Acid-Base Chemistry
16.1-16.2The Nature of Acids and Bases
General info:
Acids: sour taste (ex: vinegar, citric acid)
Bases: bitter taste, slippery, alkalis
Arrhenius Concept
Arrhenius acid
Arrhenius base
produces H+
produces OH-
“flaw” – only allows for bases that have hydroxide ion
Bronsted-Lowry Model
Bronsted-Lowry acid
Bronsted-Lowery base
donates H+
accepts H+
Example: HCl + H2O  H3O+ + ClHCl = B-L acid
H2O = B-L base
H3O+ = hydronium ion
General Form:

HA(aq) + H2O(l)  H3O+(aq) + A-(aq)
acid
base
conjugate conjugate
acid
base
This represents a struggle for the proton between 2 bases: H2O and Ao If H2O stronger than A- , a greater affinity for the proton, the equilibrium position will be
far to the right
o If A- stronger than H2O, the equilibrium position will lie further to the left
Ka= [H3O+][A-] = [H+][A-]
[HA]
[HA]
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
,where Ka = acid dissociation constant
remember that we left pure liquids and solids out of the equilibrium equation previously; for dilute
solutions, we will assume that the concentration of liquid water stays virtually the same
o that’s why water is not in the Ka equation above
o can use HA(aq) qe H+(aq) + A-(aq) for simple dissociation
used only for when a proton is removed from HA – you can write Ka for any acid knowing this
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Acid Strength
The strength of an acid is defined by the equilibrium position of its dissociation reaction:
HA(aq) + H2O(aq)  H3O+(aq) + A-(aq)
Strong acid: equilibrium position lies far to the right
 Almost all of HA is dissociated at equilibrium
 Strong acid yields weak conjugate base (low affinity for a proton)
 An acid whose conjugate base is a weaker base than H2O
 Common examples: sulfuric, hydrochloric, nitric, and perchloric acids
 Ka is large
Weak acid: equilibrium position lies far to the left
 Most of HA is still present at equilibrium
 Weak acids yield strong conjugate base (high affinity for a proton)
 An acid whose conjugate base is much stronger than H2O
 Ka is small
Diprotic acid: has 2 acidic protons –
Example: H2SO4
Most acids are oxyacids: the acidic proton is attached to an oxygen atom
Organic acids have a carbon atom backbone and commonly contain a carboxyl group (CO2H)
Monoprotic acid: has 1 acidic proton
Example: HCl
16.3 Water as an acid and a base
Amphoteric: a substance that can act as either an acid or a base (water is common)
Autoionization of water: 2 H2O  H3O+(aq) + OH-(aq)
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Kw = [H3O+][OH-] = [H+][OH-] ,where Kw is the ion-product constant (or the dissociation constant)
[H3O+] = [OH-] = 1.0 x 10-7 M @25C
Kw = [H+][OH-] =

no matter what water contains the product of [H+] & [OH-] always equal 1.0 x 10-14
@25C
--neutral: [H+] = [OH-]
--acidic: [H+] > [OH-]
--basic: [H+] < [OH-]
Sample Exercise 14.3 – Calculate [H+] and [OH-] as required for each of the following solutions at
25C, and state whether the solution is neutral, acidic, or basic.
a. 1.0 x 10-5 M OH-
b. 1.0 x 10-7 M OH-
c. 10.0 M H+
Sample Exercise 4 – At 60C, the value of Kw is 1 x 10-13.
a. Using LeChatelier’s principle, predict whether the reaction 2 H2O  H3O+(aq) + OH-(aq) is
exothermic or endothermic
b. Calculate [H+] and [OH-] in a neutral solution at 60C.
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16.4 The pH Scale
Take your own notes over this section.
Key Equations:
pH = -log [H+]
[H+] = antilog (-pH)
POH = -log [OH-]
[OH-] = antilog (-pOH)
Sample Problem: Suppose the hydronium ion concentration of vinegar is 1.6 x 10-3M. Calculate the
pH.
