ACID BASE EQUILIBRIA

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CHAPTER 17
ACID BASE EQUILIBRIA
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I. INTRODUCTION
A) Acid strength is measured by the extent of the
overall reaction of the acid with water.
1) Strong acids go 100 to the right. HCl H2O
? H3O Cl-
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2) Weak acids go "slightly" to the right. HA
H2O 34 H3O A-
Why can we make H2O Kc to Ka?
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Ka is known as the acid dissociation constant and
the acid ionization constant.
CH3COOH H2O ?H3O CH3COO-
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3) Some Ka's at 25oC
4) Some questions I will want you to be able to
answer from these data are
a) Which of the above acids is the strongest?
b) Which is the weakest acid?

c) What is the strongest conjugate base?
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II. HOW CAN WE FIND Ka's?
A) The most common way to obtain a Ka is to
measure the pH of a solution prepared by
dissolving a known amount of weak acid to form a
given volume of solution.
B) An example
The pH of a 0.10 M solution of HOCl is
4.23. What is the value of Ka?
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How do we get the values of H3O , OCl- and
HOCl?
From the pH we can obtain what value?
4.23 - log H3O - 4.23 log H3O 5.9 X
10-5 H3O
Then we know the value of the OCl- ?
The value of HOCl ?
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We set up a table like we did in Chapter 15.
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Then we put the values in the appropriate
equation for Ka.
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III. DEGREE OF DISSOCIATION AND PERCENT
DISSOCIATION (OR IONIZATION).
A) The degree of dissociation is the fraction of
molecules that react with water to give ions.
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If the H in a 0.60 M solution of HF is 0.021
M, what is the degree of dissociation?
ionization is obtained by multiplying the
degree of dissociation by 100 and adding the
sign.
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C) The larger the Ka the greater the degree of
dissociation (the greater the dissociation).
The stronger the acid the greater the
ionization.
D) As the original concentration decreases the
percent dissociation increases.
How do we explain this situation?
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IV. Calculation of the Concentration of Species
in a Weak Acid Solution Using Ka - THE
APPROXIMATE METHOD
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A) What are the concentrations of all species in
a 0.10 M of acetic acid (CH3COOH)? What is the pH
of the solution? Ka for acetic acid is 1.7 X
10-5.
You will need a balanced equation and a chart.
CH3COOH H2O ??H3O CH3COO-
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Then we put the results of the chart into the
equation for Ka.
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To avoid using the quadratic equation and focus
more attention on what is happening chemically,
we THINK AS FOLLOWS Since Ka is so small, x can
be ASSUMED to be much less than 0.10. That makes
0.10 - x 0.10. Then we don't have to use the
quadratic equation to obtain x.
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The above expression for Ka becomes
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BUT YOU will have to check your assumption
To check if the assumption is ok, we will use the
5 rule as given in the text. How do we calculate
the 5?
Another check to see if the assumption is ok is
to check to see if the precision rule allows you
drop the x. In the above case 0.10 - 0.0013
0.10 (The answer can only be reported to
hundredths.)
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The answer to the question then includes the
following CH3COOH 0.10 M H2O 55.5
M H3O 1.3 X 10-3 M CH3COO- 1.3 X 10-3
M And there is one more species in the solution.
What is it and what is its concentration?
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To answer the question, we still have to
calculate the pH.
pH - log 1.3 x 10-3 2.89
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B) Calculate the H3O in a 0.100 M solution of
nitrous acid, HNO2, for which the Ka is 4.5 X
10-4. You will need an equation and a chart.
HNO2 H2O ??H3O NO2-
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Now we have to see if the assumption is ok!!
0.100 - 0.0067 gives an answer to three digits
(0.093) so that indicates the approximation is
not appropriate for this problem.
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Also, IF YOU TAKE
You cannot use the approximate method, you must
use the quadratic equation.
What answer do we obtain with the quadratic
equation?
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Work some extra credit problems.
C) Calculating concentrations of species in a
solution of a diprotic acid. (What is a diprotic
acid? 1) We will examine the stepwise ionization
of oxalic acid, a diprotic acid found in spinach
and rhubarb.
