NAME________________________________ PER ________ DATE DUE ___________________
ACTIVE LEARNING I N C HEMISTRY E DUCATION
"ALICE"
CHAPTER 22
ACID – BASE
REACTIONS
Acid–Base Equations
The pH Scale
Chemical Indicators
Acid–Base Titrations
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©1997, A.J. Girondi
NOTICE OF RIGHTS
All rights reserved. No part of this document may be reproduced or transmitted in any form by any means,
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Copies of this document may be made free of charge for use in public or nonprofit private educational
institutions provided that permission is obtained from the author . Please indicate the name and address
of the institution where use is anticipated.
© 1997 A.J. Girondi, Ph.D.
505 Latshmere Drive
Harrisburg, PA 17109
[email protected]
Website: www.geocities.com/Athens/Oracle/2041
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©1997, A.J. Girondi
SECTION 22.1
Acid – Base Reactions
When a water solution of an acid reacts with a water solution of a base, the products are generally
water and a salt compound. This type of reaction is called neutralization:
ACID + BASE ----> WATER + A SALT
HCl + NaOH ----> HOH + NaCl
acid
base
water
salt
(Salts are compounds consisting of a metal combined with a nonmetal. NaCl is only one of many salts. The particular
salt which forms in a neutralization reaction depends on which acid and base are used.)
Problem 1. Complete the formula equations below which illustrate acid–base "neutralization" reactions.
The hydrogen in the acids combines with the hydroxide in the base to form water in every case. The other
product is a salt. First, use oxidation numbers to write the correct formulas for the salts. Then, balance
each equation.
a.
_____HCl(aq) + _____LiOH(aq)
---->
__________________________________
b.
_____HNO3(aq) + _____KOH(aq)
---->
__________________________________
c.
_____HBr(aq) + _____NaOH(aq)
---->
__________________________________
When HCl and NaOH are placed in water they dissociate into ions as shown below.
HCl(g) ----> H1+(aq) + Cl1-(aq)
NaOH(s) ----> Na1+(aq) + OH1-(aq)
When the two solutions are mixed, the positive hydrogen ions and negative hydroxide ions then attract
each other, and water is formed:
H1+(aq) + Cl1-(aq) + Na1+(aq) + OH1-(aq) ---> HOH(l) + Na1+(aq) + Cl1-(aq)
Notice that in the equation above, the salt, NaCl, does not actually form. This is because NaCl is soluble in
water, so it remains dissociated as aqueous sodium and chloride ions. (Soluble salts are dissociated when
in solution.) If the water in the system were evaporated, then the Na1+ and Cl 1- ions would combine to
form solid salt, NaCl.
The equation above in which dissociated reactants and products are shown as a mix of individual
ions is called the ionic equation. Notice that the sodium and chloride ions appear on both sides of the
equation in identical form. They are not reacting in any way, so we call them spectator ions. If we drop
them out of the equation, the particles that remain are called the reacting species. That ionic equation is
repeated below. Draw a slash through the "spectator ions" on both sides of this equation now.
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©1997, A.J. Girondi
Ionic Equation:
H1+(aq) + Cl1-(aq) + Na1+(aq) + OH1-(aq) ---> HOH(l) + Na1+(aq) + Cl1-(aq)
An equation that includes only the reacting species is called a net ionic equation. Write the net ionic
equation for this reaction below:
Net ionic equation:
{1}_____________________________________________________________
By using the appropriate subscripts to indicate phases and complete formulas for the compounds, the
"formula" equation can also be written as a traditional double replacement equation:
HCl(aq) + NaOH(aq) ----> HOH(l) + NaCl(aq)
Let's review the three methods for writing the equation for the reaction between HCl and NaOH:
Formula Equation:
HCl(aq) + NaOH(aq) ----> HOH(l) + NaCl(aq)
Ionic Equation:
H1+(aq) + Cl1-(aq) + Na1+(aq) + OH1-(aq) ---> HOH(l) + Na1+(aq) + Cl1-(aq)
Net Ionic Equation:
H1+(aq) + OH1-(aq) ----> HOH(l)
Following the examples given in this chapter, write the balanced formula, ionic, and net ionic
equations for the strong acid–strong base combinations in problems 2 and 3. Use oxidation numbers to
write the correct formulas for the products. Be sure that each equation is balanced. (If no spectator ions
are present, the ionic and net ionic equations are identical.)
Problem 2. Complete the equations for the reaction between HBr and KOH.
a. formula:
_____HBr(aq) + _____KOH(aq) --->
b. ionic:
_______________________________________________________
c. net ionic:
_______________________________________________________
What are the spectator ions in the reaction in problem 2 above?
_________________________
{2}________________________
Problem 3. Complete the equations for the reaction between HI and LiOH.
formula:
_____HI (aq) +
_____LiOH(aq)
----> __________________________
ionic:
_______________________________________________________
net ionic:
_______________________________________________________
Some acids contain more than one hydrogen "acidic" H and some bases contain more than one
hydroxide ion. More than one molecule of water is formed in reactions between such acids and bases.
You will gain some practice with such acids and bases in problem 4 below.
