I will try to structure our tour of chemistry in 3150:153 around three Big Questions:
1. What’s in a solution? How far does a reaction “go”?
2. What factors influence how far a reaction “goes” and how fast it gets there?
3. How do atomic and molecular structure influence observed properties of substances?
As chemistry is very integrated, it’s possible that sometimes we will think about something that deals
with more than one question at a time.
Big Question #1
What’s in a solution? How far does a reaction “go”?
Review (Chapter 4) & Section 11.5
Solutions: Components and Proportions
Air
Sterling silver
Mountain Dew
Common aspects?
 solute:
 solvent:
Quantitative Expressions (Units) of Concentration
Chemistry style for units with
denominators uses negative exponents.
For example, “meters per second”
would be written m∙s–1.
 What is the molarity of a 28% by mass solution of ammonia with a density of 0.90 g∙mL –1?
Big Question #1
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 What is the molality of a 15% by mass solution of sodium chloride? (NaCl: 58.44 g∙mol –1)
Saturation and Solubility; Equilibrium in Saturated Solutions
Big Question #1
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Chapter 15
Chemical Equilibrium
Consider 1.00 mol of N2O4 in a 1.00-liter flask at 25 °C and this equation: N 2O4(g)
What will be the concentration of NO2 when the reaction reaches completion?
2NO2(g)
Analysis reveals that after a very long time, [N2O4] = 0.85 M and [NO2] = 0.30 M.
Years later, the concentrations are unchanged at the same temperature.
“Completion” in the 3150:151 sense does not work for lots of reactions!
Many reactions achieve a dynamic chemical equilibrium.
 a state where the concentrations of products and reactants remain constant over time
 a dynamic state microscopically—not static
 a state where the rates of the forward and reverse reactions are equal
reactants
products
So, what is the equilibrium position? How far does the reaction “go”?
In general, for a A + b B c C + d D
[C]c [D]d
K=
where [X] ≡ the equilibrium concentration of X
[A]a [B]b
The value of the equilibrium constant for a reaction depends on the temperature.
[NO2 ]2
[0.30]2
For the reaction discussed above, K =
=
= 0.106 @ 25 °C
[N 2 O4 ]
[0.85]
Strictly speaking, K for the reaction comes out to 0.106 mol·L−1 using molarities, but equilibrium
constants don’t really have units. They are actually expressed in terms of activities instead of molarities.
In solutions that are dilute enough, molarity ≈ activity. The same argument applies to the use of partial
pressures in K expressions (see text) where the partial pressures are divided by a reference value (1 atm).
Furthermore, the distinction in the text between KC and KP is really artificial—for reactions that involve
both gaseous and aqueous substances, it’s possible to write a K that includes pressures and molarities.
K small: mostly reactants at equilibrium
K large: mostly products at equilibrium




K describes the ratio of products to reactants.
K can be measured experimentally.
K has predictive power—tied to how favorable a reaction is.
K can be used to predict which way a reaction would need to “go” to reach equilibrium if it’s not there.
Reaction Quotient, Q
The reaction quotient expresses the ratio of products to reactants at any time (not just at equilibrium).
Q is written the same way as K (“products over reactants”).
(NO 2 ) 2
For the reaction discussed above, Q =
where (X) ≡ the concentration of X at any time
(N 2 O 4 )
Big Question #1
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3150:153–004/801
Q vs. K

If Q < K, the reaction must shift to the right (toward products) to reach equilibrium.

If Q > K, the reaction must shift to the left (toward reactants) to reach equilibrium.

If Q = K, the reaction is at equilibrium.
K (and Q) Expressions for Heterogeneous Equilibria
Pure solids and liquids do not appear in K (or Q) expressions.
Usual excuse for solids and liquids not
appearing in K expressions:
“they are constant in concentration.”
Actual reason: the activity of a solid or
liquid equals exactly 1.
Dealing with K values

When an equation is reversed, the K value for the reverse reaction is the reciprocal of the original.
2H2O(l)
H3O+(aq) + OH–(aq)


