Download - Johnston Chapter 17 Notes
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Chapter 17
Aqueous Acid-Base
Equilibria
17.1 Proton Transfers in Water
17.2 The pH Scale
17.3 Weak Acids and Bases
17.4 Recognizing Acids and Bases
17.5 Acidic and Basic Salts and
Oxides
17.6 Factors Affecting Acid Strength
17.7 Multiple Equilibria
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17.1 Proton Transfers in Water
Brønsted-Lowry definition of acid-base reactions
Acid ± any substance that can donate a proton (H+) to another substance.
Can be neutral, cations or anions
HNO3, NH4+, H2PO4
-
Base ± any substance that can accept a proton (H+) from another
substance.
Can be neutral or anions
NH3, CO32-
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Some critical thinking: Actually, it
could be a cation, too²but rarely!Which of the following could take
two or more protons?
A. NH2CH2CH2 NH2
B. N2H4
C. CH3 NH2
D. Two of A-C.
E. Three of A-C.
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Types of acids and bases
Monoprotic acids: onlycapable of donating one
proton
HCl, HI, HS-, HPO4-,
HSO4-
Polyprotic acids: capable of donating two or more
protons
H2CO3, H3PO4,
H2PO
4
-
,H
2SO
4, H
2S
Monoprotic bases: onlycapable of accepting one
proton
Polyprotic bases: capable of accepting two or more
protons
SO42-, CO3
2-, PO43-
Amphiprotic : molecules or ions whichcan behave either way (either an acid
or a base)
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Types of acids and bases
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Dissociation of Water Water: solvent for most acid/base reactions
Hydronium ion: hydrated hydrogen ion
2H2O (l) H3O+ (aq) + OH-(aq)
Referred to as autoionization ± self-ionization
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D issociation of Water
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Water and Equilibrium We already talked about the autoionization
of water
2H2O (l) H3O+ (aq) + OH- (aq)
So, what does the K for this autoionization
look like?
3w 0]][[K
v!! OH O H
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An older²and hence better²way
of expressing this«
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Some terminology for K w ...
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K w would be expected to _____ with
increasing T.
A. decrease
B. increase
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K w increases with increasing T since...
% (H > and (S < .
& (H < and (S < .
C. (H > and (S > .
D. (H < and (S > .
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³ x´ moles of a weak acidneutralize fewer moles of a strong
base than would ³ x´ moles of a
strong acid .
A. True
B. False
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Example 17 - 1Determine the ion concentration in a 5. x 1 -2
M aqueous solution of HClO4, a strong acid.
What are the ion concentrations in .5 L of an aqueous solution that contains 5. g of NaOH?
Example 17 - 2
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17-1: 0.050M HClO4« This is a strong acid.
This means that it is a strong electrolyte.
Hence, it is 1 % dissociated and
H+ = . 5 M.
If you missed this one, think!!!
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17-2: 5.00 g NaOH(aq) in 0.500L
solution«
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N ote«
Having to calculate H+ for basic solutionsusually requires an extra step.
Basic solutions in general often require morework!
This is because we usually think in terms of [H+]² as we shall see when we get to pH.
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17.2 The pH Scale Methods for expressing the hydronium and
hydroxide ion concentrations
pH = - log H3O+ and pOH = -log OH-
pH + pOH = 14
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At a higher temperature, pK wwould be less than 14.00.
A. True
B. False
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Features of the pH Scale1. pH = 7. defines a neutral solution. Acid
solutions have pH < 7 and basic solutions have
pH > 7. (Note: We assume 25rC here!)
2. The more acidic the solution, the lower its pH.
3. A change in pH of one unit reflects a tenfoldchange in the H+ ion concentration.
4. The immense range of concentrations from >1M to < 1 -14 M is compressed into a moreconvenient range, from ~ -1 to ~ 15
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Example 17 - 3What are the concentrations of hydronium and
hydroxide ions in a beverage whose pH =
3. 5?
Thi s i s easy if you k now how to use your
calculator! Also, t hi s i s tr ivial grade-school mat h.
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Re
l ev
ant Equation...
1
pH
H
« » !- ½
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Alternate way for [OH
]
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Other µp¶ scales Remember when we studied equilibrium
constants?
We also discussed equilibrium constants for acidsand bases.
These constants are usually quite small, so we canalso use the logarithmic function on K a and K b
p K a = - log K a
p K b = - log K b
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Generalizatio n (for a general qua n tity, X)...