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H2C2O4 H2O ?? H3O HC2O4- Ka1 5.90 x 10-2
HC2O4- H2O?? H3O C2O42- Ka2 6.40 x 10-5
Now what do we do with this information? If the
Ka's are far apart, we can make the problem
easier. We first compare the Ka's to see which
one is more important in delivering H3O's to the
solution.
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In this problem Ka1 is much greater than Ka2. We
will assume that all the H3O's came from the
first equation, when calculating the H3O,
HC2O4-, and the C2O42- in the 0.10 M H2C2O4
solution.
You will need an equation and a chart.
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H2C2O4 H2O??H3O HC2O4- Ka1 5.90 x 10-2
We now put these values in the general equation
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Ka is fairly large so the approximate method
cannot be used. You must use the quadratic
equation to obtain the value of x. The above Ka
relation becomes x2 5.90 X 10-2x - 5.9 X 10-3
0 The positive root gives x _____ H3O
HC2O4-
We then proceed to the second ionization, where
we make a second chart corresponding to the
second reaction.
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HC2O4- H2O ??H3O C2O42- Ka2 6.40 x 10-5

We then place these values in the equation for
Ka2.
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Since Ka2 is small we can assume that y ltlt 0.053.
When we do this the above equation becomes
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We obtain the following values for the
concentrations of the following species in
solution H3O 0.053 M
HC2O4- 0.053 M
C2O42- 6.4 X 10-4 M
H2O 55.5 M
H2C2O4 0.10 - 0.053 0.05 M
OH- 1.00 X 10-14 divided by 0.053 is 1.9
X 10-13 M
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V. Base Ionization Equilibrium A) Weak bases
have a Kb.
NH3 H2O ??NH4 OH-
B) Some other weak bases are what can be called
organic relatives of ammonia.
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CH3NH2 H2O ??CH3NH3 OH-
Kb 4.4 X 10-4
(CH3)2NH H2O ??(CH3)2NH2 OH- Kb 5.1 X
10-4
(CH3)3N H2O ??(CH3)3NH OH- Kb 7.4 X 10-5
C6H5NH2 H2O ??C6H5NH3 OH-Kb 4.2 X 10-10
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C) Calculate the pH of a 1.0 M solution of
methylamine, CH3NH2.
What will be your guess at the answer? _____
We need the equation, a chart and the set-up for
Kb . CH3NH2 H2O ??CH3NH3 OH-
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We then assume that x ltlt 1.0 and we obtain the
following expression
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We then obtain the OH- CH3NH3 2.1 X 10-2
M. We have to check the assumption
OH- 2.1 X 10-2 M pOH - log OH- - log
2.1 X 10-2 1.68 This is not the answer to the
question. Remember the question is about pH, not
pOH.
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So the pH 14.00 - 1.68 12.32
Some extra credit problems for you to do in class.
VI. Acid and Base Properties of Salts A) I want
you to be able to predict whether the solution of
a salt is acidic, basic or neutral.
B) How will you be able to do this?
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C) You will have to recall the 6 strong acids and
6 strong bases, and call all others are WEAK.
1) Salts which have cations from strong bases and
the anions of strong acids have no effect upon
the H when dissolved in water.
EXAMPLE NaCl
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Na is the cation of NaOH, a very strong base,
therefore Na is a very weak conjugate acid. It
does not react with water to take an OH- from it.
Cl- is the anion of HCl, a very strong acid,
therefore Cl- is a very weak conjugate base. It
does not react with water to take an H from it.
We expect that the water solution of NaCl to be
neutral.
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Na H2O ---gt N.R.
Cl- H2O ---gt N.R. Neutral solution
results.
We have not changed the concentration of H3O or
the concentration of the OH- ions by inserting
NaCl into the water.
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2) Salts which have cations from strong bases and
the anions of weak acids produce a basic solution
when dissolved in water.