H3O+(aq) + OH–(aq)
K = 1.0 × 10–14
2H2O(l)
When an equation is multiplied by a coefficient, the K value is raised to that power.
2KClO3(s)
2KCl(s) + 3O2(g)
KClO3(s)
KCl(s) + 2O2(g)
K = 6.5 × 107
3
When equations are added, the overall equation’s K value is the product of the individual K values.
Ag+(aq) + 2CN–(aq)
Zn(CN)2(s)
Ag+(aq) + Zn(CN)2(s)
Big Question #1
Ag(CN)2–(aq)
K = 5.3 × 1018
Zn2+ (aq) + 2CN–(aq)
K = 3.0 × 10–16
Ag(CN)2–(aq) + Zn2+(aq)
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Influencing the Equilibrium Position
LeChatelier’s Principle: A change in one or more variables describing a system at equilibrium produces
a shift in the equilibrium position that counteracts the effect of the change.

Changes in Concentration
o equilibrium position changed
o K unchanged
e.g., Fe3+(aq) + SCN–(aq)

Fe(SCN)2+(aq)
Changes in Pressure
o equilibrium position usually changed
o K unchanged
1. Add/remove gaseous reactant/product
 same principle as for concentration change; equilibrium shifts to “use up” some of
added substance or produce some of removed substance
2. Add inert gas (not involved in reaction)
 total pressure increases, but no change in equilibrium position
3. Change volume of container
 equilibrium position changes to favor side of reaction with fewer moles of gas if V
decreases, or to side of reaction with more moles of gas if V increases.
[NH 3 ]2
e.g., N2(g) + 3H2(g) 2NH3(g)
K=
[N 2 ][H 2 ]3
suppose that at some temperature, K = 1.0 and all [] = 1.0. Cut the volume in
2
 1.0 mol 


0.5 L 

half…now Q =
= 0.25 < 1.0. Q < K, so shift right
3
 1.0 mol   1.0 mol 



 0.5 L   0.5 L 
 If the gaseous moles are the same on both sides of the reaction, a volume change will
not cause any change in equilibrium position.
[I ][Br ]
e.g., 2IBr(g) I2(g) + Br2(g)
K = 2 22
[IBr]
suppose that at some temperature, K = 1.0 and all [] = 1.0. Cut the volume in
 1.0 mol   1.0 mol 



0.5 L   0.5 L 
half…now Q = 
= 1.0. Q = K, so no change in equil. position
2
 1.0 mol 


 0.5 L 
Big Question #1
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
Changes in Temperature
o equilibrium position changed
o K also changed
e.g.,
N2O4(g)
2NO2(g)
ΔH° = +58 kJ
endothermic—heat is like a reactant
When warmed, it’s as if a reactant is being added, so reaction shifts toward products.
Application: Haber-Bosch Process
N2(g) + 3H2(g)
2NH3(g)
ΔH° = –92 kJ
exothermic—product favored at lower T
But, reaction is very S-L-O-W at low T
What other factors can be manipulated to maximize product yield?
Big Question #1
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Section 16.10
Aqueous Solubility Equilibria
Knowing that we do not include pure liquids or solids in K expressions, write K expressions for each of
the following reactions of “insoluble” compounds slightly dissolving in water.
Ag+(aq) + Cl–(aq)

AgCl(s)

HgI2(s)

Cr(OH)3(s)

Ca3(PO4)2(s)
Hg2+(aq) + 2I–(aq)
Cr3+(aq) + 3OH–(aq)
3Ca2+(aq) + 2PO43–(aq)
The K expressions written for “insoluble” compounds dissolving in water are called KSP expressions,
after the name “Solubility Product.” Note that there is no denominator in any of the expressions—only
the product of the terms for the cations and anions. In general, the smaller the value of KSP is, the less
soluble a compound is. It’s not uncommon for KSP to have values of 10–20, 10–30, or even 10–50 or less
for certain compounds.
Consider mercury(II) iodide in otherwise pure water at 25 °C. KSP = 3.0
HgI2(s)
10–29.
Hg2+(aq) + 2I–(aq)
What would happen to the solubility of this compound if iodide ion were added (from a soluble
compound like sodium iodide)? Hint: consider LeChatelier’s Principle.

