CHM 101 GENERAL CHEMISTRY
FALL QUARTER 2008
Section 2
Lecture Notes – 10/27/2008
(last revised: 10/27/08, 9:30 PM)
4.5
Precipitation Reactions: In this section and in the remainder
of Chapter 4, we will be concerned mostly with what happens, if
anything, when two solutions, each containing a different
electrolyte, are mixed. The most simple possibility is that
nothing happens. The next
most simple is that a
substance forms that is not
soluble in water, and it
precipitates out of the
mixture.
•
An Example of a
Precipitation Reaction:
The first example from the
text is a nicely illustrated
demonstration of what
happens when an aqueous
solution of potassium
chromate (K2CrO4 (aq)) is
added to an aqueous
solution of barium nitrate
(Ba(NO3)2 (aq)). In Figure
4.13, we see that when the
yellow solution of K2CrO4
(aq) in the graduated
cylinder is added to the
colorless solution of
Ba(NO3)2 (aq) in the beaker, a yellow precipitate forms.
•
What is the Chemistry? As aspiring chemists, we cannot
content ourselves with a verbal description of what we see when
we combine these two solutions. We need to determine the
chemistry that takes place, and we need to be able to write an
equation for chemical reaction that has obviously happened.
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•
The Reactants: Let us start by considering what species are
present in the two solutions before they are mixed. We know
that solid barium nitrate is a salt containing barium ions (Ba2+)
and nitrate ions (NO3-) arranged in a crystal lattice and held
together by electrostatic forces. When it is dissolved in water,
the lattice breaks up, the ions become hydrated, and they
disperse evenly throughout the solution.
H2O
Ba(NO3)2 (s) ——> Ba(NO3)2 (aq) ——> Ba2+ (aq) + 2NO3-(aq)
Similarly, we know that the potassium chromate is a salt
containing potassium ions (K+ (aq)) and chromate ions (CrO42(aq)) in a crystal lattice held together by electrostatic forces.
When it is dissolved in water, its lattice breaks up, the ions
become hydrated, and they also disperse evenly throughout the
solution.
H2O
K2CrO4 (s) ——> K2CrO4 (aq) ——> 2K+ (aq) + CrO42-(aq)
Figure 4.14 from your text pictures the two solutions and
diagrams the ions contained in each:
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•
The Reaction and its Products: When we mix the two
solutions, we can identify the four different kinds of ions as
possible reactants and we can describe the products as a yellow
solid in contact with an aqueous phase:
2K+ (aq) + CrO42-(aq) + Ba2+ (aq) + 2NO3-(aq) ——>
yellow solid + aqueous phase
The question is, what actually reacts with what to form the
yellow solid. We know that it must contain both cations and
anions, because it cannot have any net charge. There are 4
different combinations of a cation with an anion from the above
set of reactants. Two of them, Ba(NO3)2 and K2CrO4 can be
eliminated because they were the soluble salts in the two
original solutions. This leaves the following candidates:
o BaCrO4
o KNO3
Which of these is most likely to precipitate out of aqueous
solution as a yellow solid? We know that our initial Ba(NO3)2
solution had no color, and if we have worked at all with
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potassium salts, we know that most of them are also colorless.
