
1.0 L
Equilibrium [H3O+] = 10 3.19  6.5  10 4
mol
L
mol
L
C6H4NH2COOH(aq) + H2O(l)  H3O+(aq) + C6H4NH2COO‾(aq)
I
0.034
C -6.5 x 10-4
E
0.034
0
+6.5 x 10-4
6.5 x 10-4
6.5  10 


4 2
Ka
0.034
 1.2  10 5
0
+6.5 x 10-4
6.5 x 10-4
Chapter 17, Section 17.3
• Ion product constant for water, Kw
2 H2O(l)  H3O+(aq) + OH‾(aq)
or
H2O(l)  H+(aq) + OH‾(aq)
K w  1.0  10 14  H3O   OH  
or
• True for all aqueous solutions
• For pure water,
[H3O+] = [OH-] = 1.0 x 10-7 mol/L
H   OH  
Chapter 17, Section 17.3
• Recall from Chem 20:
pH = -log [H3O+]*
pOH = -log [OH‾]
pH + pOH = 14.00 (a log form of Kw)
[H3O+]* = 10  pH
[OH‾] = 10  pOH
* H3O+ or H+
• pH Review Questions – WS 63
Recall special deal with significant
digits and pH
Chapter 17, Section 17.3
•
Another special of Kc: Kb
• B(aq) + H2O(l)  HB+(aq) + OH‾(aq)
HB   OH  
Kb 
B 
example: NH3(aq)
NH3(aq) + H2O(l)  NH4+(aq) + OH‾(aq)
 NH4   OH  
Kb 
NH3 
Chapter 17, Section 17.3
• Practice Problems 19 c, 20c, 19 d,
20d, 22, page 698
22. Since pH = 10.26, codeine is a base
pOH  14.00  10.26  3.74
OH    10  pOH  10 3.74  1.8  10 4 mol L
C18H21NO3(s) + H2O(l)  OH‾(aq) + HC18H22NO3+(aq)
I
0.020
C -1.8 x 10-4
E
0.020
0
+1.8 x 10-4
1.8 x 10-4
1.8  10 


0
+1.8 x 10-4
1.8 x 10-4
4 2
Kb
0.020
 1.7  106
Chapter 17, Section 17.3
• Finding pH of weak acids or bases
Weak acid:
Example: Practice Problem 24, page 699
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COO‾(aq)
I
0.83
0
0
C
-x
+x
+x
E 0.83 – x
x
x
Ka = 1.8 x 10-5 (Data Booklet)
H3O   CH3COO  
pH   log  0.0039  2.41
Ka 
CH3COOH 
x
0.83  x
x2

0.83
1.8  10 5 
1.8  10 5
2
x  0.83  1.8  10 5  0.0039 mol L
can approximate
since 0.83 > 1000
times 1.8 x 10-5
this will always be true
in your questions
Chapter 17, Section 17.3
• Shortcut*: pH of a weak acid
 H3O    K a WA
pH   log  H3O  
WA = weak acid
2 notes about shortcut:
Never use to find Ka or [WA] – this is an
approximation; no need to solve a quadratic
to get Ka or [WA]; don’t approximate
You cannot eliminate all ICE boxes; only those
solving for H3O+ using Ka
WS 65
Chapter 17, Section 17.3
• Weak Base:
Example: Practice Problem 28, page 701
NH3(aq) + H2O(l)  NH4+(aq) + OH‾(aq)
I 0.105
C -x
E 0.105 - x
 NH 4   OH  
Kb 
NH3 
0
+x
x
pOH   log 0.0014  2.86
pH  14.00  2.86  11.14
x2
1.8  10 
0.105  x
x2
5
1.8  10 
(approximation)
0.105
x  0.0014
5
0
+x
x
Chapter 17, Section 17.4
• Shortcut*: pH of a weak base
OH    K b WB 
pOH   log OH  
pH  14.00  pOH
WB = weak base
Same cautions as for weak acid shortcut!
Try Practice Problem 29, page 701
Chapter 17, Section 17.3
Practice Problem 29, page 701
N2H4(l) + H2O(l)  N2H5+(aq) + OH-(aq)
I
5.9x10-2
C
-x
E 5.9x10-2 – x
1.3  10
6
1.3  10 6
0
+x
x
0
+x
x
x2

5.9  10 2  x
x2

5.9  10 2
1.3  10 5.9  10   2.8  10
pOH   log  2.8  10   3.56
okay
x  K b  WB  
6
4
pH  14.00  3.56  10.44
2
4
mol
L
Chapter 17, Section 17.3
• In previous example, you were given
Kb for ammonia
This wasn’t really necessary
• For any conjugate acid-base pair in
your Data Booklet, Ka x Kb= Kw
Data Booklet Ka for NH4+ = 5.6 x 10-10
K w 1.0  10 14
5