Sample Problem: The pH of sea water if 8.30. Calculate pOH, [H+] and [OH-].
Other “p” scales – pOH - -log [OH-] and pKa = -log Ka
16.5 Calculating the pH of Strong Acid Solutions
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Always write the major species
Remember that strong acids dissociate completely in water
Determine whether water or the acid is contributing more H+
Sample Problem: Calculate the pH of 0.10 M HNO3.
Sample Problem: Calculate the pH of 1.0 x 10-10 M HCl.
Strong Base Solutions
 Calculate the pOH then subtract from 14 to get pH
Sample problem: What is the pH of a 0.011 M solution of Ca(OH)2?
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14.1 Calculating the pH of Weak Acid Solutions
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Look at what you have, understand where you are in the problem, and THINK!!!
Always write the major species.
Remember that weak acids dissociate to a small extent in water.
Write the balanced equation for all species with H+.
Use the equilibrium constants to determine whether water or the acid is contributing more H+.
Write the equilibrium expression
ICE it!
Use the 5% if appropriate – double-check it at the end!
Calculate the pH or [H+].
Sample Problem: Calculate the pH of a 0.020 M solution of benzoic acid if the Ka = 6.3 x 10-5.

Calculation of pH in a Mixture of Weak Acids
Sample Problem (14.9): Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2 x 10-10) and
5.00 M HNO2 (Ka = 4.0 x 10-4). Also calculate the concentration of cyanide ion in this solution at
equilibrium.
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Percent Dissociation / Percent Ionization: how much of the weak acid has dissociated into solution
% dissociation = amount dissociated (n/L) x 100
initial concentration (n/L)
Sample Problem (14.10): Calculate the percent dissociation for acetic acid (Ka = 1.8 x 10-5) for 1.00 M
and 0.1 M solutions.

In general, the more dilute the weak acid solution, the greater is the percent dissociation.
Sample Problem (14.11): Lactic acid is a waste product that accumulates in muscle tissue during
exertion,
leading to pain and feeling of fatigue. In a 0.100 M aqueous solution, lactic acid is 3.7% dissociated.
Calculate the value of Ka for this acid.
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14.2 Bases
Remember that bases to do not have to contain OHStrong Bases:
Completely dissociate in water
Group I and Ba and Sr
Have a large Kb
Weak Bases:
Do not dissociate to much extent in water
All others
Have a small Kb
Kb problems work very much the same as Ka problem – use the same steps for solving
Kb = [BH+] [ OH-]
[B]
Sample Problem (14.13): Calculate the pH of a 15.0 M solution of ammonia (Kb = 1.8 x 10-5).
Sample Problem (14.14): Calculate the pH of a 1.0 M solution of methylamine (Kb = 4.38 x 10-4).
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14.3 Polyprotic Acids
Acids always dissociate stepwise
For a typical weak polyprotic acid, Ka1 > Ka2> Ka3
Refer to Table 14.4 in Zumdahl.
Generally, for Polyprotic acids in water, the first step is the only important contribution to the [H+]
However, the 2nd step for dilute sulfuric does contribute significantly to [H+]
14.4 Acid-Base Properties of Salts
Salt: ionic compounds
We are looking to see what species has the highest affinity for the H+
Neutral Solution:

strong acid + strong base  salt + water
cations of strong bases and anions of strong acids have no affect on the pH of the solution
Basic Solution:
if the anion of the salt is the conjugate base of a weak acid
Sample Problem (14.18): Calculate the pH of a 0.30 M NaF solution. The K a value for HF is 7.2 x 10-4.
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
base strength in aqueous solutions
HCN + H2O  H3O+ + CN-
Ka = 6.2 x 10-10
CN- + H2O  HCN + OH-
Kb = Kw / Ka
Acidic Solution:
if the anion of the salt is not a base and the cation is the conjugate acid of a weak
base
Sample Problem (14.19): Calculate the pH of a 0.10 M NH 4Cl solution. The Kb value for NH3 is 1.8 x 10-5.