common ion effect: addition of an ion already present in a solution at equilibrium results in the
equilibrium position shifting away from producing more of that ion.
Consider chromium(III) hydroxide in otherwise pure water at 25 °C.
Cr(OH)3(s) Cr3+(aq) + 3OH–(aq)
What would happen to the solubility of this compound if a base (OH –) were added? Hint: consider
LeChatelier’s Principle.
Conversely, what would happen to the solubility if an acid (H 3O+) were added? Hint: again, consider
LeChatelier’s Principle.
For insoluble compounds with like stoichiometry (e.g., AgI, AgBr, AgCl or Ca 3(PO4)2, Sr3(PO4)2,
Ba3(PO4)2), the value of KSP corresponds to the relative solubility. The smaller the value of KSP, the less
soluble the compound. Thus compounds can be separated by differences in their KSP values as an ion
common to the compounds is changed in concentration.
Big Question #1
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Chapter 16 (sections 1–9)
Acids, Bases, & Acid-Base Equilibria
Many models have been proposed over hundreds of years to describe acids and bases.
Observations about acids:
 taste sour (DO NOT taste lab chemicals!)
 tend to react with metals to produce H2 gas
 turn blue litmus red
 tend to react with bases to produce water and a salt
Observations about bases:
 feel slippery and often taste bitter (DO NOT touch or taste lab chemicals!)
 turn red litmus blue
 tend to react with acids to produce water and a salt
Common household acids include aspirin, the juices of citrus fruits, vinegar
Common household bases include milk of magnesia, ammonia, lye
Brønsted-Lowry Definition/Model of Acids and Bases
(early 20th C., two scientists working separately who got the same results)
 Brønsted acid: a substance that donates H+ to another substance, i.e., a proton donor.
All Arrhenius acids are also Brønsted acids, but not vice versa.
 Brønsted base: a substance that accepts H+ from another substance, i.e., a proton acceptor.
Arrhenius bases contain the Brønsted base OH–.
Note that water can react as an acid or a base—this behavior is referred to as amphoterism (adjective:
amphoteric).
Most acid-base equilibria we discuss will be modeled using the Brønsted definition.
Big Question #1
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Conjugate Acid-Base Pairs
Consider the forward and reverse reactions:
NH3(aq) + H2O(l)
NH4+(aq) + OH–(aq)
NH4+(aq) + OH–(aq) (forward)
NH3(aq) + H2O(l)
(reverse)
Members of a conjugate acid-base pair differ by a proton (H+). e.g., A conjugate acid has one more H
atom and one more positive charge (or one fewer negative charge) than the base it came from.
Acid-base reactions involve a competition between the bases for the proton (H +): where the proton ends
up depends on the relative strengths of the acids and bases, i.e., on K for the equilibrium process.
(Recall “strong” vs. “weak” acids and bases in Chapter 4; we will revisit this concept in much greater
depth shortly.)
Acid-Base Properties of Water
2H2O(l) H3O+(aq) + OH–(aq)
KW = [H3O+][OH–]
Measurements of “pure” water show that
[H3O+]W = [OH–]W = 1.0
 KW = 1.0
In any aqueous solution,
10–7 M at 25 °C
10–14 at 25 °C.
KW = [H3O+]TOT[OH–]TOT
Auto-ionization of water is already always happening in water. When an acid or base is added, another
equilibrium joins in. Then there are two or more equilibria!
Since the product [H3O+]TOT[OH–]TOT is constant (= KW), the larger [H3O+] is, the smaller [OH–] is (and
vice versa). In general:
If [H3O+] = [OH–], the solution is NEUTRAL
If [H3O+] > [OH–], the solution is ACIDIC
If [H3O+] < [OH–], the solution is BASIC
Big Question #1
9
You may know of a definition of
“neutral” that involves the number 7.
That is correct only at 25 °C. The
definition above is general to all
temperatures.
3150:153–004/801
pH scale
[H3O+] varies over a very wide range. The pH scale simplifies expressing acidity/basicity.
 1 
pH = –log[H3O+] = log 
+ 
 [H 3O ] 
.
(really an approximation; just as we did for K expressions we will use molar concentrations
instead of activities. The true definition uses the activity of H 3O+, not molarity.)
“Strong enough for a man;
pH balanced for a
woman”
pH frequently ranges between 0 and 14, but can be negative (a very acidic solution) or greater than 14 (a
very basic solution).
Regarding sig figs and logs: the number of sig figs appears after the decimal point in the log.
e.g., a solution that has [H3O+] = 0.013 M has pH = –log(0.013) =
“p function” ≡ negative log10 
pOH = –log[OH–]TOT
.
So, using the properties of logs,
[H3O+][OH–] = KW
log([H3O+][OH–]) = log KW
+
–
+
–
 at 25 °C,
log[H3O ] + log[OH ] = log KW
If pH = 7.00, the solution is NEUTRAL
–log[H3O ] –log[OH ] = –log KW
If pH < 7.00, the solution is ACIDIC
pH + pOH = 14.00 (at 25 °C)
If pH > 7.00, the solution is BASIC
Strength of Acids and Bases
For any acid HA in water, an equilibrium is established:
[H 3O + ][A  ]
HA(aq) + H2O(l)
H3O+(aq) +
A–(aq)
Ka =
[HA]
acid
base
conj. acid
conj. base
 By convention, the acids considered “strong” have Ka > 1 while “weak” acids have Ka < 1.
Recall that a strong acid is 100% dissociated in solution (Chapter 4).
 For very strong acids, the equilibrium position lies far enough to the right that the reaction is
essentially “complete” and the equation is usually written with a single arrow ().
e.g., HCl (a “selected” strong acid in Table 4.3), Ka ≈ 1 106
HCl(aq) + H2O(l) H3O+(aq) + Cl–(aq)
[H O + ][Cl ]
Ka = 3
≈ 1 106. Ratio of products to reactants is very large.
[HCl]
Therefore, we usually write HCl(aq) + H2O(l)  H3O+(aq) + Cl–(aq) with a single arrow.
Similarly for bases, an equilibrium is established:
[BH + ][OH  ]
B(aq) + H2O(l)
BH+(aq) + OH–(aq)
Kb =
[B]
base
acid
conj. base
conj. acid
 same arguments as above for acids (strong vs. weak, single vs. equilibrium arrows)
Big Question #1
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Molecular Properties and Acid Strength (Big Question #3 Alert!)
Recall (again) that in Chapter 4 you learned several selected strong/weak acids/bases. Now we can
consider these general trends:
Strong Acids
 HCl, HBr, HI
 Oxoacids where the number of
O atoms exceeds the number of
ionizable H atoms by 2 or more,
e.g., …
o HNO3
o H2SO4
o HClO4
Strong Bases
 Soluble metal oxides &
hydroxides (consider solubility
rules, Chapter 4), e.g., …
o Na2O
o Ba(OH)2
o CsOH
o SrO
Weak Acids
 HF
 Oxoacids where the number of O atoms exceeds the number
of ionizable H atoms by 1 or fewer, e.g., …
o HNO2
o H2SO3
o HOCl
 Acids with H not bonded to O or halogen (group 7A), e.g., …
o H2S
o HCN
 Carboxylic acids (R–COOH), e.g., …
[13.7]
o CH3COOH
Weak Bases
 NH3
 Amines (R3N, R2NH, RNH2), e.g., …
o (CH3)3N
o CH3CH2NH2
[13.5]
Explanations for a couple of the above
Nonmetal Hydrides
Oxoacids
 Same number of O atoms, different central atom
o Acid strength increases with increasing electronegativity of
central atom (O–H bond polarized)
HOI: Ka = 2.3 10–11
HOBr: Ka = 2.3 10–9
HOCl: Ka = 2.9 10–8