1.8

10
• Kb for NH3
K a 5.6  10 10
• Discuss question 37, page 703, do
Review 4a, 4c, page 704
Chapter 17, Section 17.3
• 4a
Conjugate base: C6H7O6‾(aq)
K w 1.0  10 14
10
Kb 


1.1

10
Ka
9.1 10 5
4c
Conjugate base: HCOO‾(aq)
K w 1.0  10 14
11
Kb 


5.6

10
Ka
1.8  10 4
Chapter 17, Section 17.4
• Acid-base indicators: conjugate acidbase pairs ( HIn(aq)/In‾(aq) ) where
acid and its conjugate base are
different colours, for example:
bromothymol blue
HBb yellow/ Bb‾ blue
formulas organic and
complex (see p. 705)
so abbreviations used
• Chart on page 10 of Data Booklet
• Indicators chosen for titrations to
change colour over appropriate pH
range
Chapter 17, Section 17.3
• Note: because of Ka’s on indicator
chart and acid/base chart of Data
Booklet you could be expected to
write an equation and predict colour
of an indicator in a solution
Example: what would be expected if
bromocresol green in its base form,
Bg‾(aq), was added to a benzoic acid
solution?
Chapter 17, Section 17.3
• As before, make your list:
Bg‾(aq), C6H5COOH(aq), H2O(l)
SB
SA
A, B
C6H5COOH(aq) + Bg‾(aq)  HBg(aq) + C6H5COO‾(aq)
Ka of HBg(aq) (1.3 x 10-5) is slightly lower than
that of C6H5COOH(aq) (6.3 x 10-5), therefore
equilibrium will favour products and colour of
HBg/Bg- will be green or yellow
Chapter 17, Section 17.4
• Titration curves: review of Chem 20
acid with base
strong
acid /
strong
base*
*Chem
20
base with acid
VNaOH or VHCl (mL)
Chapter 17, Section 17.4
•
strong
acid
weak
acid
polyprotic acid
titrated with strong
base
Note: only quantitative
reactions have
detectable endpoints
weak acid and strong
acid titrated with
strong base
Chapter 17, Section 17.4
• titration of weak acid versus strong:
favours
products
e.g. CH3COOH(aq) + OH-(aq)  H2O(l) + CH3COO-(aq)
WA with SB
SA with SB*
pH
WA with WB
>7
7
Explains why pH > 7 for
equivalence point of WA with SB
volume of base
Chapter 17, Section 17.4
• titration of weak base versus strong:
pH
7
<7
WB with WA
SB with SA*
WB with SA
volume of acid
Chapter 17, Section 17.4
• polyprotic acid with strong base
Figures 17.14
and 17.15,
pages 709-10
pH
• polyprotic base with strong acid
Chapter 17, Section 17.4
• Buffer regions
Figures 17.12
and 17.13,
pages 707-8
In buffer regions there is a mixture of a
weak acid and its conjugate base
Addition of small amounts of acid or base
has little effect on the pH
Chapter 17, Section 17.4
•
x
CH3COOH(aq) only
50/50 mixture of CH3COOH(aq)
and CH3COO‾(aq)
(page 712)
Example: CH3COOH(aq)/CH3COO‾(aq) buffer
Chapter 17, Section 17.4
• example continued:
Buffer equilibrium:
CH3COOH(aq) + H2O(l)  H3O+(aq) + CH3COOˉ(aq)
Add acid:
[H3O+] , equilibrium shifts left to
consume added H3O+(aq) and convert
some CH3COOˉ(aq) into CH3COOH(aq)
Add base: since,
essentially
OHˉ(aq) + H3O+(aq) 100% H2O(l),
the buffer equilibrium will shift
right as H3O+(aq) is consumed
Chapter 17, Section 17.4
• Buffer demo: pH 7 versus water as
concentrated HCl(aq) is added
Also show buffer capacity as more
HCl(aq) is added
Chapter 17, Section 17.4
• Buffers in your body: read pages 714717
Human blood buffered to pH = 7.4
Most important buffer in human
blood: HCO3ˉ(aq)/CO32-(aq)
Without buffering, the amount of
acid in a glass of orange juice could
kill you
Chapter 17, Section 17.4
• In internal cell fluid, most important
buffer is H2PO4ˉ(aq)/HPO42-(aq)
acidosis – decreased blood pH
respiratory
metabolic
alkalosis – increased blood pH
Chapter 17, Section 17.4
•