Thus the yellow color of the K2CrO4 solution must have been
from its chromate ions, so by elimination, the yellow solid must
be BaCrO4. Another thing we should know is that nearly all
potassium salts and nitrate salts are water soluble, so we can
identify the colorless aqueous phase that is in contact with the
solid BaCrO4 as a solution of KNO3. Thus we can now write the
equation for the reaction:
2K+ (aq) + CrO42-(aq) + Ba2+ (aq) + 2NO3-(aq) ——>
BaCrO4 (s) + 2K+ (aq) + 2NO3-(aq)
Notice, however, that we have two potassium ions and two
nitrate ions on each side of the equation. These ions do not take
part in the reaction, so we can leave them out. (We call them
spectator ions.) This allows us to simplify the equation to:
CrO42-(aq) + Ba2+ (aq) ——> BaCrO4 (s)
Figure 4.15 from the text give us a good illustration of this
precipitation reaction. Parts a and b show what happens on the
ionic level, while part c is a photograph after the reaction has
taken place:
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•
The Reaction between Silver Nitrate and Potassium
Chloride: The second example in the text is the reaction that
takes place when silver nitrate (AgNO3 (aq)) and potassium
chloride (KCl (aq)) are mixed. Figure 4.16 show a beaker of KCl
solution just after the addition of some AgNO3 solution. So let us
figure out what is happening. We start as we did for the first
example, with the equation:
AgNO3 (aq) + KCl (aq) ——>
white solid + aqueous phase
We rewrite the equation to show the reactant ions:
Ag (aq) + NO3- (aq) + K+ (aq) + Cl- (aq) ——> white solid + aqueous
phase
+
The white solid is one of the following two salts:
o AgCl
o KNO3
But it cannot be KNO3, because we identified that as the soluble
salt that forms in the reaction of K2CrO4 (aq) with Ba(NO3)2 (aq).
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Therefore it must be AgCl, and the ionic reaction for its
formation can be written:
Ag+ (aq) + Cl- (aq) ——> AgCl (s)
We have omitted the spectator ions, K+ (aq) and NO3- (aq),
because they do not participate in the reaction.
•
Solubility Rules for Salts: You may have noticed that we used
information from the first example to help us determine the
nature of the reaction in the second example. We cited the fact
that KNO3 is water soluble. Chemists have studies the water
solubility (or lack thereof) for a great number of salts and have
compiled some simple rules whether particular salts of the more
common ions are water soluble. Table 4.1 in your text is such a
set of rules. In CHM 101, we will use the following version of
these rules. (This is also posted on Marty Wallace’s website as a
PDF file.):
o Most nitrate (NO3-) salts are soluble. (Memorize this!)
o Most salts containing alkali metal ions (Li+, Na+, K+, Rb+, &
Cs+) or the ammonium ion (NH4+) are soluble. (Memorize
this!)
o Most chloride, bromide, and iodide (Cl-, Br-, & I-) salts are
soluble. Exceptions for CHM 101: The chloride, bromide,
and iodide salts of Ag+, Pb2+, and Hg22+ are insoluble.
o Most sulfate (SO42-) salts are soluble. Exceptions for CHM
101: BaSO4, PbSO4, Hg2SO4, and CaSO4 are insoluble.
o Most hydroxide (OH-) “salts” are insoluble, unless Rule #2
applies.
o Most sulfide, carbonate, chromate, and phosphate (S2-,
CO32-, CrO42-, & PO43-) salts are insoluble, unless Rule #2
applies.
As with many other sets of rules, these have a number of
exceptions. Moreover, the absolutes, soluble and insoluble, are
extremes on a continuum from very, very soluble to somewhat
soluble, to hardly soluble at all. Nevertheless, we will use them
for now, but as my Freshman Chemistry professor at Harvard
used to say, we will have to take them with a grain of salt.
•
Predicting Reaction Products: Now we turn to Sample
Exercise 4.8 (pp. 144-5) from your text. What are the reaction
products (if any)? We will work them on the whiteboard:
a) KNO3 (aq) & BaCl2 (aq)
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b) Na2SO4 (aq) & Pb(NO3)2 (aq)
c) KOH (aq) & Fe(NO3)2 (aq)
4.6
•
Describing Reactions in Solution
Let us briefly return to the first example from the previous
section. Recall that when we determined that the precipitate was
BaCrO4, we could have written the formula equation for the
reaction as:
K2CrO4 (aq) + Ba(NO3)2 (aq) ——> BaCrO4 (s) + 2KNO3 (aq)
We actually did write the formulas for the three soluble salts in
terms of their component ions, obtaining the complete ionic
equation:
2K+ (aq) + CrO42-(aq) + Ba2+ (aq) + 2NO3-(aq) ——>
BaCrO4 (s) + 2K+ (aq) + 2NO3-(aq)
Then we eliminated the spectator ions, K+ (aq) and NO3-(aq),
two of each on each side of the equation:
CrO42-(aq) + Ba2+ (aq) ——> BaCrO4 (s)
What remains is the net ionic equation, including only those
components directly involved in the reaction. Let’s summarize
these three types of equations:
o Formula Equation: The Formula Equation gives the
overall stoichiometry for the reaction, but it does not
necessarily show the actual forms of the reactants and
products.