Nonmetal hydride acid
strength increases with
increasing ∆EN across a
period and decreasing bond
strength down a group.
Big Question #1
Different number of O atoms, same central atom
o Acid strength increases with increasing number of O atoms
(more electronegative O atoms pull electron density to
themselves, polarizing O–H bond)
HOCl: Ka = 2.9 10–8
HOClO: Ka = 1.1 10–2 (aka “HClO2”)
HOClO2: Ka = ~ 5.0 102
HOClO3: Ka = 1 108
Presence of Electron-Withdrawing Groups/Atoms
 Electron-withdrawing groups/atoms frequently increase acidity for
the same reason as additional O atoms increase acidity of oxoacids
with the same central atom—increased polarization of the O–H bond.
CH3COOH: Ka = 1.8 10–5
CCl3COOH: Ka = 2.2 10–1
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We will discuss several (four to be exact) different cases of equilibria of acids and bases.
(1)
(2)
HA(aq) + H2O(l)  H3O+(aq) + A–(aq)
2H2O(l)
H3O+(aq) + OH–(aq)
Ka big
KW small
[H3O+]TOT = [H3O+]HA + [H3O+]W
[H3O+]W small compared to [H3O+]HA. Can we neglect [H3O+]W? Perhaps…
If [H3O+]W ≤ 0.05[H3O+]TOT, then we can neglect [H3O+]W.
Formality (F) [Formal/analytical concentration]
concentration of solution as prepared
(number of formula masses of initially
undissociated solute per liter of solution,
regardless of subsequent dissociation)
Strong Acid Rules
1. If Facid > 4.5 10–7 M, then
a. water dissociation unimportant
b. [H3O+]TOT ≈ Facid
2. If Facid < 4.5 10–7 M, then
a. water dissociation important
b. solve using:
[H3O+]TOT2 – Facid[H3O+]TOT – KW = 0
Similar analysis applies for strong “monobasic” bases in water (e.g., NaOH):
Strong Base Rules
1. If Fbase > 4.5 10–7 M, then
a. water dissociation unimportant
b. [OH–]TOT ≈ Fbase
2. If Fbase < 4.5 10–7 M, then
a. water dissociation important
b. solve using:
[OH–]TOT2 – Fbase[OH–]TOT – KW = 0
(1)
HA(aq) + H2O(l)
(2)
2H2O(l)
H3O+(aq) + A–(aq)
Ka
H3O+(aq) + OH–(aq)
KW
[H3O+]TOT = [H3O+]HA + [H3O+]W
Both equations involve equilibria that favor reactants. Will [H3O+]W be small compared to [H3O+]HA?
i.e., can we neglect [H3O+]W? Just as for strong acids, it depends on the relative amounts produced by
the two reactions…
Big Question #1
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Weak Acid Rules
1. If KaFacid > 1.0 10–13, then
a. water dissociation unimportant
b. solve equilibrium problem
2. If KaFacid < 1.0 10–13, then
a. water dissociation important
b. solve using:
[H3O+]TOT =
K a Facid + K W
The treatment is analogous for weak bases:
B(aq) + H2O(l)
Weak Base Rules
2H2O(l)
BH+(aq) + OH–(aq)
Kb
H3O+(aq) + OH–(aq)
KW
1. If KbFbase > 1.0 10–13, then
a. water dissociation unimportant
b. solve equilibrium problem
2. If KbFbase < 1.0 10–13, then
a. water dissociation important
b. solve using:
[OH–]TOT =
K b Fbase + K W
Solving the equilibrium problem of a weak acid/base in water may be more complicated than it
first looks. The exact solution involves a system of four (nonlinear) equations in four unknowns!
To simplify life, we can use a couple of chemically reasonable and mathematically convenient
approximations.
Example: Calculate the pH of a solution prepared by dissolving 0.10 mol of acetic acid in enough water
to make 1.0 L of solution. Ka(CH3COOH) = 1.8 10–5.
STEP 0. Check rules. Must water
dissociation be included?
STEP 1. Write balanced chemical
equation and K expression; evaluate Q.
STEP 2. Evaluate magnitude of K.
Big Question #1
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3150:153–004/801
STEP 3. Add initial concentrations and
changes in concentration (∆c) to get
equilibrium concentrations.
STEP 4. Substitute into K expression.
STEP 5. Solve.
STEP 6. Check assumptions. If %
change is 5% or less, assumption is OK.
STEP 7. Check.
Big Question #1
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What if the 5% Rule assumption fails?
 Use quadratic formula
 Use successive approximations
e.g., What is the pH of a 0.10 F solution of chlorous acid? Ka(HClO2) = 1.1
HClO2(aq) + H2O(l)
H3O+(aq) + ClO2–(aq)
+

