Elise Hyser and Amanda Homan
Acids and Bases
Neutralization
Strong Bases
Strong Acids
Weak Bases
Weak Acids
pH
Titration Curves
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Titration Curve: A graph of pH (of an acid or base) as a function of the
volume of base (or acid) added to neutralize the substance
Important Terms and Concepts to Know:
-Equivalence Point: Point in a titration where the added solute reacts
completely with the solute present in the solution.
-Ka=Acid-Dissociation Constant
-Kb=Base-Dissociation Constant
-pH=-log[H+]
-pKa=-log[Ka]
-Kb=1.0x10-14 / Ka
-ALWAYS WRITE OUT CHEMICAL EQUATIONS WHEN
COMPLETING A TITRATION PROBLEM!
1. ) Solution with
weak acid+strong
base
4.) Use Ka,, [HX] ,
and [X-] to
calculate [H+]
2.) HX+OH-
5.) Arrive at pH
X-+H2O
3.) Calculate [HX]
and [X-] after rxn
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Calculation of a solution’s pH when 45.0 mL of .100 M NaOH is
added to 50.0 mL of .100 M HC2H3O2 (Ka=1.8x10-5):
.0500L x (.100 mol HC2H3O2 /1L soln) =5.00x10-3 mol HC2H3O2
.0450L x (.100 mol NaOH /1L soln) =4.50x10-3 mol NaOH
Total volume of solution: 45.0mL +50.0mL=.0950L
[HC2H3O2 ] =.50x10-3 mol /.0950L =.0053 M
[C2H3O2-] = 4.50x10-3 mol /.0950L=.0474 M
Ka= [H+] [C2H3O2-] / [HC2H3O2 ] =1.8x10-5
[H+]= Ka x[ [HC2H3O2 ] / [C2H3O2-]] =2.0x10-6M
pH=-log[2.0x10-6]=5.70
Solubility of Compound (g/L)
Molar Concentration of Ions
Molar Solubility of Compound (mol/L)
Ksp (Solubility Product Constant)
http://www.chemguide.co.uk/physical/acidbaseeqia/summary.gif
www.sparknotes.com
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H2SeO3 + H2O == H3O+ +HSeO3HSeO3- +H2O == SeO32- + H3O+
SeO32- +H2O == HSeO3- + OH-
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Solution involving the weak acid=higher initial
pH than solution of a strong acid of equal
concentration
pH change at “rapid-rise” portion of curve near
equivalence point is smaller for weak acid than
it is for strong acid
pH at equivalence point is over 7.00 for weak
acid-strong base titration
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Thymol Blue 1.2-2.8
Methyl yellow 2.9-4.0
Methyl orange 3.1-4.4
Methyl red 4.4-6.2
Phenol red 6.4-8.0
Phenolphthalein 8.0-10.0
Alizarin yellow 10.0-12.0
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pH Meter: Electronic instrument used to
measure the acidity or basicity of a liquid
(consists of a glass electrode & electronic
meter)
Potentiometric Titration: “A volumetric
method in in which the potential between two
electrodes is measured as a function of the
added reagent volume. Types of potentiometric
titrations include acid-base, redox,
precipitation, and complexometric” (Chemistry
Dictionary & Glossary).
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Involves determining concentration of a solution by
reacting it with a certain # of moles of excess
reagent.
Excess reagent is later titrated with another reagent.
Source: diracdelta.co.uk
The Common Ion Effect: The extent of ionization of a weak
electrolyte is decreased by adding a strong electrolyte to the
solution that has an ion in common with the weak electrolyte.
0.10 M HC2H3O2 solution has a pH of 2.9, while a solution containing
0.10 M HC2H3O2 and 0.01 M NaHC2H3O2 has a pH of 4.7.
Why has the addition of NaHC2H3O2 to a HC2H3O2 solution caused a
decrease in H+ concentration (increase in pH)?
When C2H3O2- (from the strong electrolyte NaC2H3O2) is added to an
acetic acid solution at equilibrium, the equilibrium condition is
displaced.
According to LeChatelier’s principle, the system reestablishes
equilibrium by removing some of the added acetate ion.
HC2H3O2 (aq)
Equilibrium
H+(aq) + C2H3O2-(aq)
C2H3O2- is added
shifts left to remove
some C2H3O2-
This simultaneously removes H+ ions from solution with the
formation of HC2H3O2 and thereby increases the pH.
 Solutions containing a weak acid and its
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•
conjugate base or a weak base and its conjugate
acid
Used to control the pH of a solution
Because of the common ion effect the mixture of
a weak acid and its conjugate base inhibits the
ionization of both
A buffer also reduces changes in pH due to the
addition of a strong acid or strong base.
Strong acid added to a buffer reacts with the
conjugate (weak) base to form more of the
conjugate (weak) acid.
Strong base added to a buffer reacts with the
conjugate (weak) acid to form more of the
conjugate (weak) base.

Ka 
-
[H ] [: A ]
[ HA ]

log K a  log (
-
[H ] [: A ]
)
[ HA ]
-

log K a  log[H
]  log
[HA]
By Definition
- log [K a ]  pK
- log[H
Therefore

]  pH
:
-
pK
a
[A ]
 pH  log
www.tiem.utk.edu
[A ]
[HA]
a
What is the change in pH when 0.010 mol solid NaOH is added to
1.0L of a buffered solution containing 0.50M acetic acid
(HC2H3O2, Ka = 1.8 x 10-5) and 0.50M sodium acetate (NaC2H3O2)
with a pH of 4.74?
Before reaction:
After reaction:
Initial:
Change:
Equilibrium:
HC2H3O2
0.50 mol
0.49 mol
+
OH0.010 mol
0
HC2H3O2 (aq) ==
0.49
-x
0.49 - x
H+ (aq) +
0
x
x
==
C2H3O2- + H2O
0.50 mol
0.51 mol
C2H3O2- (aq)
0.51
x
0.51 + x
Ka = 1.8 x 10-5 = [H+][C2H3O2-] / [HC2H3O2] = (x) (0.51 + x) / (0.49 - x)
[H+] = x = 1.7 x 10-5 M
pH = 4.76
The change in pH from adding 0.010 mol OH- to the buffered solution is only:
4.76 - 4.74 = +0.02
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The amount of acid or base the buffer
can neutralize before the pH begins to
experience a noticeable change
Depends on the amount of acid and
base from which the buffer is made
The greater the amounts of the
conjugate acid-base pair, the more
resistant the ratio of their
concentrations (and therefore pH) is to
change