o Complete Ionic Equation: The Complete Ionic Equation
shows all the ions present in the reactants, all the solid
reaction products, and all ions that remain after the
reaction takes place.
o Net Ionic Equation: The Net Ionic Equation shows only
the ions that react plus the products that form from them.
•
Writing Equations for Reactions: Now we turn to Sample
Exercise 4.9 (p.146) from your text. Write the formula equation,
the complete ionic equation, and the net ionic equation for the
following reactions. We will work them on the whiteboard:
a) Aqueous potassium chloride is added to aqueous silver
nitrate to form a precipitate of silver chloride and a
solution of potassium nitrate.
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b) Aqueous potassium hydroxide is mixed with aqueous
iron (III) nitrate to form a precipitate of iron (III)
hydroxide and aqueous potassium nitrate.
4.7
Stoichiometry of Precipitation Reactions: This isn’t really a
new topic. The good news is that you already know how to
stoichiometric calculations on precipitation reactions.
•
In Section 3.9 you learned how to start with the mass (usually in
grams) of component A, convert that to the number of moles of
A, convert that to the number of moles of B (according to the
balanced equation), and finally convert that to the mass of B.
•
In Section 3.10 you learned how to work limiting reactant
problems, i. e., how to determine which of two reactants limited
the amount of products that would form.
•
In Section 4.3 you learned about concentrations of reagents in
solution, and how if you multiplied molar concentration
(molarity) by volume, your result would be the number of moles
contained in that volume.
•
Finally in Sections 4.5 and 4.6 you learned how to determine the
nature of a precipitation reaction and how to write a net ionic
equation for the reaction.
•
Determining the Mass of Product Formed: Now you can
apply the foregoing skills to determine masses of products
formed in precipitation reactions. We will jump immediately into
some sample exercises:
o Sample Exercise 4.10 (p. 147): Calculate the mass of solid
NaCl that must be added to 1.50 L of a 0.100 M AgNO3
solution to precipitate all the Ag+ ions as AgCl (s).
o The first step is to list the ions that are initially
present when solid NaCl is dissolved in a solution of
AgNO3. The AgNO3 solution already contains Ag+ and
NO3- ions prepared by dissolving AgNO3 in water:
H2O
AgNO3 (s) ——> Ag+ (aq) + NO3-(aq) (1)
o And the reaction for the dissolution of NaCl in water
can be written:
H2O
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NaCl (s) ——> Na+ (aq) + Cl-(aq) (2)
o Thus the ions are:
Na+, Cl- (from NaCl), Ag+, NO3- (from AgNO3)
o The next step is to use Table 4.1 to determine that
AgCl is insoluble in water, while NaNO3 is water
soluble and remains in solution as (unchanged) Na+
and NO3- ions. From this information we can write
the net ionic equation:
Ag+ (aq) + Cl- (aq) ——> AgCl (s) (3)
o Since we know the molarity and the volume of the
silver nitrate solution, we can determine the number
of moles of AgNO3 it contains:
N AgNO3 = M AgNO3 × V = 0.100 mol ×1.5L = 0.150 mol (4)
L
o How do we use this information to determine the
mass of NaCl (s) needed to precipitate the Ag+ ions
from the solution? We know from Eq. (1) that the
number of moles of Ag+ ions in the silver nitrate
solution is:
N Ag+ = N AgNO3
(5)
o Equation (3) gives us:
N Cl− = N Ag+
(6)
o And Equation (2) gives us:
N NaCl = N Cl−
(7)
o Combining Eqs. (4), (5), (6), & (7), we get the
number of moles of NaCl required for the complete
precipitation of the AgCl:
N NaCl = N AgNO3 = 0.150 mol
(8)
o All we need to do to finish is to convert moles of
NaCl to mass:
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mNaCl = N NaCl × MM NaCl = 0.150 mol × 58.45 g/mol = 8.77 g
o As you gain familiarity with this type of calculation,
you will be able to write equations like Eq. (8)
directly, without explicitly writing the intermediate
steps (5), (6), & (7).