[H O ][ClO 2 ]
(I)
0.10
“0”
0
Ka = 3
C
–∆c
+∆c
+∆c
[HClO 2 ]
[E]
0.10–∆c
∆c
∆c
Ka = 1.1
10–2 =
(c)(Δc) (c)(Δc)
≈
0.10  c
0.10
Percent Dissociation
0.10(1.1102 ) =
M  % change = 33%!!!
(c')(Δc')
(c') 2
=
 c' = 0.027 M
0.10  c (0.10  0.033)
Further cycles give
(c")(Δc")
(c")2
–2
10 =
=
 c" = 0.028 M  ∆c = 0.028 M.
0.10  c'
(0.10  0.027)
 [H3O+] = 0.028 M
and pH = 1.55
Using successive approximations: Ka = 1.1
Then another round: 1.1
∆c =
10–2.
% dissociation =
10–2 =
amount dissociated
initial concentration
100%
e.g., from calculations above, 0.10 F acetic acid is 1.3% dissociated, and 0.10 F chlorous acid is 28%
dissociated. (by contrast, a 0.10 F strong acid is 100% dissociated.)
Out of 1000 acetic acid molecules in a 0.10 F solution, 13 are dissociated and 987 of them are
undissociated at any time. (1.3% or 0.013 × 1000 = 13)
Recap: Weak Monoprotic Acid Alone in Water
(I)
C
[E]
HA(aq) + H2O(l)
Facid
–∆c
Facid –∆c
H3O+(aq) + A–(aq)
“0”
0
+∆c
+∆c
∆c
∆c
[H 3O + ][A  ]
(c)(Δc) (c)(Δc)
Ka =
=
≈
∆c = [H3O+] = K a Facid
[HA]
Facid  c
Facid
[H3O+] = [A–] = ∆c, and [HA] = Facid – ∆c
Assumptions:
 Water dissociation
unimportant
(i.e., (H3O+)o = “0”)
 5% Rule satisfied
(i.e., Facid – ∆c ≈ Facid)
pH = –log[H3O+] (for a solution of an acid, should be <7 no matter how weak or dilute)
Consider the equation from above for a solution of either a very
weak acid or a very dilute solution: [H3O+]TOT = K a Facid + K W
Pure water: Facid = 0 so [H3O+]TOT = K W (1.0 × 10–7 M @ 25 °C)
Concentrated enough or strong enough acid to neglect water
dissociation: drop KW so [H3O+] = K a Facid (as in box above)
BothQuestion
bullets above
Big
#1 are special cases of the first equation
15 really.