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You can also have acid solutions caused by highly charged metal ions such as aluminum; in
general, the higher the charge, the stronger the acidity of the hydrated ion
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You can have the situation where both cation and anion can affect the pH of the solution; in
this case, you must make a comparison of the Ka and Kb to predict whether the solution is
acidic or basic (Table 14.5 in Zumdahl)
Sample Problem (14.21): Predict whether an aqueous solution of each of the following salts will be acidic, basic, or
neutral.
a. NH4C2H3O2
b. NH4CN
c. Al2(SO4)3
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14.9 The Effect of Structure on Acid-Base Properties
H-X
- strength of bond
- polarity of bond
H-O-X
- HOCl, HO4Cl, HO5Cl
- O draws electrons away form the Cl atom and O-H bond  polarizes
and
weakens O-H bond  proton produced
- strength increase with an increase in the number of oxygen molecules
attached to the central atom
- good correlation between electronegativity of X and acid strength
Metal ions
- the greater the charge, the more acidic
14.5 Acid-Base Properties of Oxides
Anhydride: an oxide of a metal or non-metal that combines with water to give either an acid or a base
Acid anhydride: SO3 + H2O  H2SO4
Basic anhydride:
Na2O + H2O  2 NaOH
14.11 The Lewis Acid-Base model
Lewis acid
Lewis base
electron-pair donor
electron-pair acceptor
Acid: has an empty orbital it can use to accept (share) and electron pair from a molecule that has a one pair of
electrons (Lewis Base)
Lewis model encompasses the Bronsted-Lowry theory but the converse is not true
Key point: it cover reactions not involving Bronsted-Lowry acids
15.1 Solutions of Acids or Bases Containing a Common Ion
common ion effect
 shift in equilibrium position that occurs because of the
addition of an ion already involved in the equilibrium
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reaction
Sample Problem 15.1: A 1.0 M solution of HF dissociates to give 2.7 x 10-2M of H+. Calculate [H+] in a
solution containing 1.0M HF (Ka = 7.2 x 10-4) and 1.0 M NaF.
15.2 Buffered Solutions
buffered solution
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resist a change in its pH when acid / base added
made of an acid-base conjugate pair by mixing the
weak acid or base with a salt of that acid or base
Sample Problem 15.2 - A buffered solution contains 0.50 M acetic acid (Ka = 1.8 x 10-5) and 0.50M sodium
acetate. Calculate the pH of this solution.
Sample Problem 15.3 - Calculate the pH change that occurs when 0.010 mole solid NaOH is a to 1.0 L of the
buffered solution. Compare that to the change in pH when the NaOH is added to 1.0 L of water.
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Original
buffered
solution pH
Step 1
Do stoichiometry
calculation to determine
new concentrations.
Assume reaction with
H=/OH-goes to
completion.
Modified
pH
Step 2
Do
equilibrium
calculation
How does a buffer work?
Henderson-Hasselbalch equation
pH = pKa + log [A-] = pKa + log [base]
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[HA]
[acid]
For a particular buffering system, all solutions that have the same ratio [A-] / [HA] will have the same pH.
Sample Problem 15.4 - Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 x 10-4) and 0.25
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Usually buffers are made so that the concentrations of conjugate pairs are roughly equal. pH of a buffer  pKa.
Adding water to a buffer does not change the pH of the buffer, only the buffer capacity is affected.