•
A System for Solving Stoichiometry Problems for
Reactions in Solution: The text gives the following step by
step procedure for solving solution stoichiometry problems:
o Identify the species present in the combined solution, and
determine what reaction occurs.
o Write the balanced net ionic equation for the reaction.
o Calculate moles of reactants.
o Determine which reactant is limiting.
o Calculate the moles of product(s), as required.
o Convert to grams or other units, as required.
•
Determining the Mass of Product Formed (Sample Exercise
4.11(p. 148):
o When aqueous solutions of Na2SO4 and Pb(NO3)2 are
mixed, PbSO4 precipitates. Calculate the mass of PbSO4
formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of
0.0250 M Na2SO4 are mixed. We will work this out on the
whiteboard.
4.8
•
Acid-Base Reactions:
Definitions – Acids and Bases: We will start by revisiting the
definitions of acids and bases.
o Arrhenius’ Definitions:
1. Acid: An acid produces H+ ions when dissolved in
water.
2. Base: A base produces OH- ions when dissolved in
water.
These definitions are fundamentally correct, and they
serve us well for dealing with strong acids and strong
bases. However, they are somewhat inconveniently
limiting when we consider weak acids and weak bases.
Consider the weak base, ammonia (NH3). What is the
source of the OH- ions that it generates when dissolved in
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water? It is not to be found in gaseous ammonia. The fact
is that an ammonia molecule must combine with a water
molecule in order to generate a hydroxide:
NH3 (aq) + H2O (l) = NH4+ (aq) + OH- (aq)
In other words, in order for an ammonia molecule to
generate a hydroxide ion in aqueous solution, it must
accept a proton from a water molecule. Then what is left of
the water molecule is the desired hydroxide ion.
The problem is not so severe for weak acids. Consider the
weak acid, acetic acid (HC2H3O2). Acetic acid will generate
protons when dissolved in water:
HC2H3O2 (aq) = H+ (aq) + C2H3O2- (aq)
However, as we learned earlier, only a small fraction of
acetic acid molecules will form ions. In 0.1 M acetic acid,
about 99% of the acetic acid remains as undissociated
neutral acetic acid molecules; only about 1% will ionize to
form hydrogen ions and acetate ions.
o Picturing the Reaction of a Weak Acid with a Strong
Base: If equimolar aqueous solutions of a weak acid and a
strong base are mixed, the reaction will go to completion.
All of the base will consume all of the acid. For example, if
acetic acid is reacted with sodium hydroxide, we can write
the overall reaction as:
HC2H3O2 (aq) + NaOH (aq) = NaC2H3O2 (aq) +H2O (l)
If we strictly apply the Arrhenius definition of an acid, we
must imagine that the initial 1% of hydrogen ions from the
acetic acid react with hydroxide ions. Then more acetic
acid molecules ionize, and those hydrogen ions react with
more hydroxides, and so on until all the acetic acid has
ionized and all the resulting hydrogen ions have reacted
with hydroxides. Wouldn’t it be simpler if we said that
acetic acid reacts directly with hydroxide ion by donation of
a proton to hydroxide?