Occasionally someone argues that they and
their calculator can solve a weak acid
equilibrium problem exactly without the
simplifying assumptions/approximations.
Good for them—they should still be able to do
it the way above though.
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Diprotic Acids (FYI—probably not done in detail in class)
Diprotic acids have two ionizable protons that undergo successive ionization.
[H O + ][HA  ]
H2A + H2O H3O+ + HA–
K a1 = 3
[H 2 A]
[H3O + ][A 2 ]
[HA  ]
In general, Ka1 >> Ka2. (Harder to remove H+ from something already negatively charged.) In this case,
the two ionization reactions can be treated as occurring stepwise and we can proceed as follows:
HA– + H2O
H3O+ + A2–
K a2 =
The calculation of [H3O+], [HA–], and [H2A] start exactly the same way as for monoprotic acids.
From the first dissociation (Ka1),
o [H3O+]1 = [HA–]1 = ∆c (solved using assumptions, successive approximations, or quadratic
formula as needed) and


o [H2A] = Facid – [H3O+] (i.e., Facid –∆c as above).

Then, the second ionization step can be considered:
(I)
C
[E]
HA–(aq) + H2O(l)
[HA–]1
–∆c
[HA–]1 –∆c
H3O+(aq) + A2–(aq)
[H3O+]1
0
+∆c
+∆c
[H3O+]1 + ∆c ∆c
+
+
[H3O + ][A 2 ] [H3O ]1  c   c  [H3O ]1  c 
K a2 =
=
≈
[HA  ]
[HA  ]1  c
[HA  ]1
So,
[H3O+]TOT ≈ [H3O+]1
[HA–]TOT ≈ [HA–]1
[H2A] = Facid – [H3O+]
[A2–] = Ka2
∆c = Ka2
Same [H3O+], [HA–], and [H2A] as if only the first ionization
were considered. The second ionization…
(1) uses up so little HA– and (2) produces so little additional H3O+
…that it doesn’t contribute significantly to those quantities.
conjugate pair
HA(aq) + H2O(l)
HA & A–
A–(aq) + H2O(l)
H3O+(aq) + A–(aq)
HA(aq) + OH–(aq)
Ka (acid)
Kb (conj. base)
For any conjugate acid-base pair, KaKb = KW
As an acid gets stronger, its conjugate base gets weaker (and vice versa)
Salts containing ions that hydrolyze (i.e., ionize in water)
e.g., CH3COONa contains CH3COO–, the conjugate base of CH3COOH (a weak acid),
and Na+, the cation of NaOH (a strong base).
CH3COO– is basic. Na+ is not acidic.  CH3COONa solution is basic.
e.g., NH4Cl contains NH4+, the conjugate acid of NH3 (a weak base),
and Cl–, the conjugate base of HCl (a strong acid).
NH4+ is acidic. Cl– is not basic.  NH4Cl solution is acidic.
Big Question #1
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Quantitative treatment: Salt of a Weak, Monoprotic Acid in Water
(1)
A–(aq) + H2O(l)
(2)
2H2O(l)
HA(aq) + OH–(aq)
Kb(A–) =
H3O+(aq) + OH–(aq)
KW
[HA][OH  ]
[A  ]
Salt of Weak Acid Rules
1. If KbFsalt > 1.0 10–13, then
a. water dissociation unimportant
b. solve equilibrium problem
2. If KbFsalt < 1.0 10–13, then
a. water dissociation important
b. solve using: [OH–]TOT =
K b Fsalt + K W
e.g., what is the pH of a 0.10 M solution of sodium acetate? (Ka of acetic acid = 1.8
The treatment is analogous for salts of weak bases:
BH +(aq) + H2O(l)
Salt of Weak Base Rules
2H2O(l)
–13
1. If KaFsalt > 1.0 10 , then
a. water dissociation unimportant
b. solve equilibrium problem
2. If KaFsalt < 1.0 10–13, then
a. water dissociation important
b. solve using: [H3O+]TOT =
10–5.)
H3O+(aq) + B(aq)
Ka
+
–
H3O (aq) + OH (aq) KW
K a Fsalt + K W
Salts containing highly charged metal cations e.g., AlCl3, Fe(NO3)3, CrCl3
Al(H2O)63+ + H2O
H3O+ + Al(H2O)5(OH)2+ Ka = 1.0
2+
OH2
+ H 2O
+
H 3O +
H2O
OH2
Al
H2O
O
OH2
Big Question #1
10–5 (similar Ka to acetic acid!)
17
H
3150:153–004/801
Reminder about effect of molecular structure on oxoacid strength
More electron-withdrawing character in the rest of the molecule will polarize the O–H bond more
greatly, causing the acid strength to increase.
On rest of molecule…
more oxygen atoms
more electronegative central atom
O–H bond more polarized;
added electron-withdrawing groups
increase in acid strength
highly charged aquated metal ion
Salts, Conjugate acid-base pairs, and Drugs
Many drugs are formulated as salts since water solubility of a salt is often greater than that of a neutral
covalent compound. e.g., “pseudoephedrine HCl” means that the compound is present as the chloride
salt of its conjugate acid. Many compounds with amine groups are prepared as “hydrochlorides” this
way. Many weakly acidic compounds are prepared as the sodium salts of their conjugate bases (e.g.,
naproxen sodium).
WARNING! DO NOT do the following acid-base chemistry at home. It is illegal and harmful.
+ HCl
+ Cl
–
Salt of a Weak Acid and a Weak Base
e.g., Would a solution of methylammonium benzoate, CH 3NH3C6H5COO, be acidic or basic?
contains CH3NH3+ and C6H5COO– ions
CH3NH3+ = conjugate acid of CH3NH2 (Kb = 4.4 10–4)
K
1.0 1014
Ka of CH3NH3+ = W =
= 2.3
Kb
4.4 104
C6H5COO– = conjugate base of C6H5COOH (Ka = 6.3 10–5)
K
1.0 1014
Kb of C6H5COO– = W =
= 1.6
Ka
6.3 105
The same principle would apply to
determining whether an amphoteric
species (such as HCO3–) is acidic or
basic in solution. Calculate its Ka and
Big
Kb, and
theQuestion
larger value#1
would rule.
18
10–11
10–10
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
buffer solution: resists change in pH when small amounts of acid or base are added
It’s easy (relatively) to transform the Ka expression to relate pH to the ratio of A– to HA:
[H O + ][A  ]
[A  ]
Ka = 3
log Ka = log [H3O+] + log
[HA]
[HA]
Ka = [H3O+]
[A  ]
[HA]
– log Ka = – log [H3O+] – log