EXAMPLES NaC2H3O2 sodium acetate NaC2H3O2 in
water dissociates 100 to Na ions and CH3COO-
ions.
Na is the cation of NaOH, a very strong base,
therefore Na is a very ________ ________
_______________ . It does not react with water to
take an OH from it.
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CH3COO- is the anion of the weak acid, acetic
acid, therefore its conjugate base is stronger
and will react with water to produce more OH-
ions.
Na H2O ----gt N.R. CH3COO- H2O ??CH3COOH
OH-
Do you see how a basic solution results from the
action of the conjugate base with water?
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Another example is NaF. What happens with it?
It dissociates 100 into ________ions and
________ions.
Which of these ions can react with water?
What kind of solution will it make?
Will the pH of the solution be greater or less
than 7?
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3) A salt which has the cation from a weak base
and the anion of a strong acid produce an acidic
solution. EXAMPLE
NH4Cl
NH4Cl ----gt NH4 Cl- Cl- H2O ----gt N.R.
(Why?) NH4 H2O ??NH3 H3O
We see that the concentration of the H3O
increases so the pH of the solution will be
__________________ 7.
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4) Salts derived from a weak acid and a weak base
have both ions reacting with water (undergoing
hydrolysis).
a) Whether the solution is acidic or basic
depends on the relative acid - base strengths of
the two ions.
To determine this, you will need to compare the
cation's Ka with the Kb of the anion.
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b) If the Kb of the anion gt the Ka of the cation,
the solution is basic.
c) If the Kb of the anion lt the Ka of the cation,
the solution is acidic.
d) If the Kb of the anion the Ka of the cation,
the solution is neutral.
B) How do we obtain Ka's and Kb's for ions?
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What is the Kb for the CH3COO- ion?
We start with CH3COONa.
It dissociates 100 CH3COONa ----gt CH3COO- Na
Na H2O ----gt N.R. CH3COO- H2O ??CH3COOH
OH-
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Generally the values for the Ka's and Kb's for
ions are not given in books. How can we obtain
them if this is the case?
What will be given in the text and how can we use
it?
The Ka for the weak acid will be given. How is
that related to the Kb for the conjugate base?
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Compare Ka with Kb.
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If we look closely, we see that part of Kb is
upside down from the Ka. So let's invert it.
BUT this still isn't equal to Kb. We have to
eliminate H3O from the denominator and get OH-
in the numerator. How can we do this?
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So for the acetate ion, the value of Kb is
obtained by the following division
This is a small number, but large enough to
affect the pH of the solution.
We can conclude from this that Kw KaKb
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What is the pH of a 0.10 M solution of CH3COONa?
CH3COONa ----gt CH3COO- Na Na H2O ----gt
N.R. CH3COO- H2O ??CH3COOH OH-
We have the equation, now we need a chart.
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To solve for x we put these values in the
equation for Kb for the ion.
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That is not the answer yet, remember the question
asks for the pH of the solution. So we have a
further calculation to do.
Notice that we are very close to having to use
the 1 X 10-7 to add to the OH- ions. For this
class we won't do it, but for a more precise
answer you would put it in.
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C) What is the Ka for the NH4 ion? The salt of
NH4 with non-complicating factors is NH4Cl (or
Br-, I-, NO3-).
We need the equation for the reaction of NH4
with water.
NH4 H2O ??NH3 H3O
Ka for the anion is going to equal what?
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Now that we know the Ka for the NH4 we should be
able to calculate the pH of a 0.10 M solution of
NH4Cl.
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What kind of solutions (acid or base) will we get
if we dissolve CuCl2 in water AlCl3?
Show by appropriate reactions of the hydrolysis
of the reacting ions.
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VII. THE COMMON ION EFFECT
A) The Common Ion Effect is the shift in an ionic
equilibrium caused by the addition of a solute
that provides an ion that takes part in the
equilibrium.
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1) Calculate the pH of a 0.50 M acetic acid
solution.
CH3COOH H2O ??H3O CH3COO-
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2) What is the pH of a solution which is 0.50 M
in CH3COOH and 0.50 M in CH3COONa?