HC2H3O2 (aq) + OH- (aq) = C2H3O2- (aq) +H2O (l)
o Brønsted and Lowry’s Definitions:
1. Acid: An acid is a proton donor.
2. Base: A base is a proton acceptor.
These definitions are more general than those of Arrhenius
and they resolve the difficulties we have just discussed.
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•
Completeness of Acid – Base Reactions: For reactions of
equimolar quantities:
1. A strong acid will react completely with a strong base.
2. A strong acid will react completely with a weak base.
3. A weak acid will react completely with a strong base.
•
Stoichiometry Calculations for Acid – Base Reactions: The
text gives this step-by-step procedure for performing acid – base
calculations. It is very similar to the procedure that we just
applied to precipitation reactions:
1. List the species present in the combined solution before
any reaction occurs, and determine what the reaction
will be.
2. Write the balanced net ionic equation for this reaction.
3. Calculate the moles of reactants. For reactions in
solution, use the volumes of the original solutions and
their molarities.
4. Determine the limiting reactant where appropriate.
5. Calculate the moles of the required reactant or product.
6. Convert to grams of volume (of solution) as required.
•
Neutralization Reactions: These are reactions where just
enough acid is added to some base (or just enough base is
added to some acid) to react completely. Since there is no
excess either of acid or base, we say the resulting solution is
neutralized.
o Sample Exercise 4.12 (p. 150): What volume of 0.100 M
HCl solution is required to neutralize 25.0 mL of 0.350 M
NaOH?
After the solutions are mixed, but before any reaction
takes place, the ions in solution are:
H+ (aq), Cl- (aq) (from HCl), Na+ (aq), OH- (aq) (from NaOH)
Since NaCl is water soluble, we do not expect any reaction
between the Na+ (aq), and the Cl- (aq), but we do expect
that H+ (aq) and OH- (aq) will react:
H+ (aq) + OH- (aq) ——> H2O (l)
The reaction is already balanced, but if it weren’t, we
would balance it now.
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Since this is an exact neutralization reaction, the numbers
of moles of hydrogen ion and hydroxide ion (before
reaction) are equal, and there is no need to ask which is
the limiting reactant:
N H + = N OHBut these are the same as the numbers of moles of HCl
and of NaOH:
N HCl = N NaOH
We can rewrite this in terms of the molarities and volumes
of the NaOH and HCl solutions:
M HCl × VHCl = M NaOH × VNaOH
Since we know the molarities of the two solutions and the
volume of the NaOH solution, we can rearrange this
equation to solve for the volume of HCl:
VHCl =
M NaOH × VNaOH 0.350 mol/L × 0.025 L
=
= 0.0875 L
M HCl
0.100 mol/L
o Sample Exercise 4.13 (p. 151): In a certain (peculiar?)
experiment, 28.0 mL of 0.250 M HNO3 and 53.0 mL of
0.320 M KOH are mixed. Calculate the amount of water
formed in the resulting reaction. What is the concentration
of H+ or OH- ions (whichever is in excess) after the
reaction goes to completion?
After the solutions are mixed, but before any reaction
takes place, the ions in solution are:
H+ (aq), NO3- (aq) (from HNO3), K+ (aq), OH- (aq) (from KOH)
Since KNO3 is water soluble, we do not expect any reaction
between the K+ (aq), and the NO3- (aq), but we do expect
that H+ (aq) and OH- (aq) will react:
H+ (aq) + OH- (aq) ——> H2O (l)
The reaction is already balanced, but if it weren’t, we
would balance it now.