15.3 Buffer Capacity
buffer capacity
 refers to the amount of acid or base that a buffer can
neutralize before its pH changes appreciably.
maximum capacity
 in general, concentrations of weak acid and its
conjugate base are kept large and approximately equal
to each other
applications of buffers
 blood is buffered to maintain a pH of 7.4
Sample Problem 15.7 – Calculate the change in pH that occurs when 0.010 n of gaseous HCl is added to 1.0 L of
each of the following solutions. The Ka of acetic acid is 1.8 x 10-5.
a. 5.00 M acetic acid and 5.00 M sodium acetate
b. 0.050 M acetic acid and 0.050 M sodium acetate
Acids & Bases
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The pKa of the weak acid to be used in the buffer should be as close as possible to the desired pH.
Sample Problem 15.8– chemist needs a solution buffered at pH 4.30 & can choose from the following acids (& their
sodium salt).
a. chloroacetic acid (Ka = 1.35 x 10-3)
b. propanoic acid (Ka = 1.3 x 10-5)
c. benzoic acid (Ka = 6.4 x 10-5)
d. hypochlorous acid (Ka = 3.5 x 10-8)
Acids & Bases
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15.4 Titrations and pH Curves
pH curve/titration curve
the progression of an acid-base titration, monitored by
plotting the pH of the solution being analyzed as a
function of the amount of titrant added
Equivalence (stoichiometric) point
the point in the titration where an amount of base has
been added to exactly react with all of the acid originally
present
Halfway point/Midway point
halfway to equivalence point
1 mole = 1mmole
1L
1 mL
Finding pH
Strong Acid-Strong-Base
1) Before equivalence point: calculate remaining H+ by dividing the mmol of H+ by the total volume in mL
2) At equivalence point: pH=7
3) After equivalence point: pH = excess OH- by dividing the mmol of OH- by the total volume in mL
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Each time you must do the calculations based on the original amounts present to cut out error
Remember that the volume used is additive (acid + base)
Near the endpoint note that the curve shows a dramatic change
 this is due to the fact that a small amount of base added will change the pH significantly because
there is little H+ present
 at the beginning, there was so much acid present that a small addition of base did not significantly
affect the pH
Example: Consider the titration of 50.0 mL of 0.200 M HNO3 with 0.100 M NaOH. We will calculate the pH of the
solution at selected point during the course of the titration where specific amounts of NaOH have been added.
Acids & Bases
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Weak Acid-Strong Base
1) Before equivalence point: To find pH is a 2-step process (each done separately)
 stoichiometry problem
 equilibrium problem
2) At equivalence pH is always greater than 7
3) After equivalence point: pH is determined by excess strong base (anion from acid doesn’t contribute
significantly)
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At half way point, [H+] = Ka
The weaker the acid, the greater the pH is at equivalence point.
The equivalence point is determined by the stoichiometry not by the pH.
Example: Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (Ka = 6.2 x 10-10) when
dissolved in water. If a 50.0 mL sample of 0.100 M HCN is titrated with a 0.100 M NaOH, calculate the pH of the solution
a. after 8.0 mL of 0.100 M NaOH has been added
b. at the halfway point
c. at the equivalence point of the titration
Weak Base-Strong Acid
Just like weak acid-strong base
Acids & Bases
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15.5 Acid-Base Indicators
The two ways to determine the equivalence point of a titration:
1. use a pH meter; plot the titration curve; the center of the vertical region indicates the equivalence point
2. use an acid-base indicator; this will mark the end point by a color change (although the end point is the
color change and the equivalence point is the stoichiometric point, choosing an appropriate indicator can
minimize the error between the two)
The Henderson-Hasselbalch equation is useful for determining the pH at which an indicator changes color.
pH = pKa + log ([In-] / [HIn])
where HIn(aq) qe H+(aq) In-(aq)
Example: Bromthymol blue, an indicator with a Ka value of 1.0 x 10-7, is yellow in its HIn form and blue in its Inform. Suppose we put a few drops of this indicator in a strongly acidic solution. If the solution is then titrated with
NaOH, at what pH will the indicator color change first be visible?
Review the pH ranges of common indicators found in Figure 15.8 of Zumdahl (pg. 734)
Acids & Bases
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