1log
log
pX X
X
!
|
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Example 17 - 4What is the pH of a .25M solution of NaOH?
Remember t hat we are deal ing wit h a base
and, t hus, t here i s a l ittle more wor k. W e
shall do t hi s two ways...
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The quickest way (maybe)...
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The other way...
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Overview of pH and pOH
Given any one of these following: H3O , OH ,
pH or pOH, the other three can be determined.
pH = log H3O H3O
= 1 pH
pOH = log OH OH = 1 pOH
H3O OH = K w pH + pOH = pK w
At 25. °C:
H3O OH = 1. x 1
At 25. °C:
pH + pOH = 14.
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P lease remember...
P lease remember that the following are the
same thing (but that one is simpler to
write)...
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17.3 Weak Acids and Bases Think back to the table of strong acids
and bases. There aren¶t too manylisted.
What about the other acids and bases? They are referred to as weak acids and
bases.
Why?
They do not totally dissociate in
solution. ± Weak acids contain the acid and water as
major species.
± Weak bases contain the base and water asmajor species
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Weak Acids Proton transfer to water is not quantitative
Therefore, there exits an equilibrium where
only a small fraction of the acid moleculeshave transferred their protons to water
In a solution, the major species are water
molecules and the acid, HA.
HA + H2O H3O+ + A- 3
HA
AO H K
a
!
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Percent Ionization
The hydronium ionconcentration is atequilibrium.
The weak acidconcentration is theinitial concentration .
initial
eq ionized
!
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Example 17 - 5The pH of a .25 M aqueous HF solution is
1.92. Calculate the K a for this weak acid.
Thi s i s an e xam ple o f gett ing K g iven t he
concentrat ions at equil ibr ium. W e shall do
t hi s now b y just showing t he ste p s ( no comments)...
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Is it usually true that, for a weak
acid (HA) dissolved in water,
H+ = A ?
A. Yes
B. No
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This is because...A. of stoichiometry.
B. the solution must be neutral.
C. because I said so!!!
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When would H+
= A
not betrue for an acid, HA.
A. HA is a strong acid.
B. HA is extremely dilute (< 1 7 M).
C. HA is extremely weak ( p K a > 12).
D. B and C true; A false.
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Weak Bases Proton transfer from water is not quantitative.
Therefore, there exists an equilibrium where only
a small fraction of the base molecules haveaccepted protons from water.
In a solution, the major species are water
molecules and the base, A.
A + H2O HA + OH
!
A
A K
b
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What about neutral bases?
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Example 17 - 7Ammonia has K b = 1.8 x 1 -5. What is the pH
of .25 M aqueous ammonia?
Remember a base i s invol ved! S o t here may
be a l ittle e xtra wor k ! Since we have seen
t hi s sort o f t hing be f ore, we can ski p t he RICE ...
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Recognizing Acids and Bases
Oxoacids ± an acid that contains an inner atom bonded to a variablenumber of oxygen atoms and acidic OH groups. HxEOy
S ome Weak Oxoacids
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Carboxylic Acids
RCO2H
All carboxylic acids are weak acids.
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Representative Polyprotic Acids
S ome Weak Acids
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Example 17 - 8
Examine the following formulas. Decide if each represents a strong acid, a weak acid
or neither. Justify your conclusions.
(a) Cl3CCO2H
(b) CH3CH2CH2OH(c) HCN
(d) HClO4
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A n
sw e
rs...(a) Cl3CCO2H
A weak acid (but stronger because of the chlorines onthe terminal carbon).
(b) CH3CH2CH2OH Not an acid at all. This is a simple alcohol.
(c) HCNA weak acid (but don¶t breathe it!)
(d) HClO4
A strong acid. In fact, the strongest known in aqueoussolution!
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How does water stack up as a weak
acid?
? A2
14
2
14
16
1. 1 @ 25
1. 1
55.5
1.8 1
a H O
K C
H OH K
H O
! v r
« » « »- ½ - ½
v!
! v
w
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Weak Bases
Water is a weak base
Another common weak base is NH3
Many other weak bases are derivatives of ammonia called amines.
Some of the N ± H bonds have been replaced with C ± H bonds.
R epresentative Organic Bases
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How does water stack up as a weak
base?
