ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA

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ADVANCED PLACEMENT CHEMISTRYACIDS, BASES, AND
AQUEOUS EQUILIBRIA
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Acids- taste sourBases(alkali)- tastes bitter
and feels slippery
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Arrhenius concept- acids produce hydrogen ions in
aqueous solution while bases produce hydroxide
ions
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Bronsted-Lowry model- acids are proton (H)
donors and bases are proton acceptors
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Lewis model- acids are electron pair acceptors
while bases are electron pair donors
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hydronium ion (H3O)- formed on reaction of a
proton with a water molecule. H and H3O are
used interchangeably in most situations.
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HA(aq) H2O(l) ? H3O(aq) A-(aq)Acid
Base Conjugate Conjugate
Acid
Baseconjugate base- everything that remains of
the acid molecule after a proton is
lostconjugate acid- base plus a proton
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Acid dissociation constant (Ka)Ka H3OA-
or Ka HA- HA
HA
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Strong acid - mostly dissociated -
equilibrium lies far to the right - a
strong acid yields a weak conjugate base (much
weaker than H2O)Weak acid- mostly
undissociated - equilibrium lies far to
the left - has a strong conjugate base
(stronger than water)
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Common strong acids -all aqueous solutions (Know
these!)H2SO4 (sulfuric)HCl (hydrochloric)HNO3
(nitric)HClO3 (chloric)HClO4 (perchloric)HI
(hydroiodic)H2CrO4 (chromic)HMnO4
(permanganic)HBr (hydrobromic)
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Sulfuric acid is a diprotic acid which means that
it has two acidic protons. The first (H2SO4) is
strong and the second (HSO4-) is weak.
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Oxyacids- most acids are oxyacids -
acidic proton is attached to OWeak oxyacids
H3PO4 (phosphoric) HNO2 (nitrous)
HOCl (hypochlorous)
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Within a series, acid strength increases with
increasing numbers of oxygen atoms. For
example HClO4 gt HClO3 gt HClO2 gt HClO and
H2SO4 gt H2SO3 (Electronegative O draws
electrons away from O-H bond)
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Acid strength increases with increasing
electronegativity of oxyacids. For example
HOClgtHOBrgtHOIgtHOCH3
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Organic acids-
O have carboxyl group -C-OH -
usually weak acids CH3COOH acetic C6H5COOH
benzoic
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Hydrohalic acids- H is attached to a halogen
(HCl, HI, etc.)HF is the only weak hydrohalic
acid. Although the H-F bond is very polar, the
bond is so strong (due to the small F atom) that
the acid does not completely dissociate.
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Weak acid strength is compared by the Ka values
of the acids. The smaller the Ka, the weaker the
acid. Strong acids do not have Ka values because
the HA is so small and can not be measured
accurately.
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Amphoteric substance- Substance that can act as
an acid or as a base. Ex. H2O, NH3,
HSO4-(anything that can both accept and donate
a proton)
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autoionization of water H2O H2O ? H3O
OH- base acid conjugate conjugate
acid
base
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Ion product constant for water (Kw) Kw
H3OOH-
Kw HOH-At
25oC, Kw 1 x 10-14 mol2/L2 because H OH-
1 x 10-7 M
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No matter what an aqueous solution contains, at
25oC, HOH- 1 x 10-14 Neutral
solution H OH- Acidic solution
H gt OH- Basic solution H lt
OH-Kw varies with temperature
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pH -log HIf H 1.0 x 10-7 M, pH
7.00
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Significant figures in pH and other log values
The number of decimal places in the log value
should equal the number of significant digits in
the original number (concentration).
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pOH -log OH-pK -log KpH and pOH are
logarithmic functions. The pH changes by 1 for
every power of 10 change in H. pH decreases
as H increases.
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pH pOH 14H antilog(-pH) OH-
antilog(-pOH)
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Calculating pH of Strong Acid Solutions
Calculating pH of strong acid solutions is
generally very simple. The pH is simply
calculated by taking the negative logarithm of
concentration of a monoprotic strong acid. For
example, the pH of 0.1 M HCl is 1.0. However,
if the acid concentration is less than 1.0 x
10-7, the water becomes the important source of
H and the pH is 7.00. The pH of an acidic
solution can not be greater than 7 at 25oC!!!!!
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Another exception is calculating the pH of a
H2SO4 solution that is more dilute than 1.0 M.
At this concentration, the H of the HSO4- must
also be calculated.
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Ex. Calculate the H and pH in a 1.0 M
solution of HCl.

• HCl is a strong monoprotic acid, therefore its
  concentration is equal to the hydrogen ion
  concentration.
• H 1.0 M
• pH - log (1.0) 0.00

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Ex. Calculate the pH of 1.0 x 10-10 M HCl.