The most important thing that needs to come to
mind when you see two similar but different
compounds is to write TWO separate equations and
draw two separate charts.
Do not ever combine the two reactants in one
equation. Crap results
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CH3COONa CH3COO- Na
CH3COOH H2O ??H3O CH3COO-
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Now we have to put the resultant values in the
expression for Ka.
If x is ltlt 0.50 then we obtain the following
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The pH has gone from 2.54 to 4.77
H3O has gone from 2.9 X 10-3 to 1.7 x 10-5
Is this reasonable taking into account all that
we have studied? What about Le Chatelier's
Principle?
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VIII. BUFFERS A) A buffer is a solution
characterized by the ability to resist changes in
pH when limited amounts of acid or base are added
to it.
B) The most important application of the common
ion effect is for making buffer solutions.
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C) Aqueous solutions of the body have a
characteristic pH. The blood, for example, has a
pH between 7.3 and 7.5. Death generally results
at pH's below 7.0 and above 7.9. The blood's
ability to control pH is remarkable in that many
body reactions produce acids. When these enter
the blood, if the blood were not buffered, its pH
would be changed drastically with any body
function, and death would result.
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D) Some important buffer systems in the blood
include the H2CO3/HCO3- (from CO2 dissolving in
the blood) and H2PO4-/ HPO42-.
E) Buffers contain either a weak acid and its
conjugate base or a weak _____ and its________
_______. The acid and its salt or the _______and
its ______are in water solution. By choosing the
correct mixture of substances we can make buffers
of selected pHs.
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F) Let's look at our common ion solution prepared
previously, and add a small amount of HCl to the
solution. CH3COOH H2O ??H3O CH3COO-
To make the arithmetic easier, we'll assume that
initially all the HCl reacts with the CH3COO- to
form CH3COOH molecules.
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Then, of course, some of the molecules react in
the forward direction
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We now put these values in the expression for Ka .
pH 4.75 compared to 4.77 before the HCl was
added. Why did the pH go down?
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NOW, I want you to compare this to water only.
What is the pH of pure water? What is the pH of
a 0.01 M water solution of HCl? How many units
does the pH change? How many times is this?
G) Biologists and biochemists often calculate the
pH of a buffer solution using the
Henderson-Hasselbach Equation.
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A buffer is made containing 0.10 M CH3COOH ( pKa
4.77 ) and 0.050 M CH3COONa. What is the pH of
this solution?
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H) In lab you were asked to make a buffer
solution of a certain pH knowing Ka. Given acid
and conjugate base solutions of the same
concentrations, the ratio of the concentrations
becomes the ratio of the _______________ of the
two samples.
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1) Calculate the ratio of concentrations of HOCl
(hypochlorous acid) and NaOCl (sodium
hypochlorite) needed to produce a buffer solution
with a pH of 7.60.
Ka for HOCl 2.9 X 10-8
HOCl H2O ??H3O OCl-
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Translated into volumes, what does this ratio
mean?
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IX. TITRATIONS
A) There are three types
1) strong acid and strong base
2) weak acid and strong base
3) strong acid and weak base
B) Those involving a strong acid and a strong
base
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1) Example titrate a 25.00 mL of 0.100 M HCl
with a 0.100 M NaOH solution. The NaOH solution
is added from the buret to the HCl in the flask.
If we monitor the titration with a pH meter, we
obtain the following results
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After 25.00 mL of base have been added, we call
that the equivalence point, because a
stoichiometric amount of the reactant has been
added. For the strong acid-strong base titration,
the pH at the equivalence point is _____ . A
neutral solution has been produced.
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How can we calculate the pH of a solution after a
certain amount of NaOH has been added to the HCl
solution?
Example What is the pH of a solution after 10.0
mL of 0.100 M NaOH have been added to 25.0 mL of
0.100 M HCl?
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C) Weak acid and strong base titration
Why isn't the equivalence point at pH 7?
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D) What would the titration curve of a titration
of strong acid with a weak base look like?
At about what pH would its equivalence point lie?
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The End