We have data to calculate the numbers of moles of both
starting reagents. It is very likely that one or the other will
be in excess. The number of moles of HNO3 is:
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N HNO3 = M HNO3 × VHNO3 = 0.250 mol/L × 0.0280 L = 0.00700 mol
And the number of moles of KOH is:
N KOH = M KOH × VKOH = 0.320 mol/L × 0.0530 L = 0.0170 mol
Thus HNO3 is the limiting reactant. The net ionic reaction
shows that 1 mole of H2O will be generated for each 1
mole of H+ that reacts. And we know that each 1 mole of
HNO3 will supply 1 mole of H+. Thus the number of moles
of water generated is:
N H 2O = N HNO3 = 0.0070 mol
The mass of this water is:
mH 2O = N H 2O × MM H 2O = 0.00700 mol × 18.02 g/mol = 0.125 g
The number of moles of unreacted OH- is:
N OH − = N KOH − N HNO3 = 0.0170 mol − 0.0070 mol = 0.0100 mol
Now we need to be careful. The volume of the final
solution is the combined volumes of the HNO3 solution and
the KOH solution. (We neglect the volume of product water
that forms in the reaction.)
V = VKOH + VHNO3 = 0.0280 L + 0.0530 L = 0.0810 L
So the molarity of OH- in the end solution is:
M OH − =
•
N OH −
V
=
0.0100 mol
= 0.123 mol/L
0.0810 L
Acid – Base Titrations: In volumetric analysis, one determines
the amount of an unknown substance by reacting it with a
carefully measured volume of a solution of a known substance
with a known concentration.
o For example, suppose one had a solution of a weak acid
with an unknown concentration. And let us also suppose
one had a solution of a strong base whose concentration is
known. One could determine the concentration of the weak
acid solution by measuring a precise volume of it (with a
transfer pipette) into an Erlenmeyer flask and adding
strong base solution to it from a burette until one reaches
the stoichiometric point (i. e., the end point) of the
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reaction where the number of moles of added base exactly
matches the number of moles of weak acid present in the
original sample of weak acid. An indicator added to the
solution in the flask signals the end point by changing its
color when the end point is achieved.
o The titration of a sample of an unknown weak acid with a
strong base is pictured in Figures 4.18a-c from the text:
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o The process is called titration, and the solution of known
substance is called the titrant. The solution of the unknown
is called the analyte. The reaction of the known with the
unknown is well characterized, so from the stoichiometry
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of the reaction, one can determine the amount of unknown
present in the original unknown sample.
o In general, the requirements for a successful titration are:
1. The exact reaction between titrant and analyte must be
known and rapid.
2. The stoichiometric (equivalence) point must be marked
accurately.
3. The volume of titrant required to reach the
stoichiometric point must be measured accurately.
o In an acid – base titration, the analyte is a base or an acid.
(It may either be strong or weak.) The titrant then is a
strong acid or a strong base, respectively. A common
indicator for acid – base titrations is phenolphthalein,
which is colorless in acidic solutions and pink-purplish in
basic solutions.
•
Neutralization Titration – Standardization of a Solution of
NaOH (Sample Exercise 4.14, p. 153): Solutions of NaOH are
commonly used to titrate
unknown acids. One might
think that one could prepare
a solution of NaOH by
accurately weighing some
solid NaOH and dissolving it
in water to prepare an
accurately measured volume
of solution. However, solid
NaOH draws water out of its
surrounding atmosphere so
fast that an accurate
weighing is not possible.
The actual procedure for
preparing a standardized
solution of NaOH is first to
make an NaOH solution that
has approximately the desired concentration. Then one takes an
accurately weighed amount of potassium hydrogen phthalate,
dissolves it in water, and titrates it to a phenolphthalein end
point with the NaOH solution. This works because potassium
hydrogen phthalate has one hydrogen (the one bonded to
oxygen in the figure) that can be donated to a strong base.
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Hence it functions as a weak acid, even though it is nominally a
salt.