• Since the H is less than 1.0 x 10-7, the H
  from the acid is negligible and the pH 7.00

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Calculating pH of Weak Acid Solutions
Calculating pH of weak acids involves setting up
an equilibrium. Always start by writing the
equation, setting up the acid equilibrium
expression (Ka), defining initial concentrations,
changes, and final concentrations in terms of X,
substituting values and variables into the Ka
expression and solving for X.
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Ex. Calculate the pH of a 1.00 x 10-4 M solution
of acetic acid. The Ka of acetic acid is 1.8 x
10-5
HC2H3O2 ? H C2H3O2-
Ka HC2H3O2- 1.8 x 10-5
HC2H3O2
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• Reaction HC2H3O2 ? H
  C2H3O2-
• Initial 1.00 x 10-4 0
  0
• Change -x x
  x
• Equilibrium 1.00 x 10-4 - x x
  x
• 
  
• 
  Often, the -x in a Ka expression
• 1.8 x 10-5 (x)(x) can be treated
  as negligible. 1.00x10-4 - x
• 1.8 x 10-5 ? (x)(x)
  x 4.2 x 10-5 1.00 x 10-4

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When you assume that x is negligible, you must
check the validity of this assumption. To be
valid, x must be less than 5 of the number that
it was to be subtracted from. In this example
4.2 x 10-5 is greater than 5 of 1.00 x 10-4.
This means that the assumption that x was
negligible was invalid and x must be solved for
using the quadratic equation or the method of
successive approximation.
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Use of the quadratic equationx2 1.8 x
10-5x - 1.8 x 10-9 0x 3.5 x 10-5 and -5.2
x 10-5 Since a concentration can not be
negative, x 3.5 x 10-5 M x H 3.5 x 10-5
pH -log 3.5 x 10-5 4.46
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Another method which some people prefer is the
method of successive approximations. In this
method, you start out assuming that x is
negligible, solve for x, and repeatedly plug your
value of x into the equation again until you get
the same value of x two successive times.
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Using successive approximation for the previous
example would go as follows
x 4.2 x 10-5

x 3.2 x 10-5

x 3.5 x 10-5


x 3.4 x 10-5



x 3.4 x
10-5 H 3.4 x 10-5 pH 4.47
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or/ with a graphing calculator?(1.8 x 10-5 x
1.00 x 10-4) 4.2 x 10-5 (not negl) ?((1.8 x
10-5)( 1.00 x 10-4 -ans)) 3.2 x 10-5
3.5 x 10-5
3.4 x 10-5 3.4 x 10-5
Use answer key on calculator for this!
Press Enter key repeatedly until you get the same
answer each time
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Calculating pH of polyprotic acids All
polyprotic acids dissociate stepwise. Each
dissociation has its own Ka value. As each H is
removed, the remaining acid gets weaker and
therefore has a smaller Ka. As the negative
charge on the acid increases it becomes more
difficult to remove the positively charged
proton.
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Except for H2SO4, polyprotic acids have Ka2
and Ka3 values so much weaker than their Ka1
value that the 2nd and 3rd (if applicable)
dissociation can be ignored. The H obtained
from this 2nd and 3rd dissociation is negligible
compared to the H from the 1st dissociation.
Because H2SO4 is a strong acid in its first
dissociation and a weak acid in its second, we
need to consider both if the concentration is
more dilute than 1.0 M. The quadratic equation
is needed to work this type of problem.
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Ex. Calculate the pH of a 1.00 x 10-2 M H2SO4
solution. The Ka of HSO4- is 1.2 x 10-2

• H2SO4 ? H
  HSO4-
• Before 1.00 x 10-2 0 0
• Change -1.00 x 10-2 1.00 x 10-2 1.00 x 10-2
• After 0 1.00 x 10-2
  1.00 x 10-2
• Reaction HSO4- ? H
  SO4-
• Initial 1 x 10-2 1x 10-2
  0
• Change -x x
  x
• Equil. 1 x 10-2 -x 1 x 10-2 x
  x

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• Ka HSO4- 1.2 x 10-2 HSO4-
• 1.2 x 10-2 (1 x 10-2 x)(x)
• (1 x 10-2 -x)
• Using the quadratic equation,
• x 4.52 x 10-3
• H 1 x 10-2 (4.52 x 10-3)
• 1.45 x 10-2
• pH 1.84

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Determination of the pH of a Mixture of Weak
AcidsOnly the acid with the largest Ka value
will contribute an appreciable H. Determine
the pH based on this acid and ignore any
others.
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Determination of the Percent Dissociation of a
Weak Acid dissociation amt. dissociated
(mol/L) x100
initial concentration
(mol/L) final H x 100initial
HA
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For a weak acid, percent dissociation (or
ionization) increases as the acid becomes more
dilute. Equilibrium shifts to the right.