? A2
14
2
14
16
1. 1 @ 25
1. 1
55.5
1.8 1
b H O
K C
H OH K
H O
! v r
« » « »- ½ - ½
v!
! v
w
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Conjugate Acid ± Base Pairs
A pair of compounds which differ only by
one proton.
± After an acid donates a proton, the remaining
species turns into a conjugate base (CB).
± After a base accepts a proton, the remaining
species turns into a conjugate acid (CA).
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HF (aq) + H2O (l) H3O+ (aq) + F- (aq)
- H +
+H +
Acid Base C on jugate
Acid
C on jugate
Base
A + B C A + C B
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Example 17 - 1
Write the chemical formula and the Lewisstructure and draw a molecular picture of
each of the following:(a) The conjugate acid of NH3
(b) The conjugate base of HCO2H
(c) The conjugate acid of HSO4
-
(W e¶ll show t he answers on t he ne xt two sl ides. )
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The answers (formulas):
(a) The conjugate acid of NH3 is NH4+.
(b) The conjugate base of HCO2H is HCO2.
(c) The conjugate acid of HSO4 is H2SO4.
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What is the conjugate acid of
water?A. H2O itself
B. H3O+
C. OH
D. H2SO4
E. Lime Jello
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What is the conjugate base of
H3O+?A. H2O
B. H3O+ itself
C. OH
D. H2SO4
E. Luciferin
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What is the conjugate base of
water?A. H2O itself
B. H3O+
C. OH
D. H2SO4
E. Lime Jello
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What is the conjugate acid of the
hydroxide ion?A. H2O
B. H3O+
C. OH
D. There can be none!
E. A sweet, singing, Eskimo Pie.
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What is the conjugate base of the
hydroxide ion?A. H2O
B. H3O+
C. OH
D. The oxide anion
E. There can be none since hydroxide only
exists in aqueous solution.
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The answers (formulas):
(a) The conjugate acid of NH3 is NH4+.
(b) The conjugate base of HCO2H is HCO2.
(c) The conjugate acid of HSO4 is H2SO4.
(Here, we are continuing from what we were
doing earlier!)
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M ore details...
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17.5 Acidic and Basic
Salts and Oxides An aqueous solution of a soluble salt
contains cations and anions.
The ions can often have acid-base properties.
Anions that are conjugate bases of weak acids make a solution basic.
Cations that are conjugate acids of weak bases make a solution acidic.
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H2O (l) + F- (aq) HF (aq) + OH- (aq)
Notice the form for K b ...
This leads to an interesting relationshipwhen K w enters in...
!
F
OH HF
K b
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K a
and K b
and K w
How are the three related?
K a K b = K w
Where K a is the ionization constant for a weak acid and K b is the ionization constant for itsconjugate base.
p K a + p K b = 14. (25.
C)
( N ote that we have already derived this!)
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At 25rC, p K b = 4.35 for a given base. What is p K a at this
temperature?
Error tolerance: s0.01
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At 25rC, K a = 4.7 v 107.Calculate p K b.
Error tolerance: s0.01
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Chapt er 16
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Important points about K a
and K b
Acid
Strength K a
Conjugate Base
Strength K b
Strong >1 Very Weak < 10-16
Weak 10-16 to 1 Weak 10-16 to 1
Very Weak < 10-16 Strong >1
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Example 17 - 11
Sodium hypochlorite (NaOCl) is the active
ingredient in laundry bleach. Typically,
bleach contains 5.0% of this salt by mass,which is a 0.67 M solution. Determine the
concentrations of all species and compute
the pH of laundry bleach.
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Steps for solution...
Write the equilibrium expressions.
Get the needed constants.
Set up and solve the equations. Determine the relevant species present and
get their concentrations.
W e shall do t hi s on t he ne xt sl ides wit hminimal comment ...
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Relevant species...
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Example 17 - 12
What are the important acid-base equilibria in
an aqueous solution of pyridinium chloride
(C5H5 NHCl)? What are the values of their equilibrium constants?
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T hus you should a n sw er...(Note the three equation forms for the 1st equilibrium!)
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17.6 Factors Affecting Acid Strength
Effect of charge ± affects the ability to
donate and accept protons (remember
opposites attract)
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Structural Factors
In order to donate a proton, a molecule must break a H ± X bond.
This becomes easier as bond strengths decrease,
and therefore the acids become stronger.
It is also affect by polarity.