In this particular example, a student weighs out a 1.3009 g
sample of potassium hydrogen phthalate (KHC8H4O4, MM =
204.22 g/mol), dissolves it in water in an Erlenmeyer flask, adds
phenolphthalein indicator, and titrates it with the solution of
NaOH that she wishes to standardize. At the end point of the
titration, she finds that she has added 41.20 mL of NaOH
solution. What is the concentration of the NaOH solution?
o The first step is to determine what species are initially
present and which of them react. The NaOH contributes
Na+ (aq) and OH- (aq) ions, and the potassium hydrogen
phthalate contributes K+ (aq) and HC8H4O4- (aq) ions. The
actual reaction that takes place is between HC8H4O4- (aq)
and OH- (aq), even though they are both negative ions:
HC8H4O4- (aq) + OH- (aq) ——> C8H4O42- (aq) + H2O (l)
The reaction is between the hydrogen phthalate ion, acting
as a weak acid (Brønsted-Lowry definition), and the
hydroxide, a strong base.
o The stoichiometry is 1:1, so she can write:
N NaOH = N OH− = N HC H O - = N KHC8H4O4
8
4
4
o Now she converts moles of NaOH to molarity ( M NaOH ) and
volume (VNaOH ) , and moles of KHC8H4O8 to mass
(mKHC8H4O4 ) and molar mass ( MM KHC8H4O4 ) :
M NaOH × VNaOH = N NaOH = N KHC8H4O4 =
mKHC8H4O4
MM KHC8H4O4
o Plugging in her data, she gets:
0.04120 L × M NaOH = M NaOH × VNaOH =
=
mKHC8H4O4
MM KHC8H4O4
=
1.3009 g
= 6.3701×10−3 mol
204.22 g/mol
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o Thus the molarity of her solution is:
M NaOH
6.3701×10−3 mol
=
= 0.1546 mol/L
0.04120 L
o She will use this NaOH solution and this molarity result in
the next example.
•
Neutralization Titration – Analysis of a Sample Containing
a Weak Acid (Sample Exercise 4.15, pp. 153-4): Our intrepid
chemist now turns to her real problem, the analysis of a sample
of the effluent from a waste treatment process. This effluent is
known to contain carbon tetrachloride (CCl4) and benzoic acid
(HC7H5O2), a weak acid that can donate one proton to a strong
base. She weighs out a 0.3518 g sample of the effluent and
shakes it with water to dissolve the benzoic acid. Then she
titrates it with 0.1546 M NaOH, and the titration requires 10.59
mL of the NaOH solution to reach a phenolphthalein end point.
What is the mass percent of benzoic acid in the original sample?
(The molar mass of benzoic acid is 122.12 g/mol.)
o The titration reaction is between NaOH and HC7H5O2 to
produce benzoate ion (C7H5O2- (aq)), so she writes the net
ionic equation:
HC7H5O2 (aq) + OH- (aq) ——> C7H5O2- (aq) + H2O (l)
o Thus the stoichiometry between benzoic acid and NaOH is
1:1, so the number of moles of benzoic acid in the sample
is the same as the number of moles of NaOH used in the
titration:
N HC7 H5O2 = N NaOH
o She expresses the number of moles of benzoic acid in
terms of its (unknown) mass ( mHC7 H 5 O 2 ) and its molar mass
(MM HC7 H5O 2 ) , and she expresses the number of moles of
NaOH in terms of the molarity ( M NaOH ) and volume
(VNaOH ) of the solution:
mHC7 H5O2
MM HC7 H5O2
= N HC7 H5O2 = N NaOH = M NaOH × VNaOH
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o Now she can plug in her data:
mHC7 H5O2
122.12 g/mol
= 0.1546 mol/L × 0.01059 L = 1.637 ×10−3 mol
o Thus she finds that the mass of benzoic acid in her effluent
sample is:
mHC7 H5O2 = 1.637 ×10−3 mol ×122.12 g/mol = 0.1999 g
o Finally, she computes the percentage of benzoic acid that
was present in her original sample:
%HC7 H5O2 =
mHC7 H5O2
msample
=
0.1999 g
= 56.82%
0.3518 g
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