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BASESThe hydroxides of Group I and IIA metals
are all strong bases. The Group IIA hydroxides
are not very soluble. This property allows some
of them to be used effectively as stomach
antacids.
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Ex. Calculate the OH-, H, and pH of a
0.0100 M solution of NaOH.

• NaOH is a strong base.
• OH- 0.0100 M
• H 1 x 10-14/1 x 10-2 1.00 x 10-12 M
• pH - log 1.00 x 10-12 12.000

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Weak bases (bases without OH-) react with water
to produce a hydroxide ion. Common examples of
weak bases are ammonia (NH3), methylamine
(CH3NH2), and ethylamine (C2H5NH2).
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B(aq) H2O(l) ? BH(aq) OH-(aq)
base acid conjugate
conjugate
acid base NH3 H2O ? NH4
OH- base acid conjugate
conjugate acid
base The lone pair on N forms a
bond with a H. Most weak bases involve
N.
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Base dissociation constant (Kb)Kb BHOH-
Kb NH4OH- B
NH3
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Determination of the pH of a weak base is very
similar to the determination of the pH of a weak
acid. Follow the same steps. Remember, however,
that x is the OH- and taking the negative log
of x will give you the pOH and not the pH!
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Ex. Calculate the OH- and the pH for a 15.0 M
NH3 solution. The Kb for NH3 is 1.8 x 10-5.

• Reaction NH3 H2O ? NH4 OH-
• Initial 15.0 --- 0
  0
• Change -x --- x
  x
• Equil 15.0-x --- x
  x
• Kb 1.8 x 10-5 NH4OH- x2 ?
  x2
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  NH3 15.0-x 15.0
• x 1.6 x 10-2 OH-
• pOH -log 1.6 x 10-2 1.78
• pH 14-1.78 12.22

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Determination of the pH of Salts
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Neutral Salts- Salts that are formed from the
cation of a strong base and the anion from a
strong acid form neutral solutions when dissolved
in water. Ex. NaCl, KNO3
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Acid Salts- Salts that are formed from the
cation of a weak base and the anion from a strong
acid form acidic solutions when dissolved in
water. Ex. NH4Cl The cation hydrolyzes the
water molecule to produce hydronium ions and thus
an acidic solution.NH4 H2O ? H3O
NH3 strong
acid weak base
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Basic Salts- Salts that are formed from the
cation of a strong base and the anion from a weak
acid form basic solutions when dissolved in
water. Ex. NaC2H3O2, KNO2 The anion
hydrolyzes the water molecule to produce
hydroxide ions and thus a basic
solution.C2H3O2- H2O ? OH-
HC2H3O2
strong base weak acid
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When determining the exact pH of salt solutions,
we can use the Ka of the weak acid formed to find
the Kb of the salt or we can use the Kb of the
weak base formed to find the Ka of the salt.
Ka x Kb Kw
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Ex. Calculate the pH of a 0.15 M solution of
sodium acetate. Sodium acetate is the salt of a
strong base (NaOH) and a weak acid (acetic acid)
and thus forms a basic solution. The acetate ion
hydrolyzes to produce acetic acid and hydroxide
ions.

• Reaction C2H3O2- H2O ? HC2H3O2 OH-
• Initial 0.15M - 0
  0
• Change -x x
  x
• Equil. 0.15- x x
  x

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• Kb HC2H3O2OH- C2H3O2-
• Kb Kw 1 x 10-14 5.6 x 10-10 Ka 1.8
  x 10-5
• 5.6 x 10-10 x2 ? x2 x
  9.2 x 10-6 0.15 - x
  0.15
• OH- 9.2 x 10-6
• pOH - log 9.2 x 10-6 5.04
• pH 14.00-5.04 8.96

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Acidic and Basic Oxides When metallic
(ionic) oxides dissolve in water they produce a
metallic hydroxide (basic solution). When
nonmetallic (covalent) oxides dissolve in water
they produce a weak acid (acidic solution).
CaO H2O ? Ca(OH)2 CO2 H2O ?
H2CO3
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Salts of Highly Charged Metals Salts that
contain a highly charged metal ion produce an
acidic solution.AlCl3 6H2O ? Al(H2O)63
3Cl- Al(H2O)63 ?Al(H2O)5(OH) 2 H
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The higher the charge on the metal ion the
stronger the acidity of the hydrated ion. The
electrons are pulled away from the O-H bond and
toward the positively charged metal ion. FeCl3
and Al(NO3)3 also behave this way.
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