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Example 17 - 13
Oxalic acid, HO2C ± CO2H, has K a1 = 1.3 x
10-2 an K a2 = 1.4 x 10-4. Formic acid
HCO2H, has K a = 1.8 x 10-4. Explain whythe first proton of oxalic acid is
substantially more acidic than the proton of
formic acid, but the second proton is less
acidic.
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A little reasoning...
The more polar a bond is,
± the less covalent it is!
The less covalent it is, ± the weaker it is! (That is, the electron density
between the atoms decreases.)
Thus polar bonds
± break more easily in polar media! (That is, polar solvents.)
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T hus, the solutio n com eth to
mynde... The COOH group is quite electron
withdrawing.
Thus, the first proton in HOOC-COOH ismore easily removed than is the proton inHCOOH! (It¶s OH bond is more polar!)
The second proton in oxalic acid is more
difficult to remove since we now have anegatively charged species, HOOC-COO.
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N ecessary point...
The simple arguments used here and in the
book presuppose that K 1a >> K 2a.
Usually, if they are three orders of
magnitude apart (or more), the simple ideas
hold!But, if they are closer together, all bets are off!
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C oncentrations of species in an
aqueous solution of a diprotic acid:
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Example 17 - 14
Carbonated water contains carbonic acid, a diproticacid that forms when carbon dioxide dissolves inwater.
CO2 (g) + H2O (l) H2CO3 (aq)
A typical carbonate beverage contains 0.050 M
H2CO3. Determine the concentrations of the ions present in this solution.
(W e sol ve t hi s wit h minimum comments... )
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The solution...
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Example 17 - 16
Potassium sulfite is commonly used as a food
preservative, because the sulfite anion undergoes
reactions that release sulfur dioxide, an effective preservative. Determine the concentrations of the
ionic species present in a solution of potassium
sulfite that is 0.075 M.
(When you sol ve f or O H
, t hings pretty much parallel what we just d id wit h H + f or a d i prot ic
acid!)
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Some points here...
We shall look at a general case first.
Again, we get a simple solution if the K -
values are separated by several orders (3 atleast²4 or more better) of magnitude.
The derivation is quite straightforward and
will be done with no comment on the slides.
The Derivation
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T he Deri vatio n ...
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F or this example...
The solution (no comments)
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http://slidepdf.com/reader/full/johnston-chapter-17-notes 98/114
The solution (no comments)...
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 99/114
C omment...
There is a ³symmetry´ between solving for
H+ in acidic solution and OH] in basic
solution. In the latter case, K b¶s replace K a¶s and we
solve for OH] directly instead of H+].
Nothing pHurther need be said unless youenjoy suffering!
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 100/114
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 101/114
Some exact solutions...
You have been given some methods that
work quite well in many cases.
However, at times, you will encounter errors with very dilute solutions or very
small K a (with H+) or very small K b (with
OH
) values.
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 102/114
Strategy with exact solutions...
Write out all equilibrium constants.
Impose conservation of mass.
Impose conservation of electrical charge.
Combine the equations, and solve for H+]
in acidic solutions (or OH] in basic
solutions; we shall not discuss this latter case).
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 103/114
Q uestio n
This problem is mean, sneaky, anddeceptive!
± Thus, I love it! Here it is:
± What is the pH of a 1.0 x 108 M aqueoussolution of HClO4?
± If you said ³ pH = 8.0,´ please put on your dunce cap! (Please remember that water itself already has H+] = 1.0 v 107 M.)
The exact solution
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 104/114
The exact solution...
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 105/114
F or 1.0 v 10-8 M HClO4...
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 106/114
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 107/114
A ³monster case´...
We shall end this discussion with an exactsolution for a triprotic acid .
Why? ± This is about the most complicated case you
will ever encounter (e. g ., H3PO4).
± Having this, the diprotic and monoprotic cases
are easily derived. Johnston has programmed hexaprotic acids
in his calculator. He is, of course, ashowoff...
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 108/114
The defining restraints on the system...
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 109/114
f g y
Algebra pHun, part I:
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 110/114
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 111/114
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 112/114
Diprotic acid...
Just set K 3 = 0 and go from there...
Monoprotic acid
8/8/2019 Johnston Chapter 17 Notes
http://slidepdf.com/reader/full/johnston-chapter-17-notes 113/114
M onoprotic acid...
Just set K 2 = 0, leaving only K 1 and K w...
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