 [A ] 


log Ka = log   H 3O 

[HA] 

[A  ]
[HA]
[A  ]
pKa = pH – log
[HA]
pH = pKa + log
[A  ]
[HA]
Henderson-Hasselbach Equation
Add OH– to a buffer: some H3O+ reacts to consume it.
HA(aq) + H2O(l)
H3O+(aq) + A–(aq)
[A  ]
[A–] ↑ ; [HA] ↓ ;
increases; pH increases slightly
[HA]
HA(aq) + H2O(l)
H3O+(aq) + A–(aq)
Add H3O+ to a buffer: some A– reacts to consume it.
[A  ]
[A–] ↓ ; [HA] ↑ ;
decreases; pH decreases slightly
[HA]
¡¡¡Relative amounts of HA and A– are important!!! Large amounts of HA and A– will keep relative
changes small when small amounts of acid or base are added.
buffer capacity: measure of the ability of a buffer to resist pH change—the amount of acid/base a buffer
can “absorb” without “significant” change in pH—usually defined as ± 1 pH unit, “buffer range”
Optimal buffering occurs when [A–] = [HA].
o When [A–] = [HA], solution will buffer equally well against addition of acid or base.
o If not equal, the buffer will be better against either acid or base than the other.
[A  ]
pH = pKa + log
, so if [A–] = [HA], then pH = pKa.
[HA]
e.g., if you want a pH 4 buffer with [A–] = [HA], you need to find an acid with pKa = 4, i.e., Ka = 10–4.
[BH  ]
[B]
+
e.g., if you want a pH 10 buffer with [BH ] = [B], you need to find a base with pKb = 4, i.e., Kb = 10–4.
B(aq) + H2O(l)
BH+(aq) + OH–(aq)
pOH = pKb + log
¡¡¡CAUTION!!!
It is NOT necessarily true that a “basic buffer” (a buffer solution with pH >7) must be made with a weak
base and its conjugate acid. For example, if you used equal concentrations of HA and A – where HA had
Ka=1.0×10–8, then the pH of that buffer would be 8.00.
Big Question #1
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

end point: where the indicator changes color in a titration
equivalence point: where exactly enough base (acid) has been added to react with the acid (base)
originally present in a titration.
Ideally, the end point and equivalence point are reached at the same time in your titration.
Weak Acid/Strong Base titration



HA(aq) + H2O(l) H3O+(aq) + A–(aq)
H3O+(aq) + OH–(aq)  2H2O(l)
HA(aq) + OH–(aq)  H2O(l) + A–(aq)
pH at equivalence point > 7.0
At equivalence point, solution contains water, the
cation of the strong base, and the anion of the
weak acid (its conjugate base). The anion of a
weak acid is basic, so the pH at the equivalence
point is > 7.
buffer region
Half-equivalence point
at “half-titration” (i.e., VBASE = 12VEP), half of the
original acid has been converted to its conjugate
base (i.e., [A–] = 12[HA]o). Thus [A–] = [HA] here,
so at this point, pH = pKa of the weak acid.
“HPr” = CH3CH2COOH, propionic acid
 Lewis acid: a substance that accepts a pair of electrons.
 Lewis base: a substance that donates a pair of electrons.
All Arrhenius/Brønsted acids are Lewis acids (ditto for bases)
Big Question #1
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