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©1997, A.J. Girondi
Problem 4. Complete the formula equations below which illustrate acid–base "neutralization" reactions.
Be sure to balance each equation after you have written the correct formulas for the products (using
oxidation numbers). The first one has been completed for you as an example:
a. _____H 2SO 4(aq) + _____Ca(OH)2(aq) ---->
______2 HOH(l) + CaSO4(aq)______
b. _____H 3PO 4(aq) + _____Ba(OH)2(aq) ---->
_____________________________
c. _____H 2CO3(aq) + _____CsOH(aq) ---->
_____________________________
d. _____HClO4(aq) + _____Sr(OH)2(aq) ---->
_____________________________
The three forms of equations (formula, ionic, and net ionic) can be written for reactions of acids
which have more than one acidic H and for bases which contain more than one OH 1- ion. For example, in
part a of problem 4 we wrote the formula equation for the reaction between H2SO 4 and Ca(OH)2. The ionic
equation for that reaction is written as:
2 H1+(aq) + SO42-(aq) + Ca2+(aq) + 2 OH1-(aq) ----> 2 HOH(l) + Ca2+(aq) +SO42-(aq)
If we drop out the spectator ions, the result is the net ionic form of the equation:
2 H1+(aq) + 2 OH1-(aq) ---->
2 HOH(l)
Since a balanced equation should always have the lowest ratio of coefficients, we reduce the above to:
H1+(aq) + OH1-(aq) ---->
HOH(l)
Problem 5. Write the three forms of the equation for the reaction between H2SO 4 and LiOH:
a. formula:
___________________________________________________
b. ionic:
___________________________________________________
c. net ionic:
___________________________________________________
Problem 6. Write the three forms of the equation for the reaction between H3PO 4 and NaOH:
a. formula:
___________________________________________________
b. ionic:
___________________________________________________
c. net ionic:
___________________________________________________
Problem 7. Write the three forms of the equation for the reaction between HBr and Ba(OH)2:
a. formula:
___________________________________________________
b. ionic:
___________________________________________________
c. net ionic:
___________________________________________________
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©1997, A.J. Girondi
ACTIVITY 22.2
The Production of a Salt by an Acid–Base Reaction
In this activity you will be producing solid NaCl by reacting an HCl solution with a solution of NaOH.
Follow the procedure carefully.
The reaction equation is:
HCl(aq) + NaOH(aq) ----> NaCl(aq) + HOH(l)
Procedure:
1. Weigh a clean dry evaporating dish and watch glass together to the nearest 0.01 g. Record the mass in
Table 22.1.
2. Obtain 10.0 mL of 1M HCl solution and pour it into the evaporating dish.
3. Obtain 10.0 mL of 1M NaOH solution and carefully add it to the HCl solution in the evaporating dish.
Caution: keep NaOH solution off of skin and clothes.
4. Cover the dish with the watch glass and, using a laboratory burner, heat the contents of the
evaporating dish to a boiling. Boil the mixture gently until it is dry.
5. When the dish has cooled, weigh the dish with contents (NaCl) and watch glass to the nearest 0.01 g.
Table 22.1
Production of Salt
1. Mass of dish and watch glass
_________g
2. Mass of dish, watch glass, and NaCl
_________g
3. Mass of NaCl formed (2 - 1)
_________g
Calculations:
1. Starting with 10.0 mL of 1.0 M NaOH solution, calculate the theoretical amount of NaCl which should
have been produced in this reaction. Do this by finishing the partially completed set-up below.
10.0 mL NaOH X
1 L NaOH
X
1000 mL NaOH
1 L NaOH
X
1 mole NaOH
=
X
g NaCl
1 mole NaCl
_______ g NaCl
2. Using the theoretical mass of NaCl above as the accepted value, and the mass of NaCl produced in
your experiment as the observed value, calculate your percentage error.
Error = __________%
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©1997, A.J. Girondi
SECTION 22.3 The Dissociation Constant Of Water, K w
Pure water is capable of conducting a very small amount of electrical current. It must, therefore,
contain a very small concentration of ions. Pure water contains ions because polar water molecules can
react with themselves! Under the proper conditions, it is possible for the partially negatively charged
(oxygen) end of one water molecule to pull a hydrogen ion away from another water molecule:
H2O
+
H2O
<====>
H3O1+
+
OH1-
The result of this reaction is the formation of a hydronium ion and a hydroxide ion. The reverse
reaction is very good, since H 3O1+ is an excellent acid and OH1- is an excellent base. So an equilibrium is
established, with far more reactants than products present. In fact, only about two water molecules in one
billion are in the form of ions at any point in time! The equation for the reaction is:
H2O(l) + H2O(l) <===> H3O1+(aq) + OH1-(aq)
Since there is an equilibrium, we can write an equilibrium expression. There will be no denominator, since
the reactant, H2O, is a pure liquid:
K w = [H3O1+] [OH1-]
The equilibrium constant for this system is given the symbol, Kw, and is called the dissociation constant of
water. Experiments have been done to determine the concentrations of H3O1+ and OH1- in water. It was
found that in pure water:
[H3O1+] = 1 X 10-7 and [OH1-] = 1 X 10-7
Therefore, the value of K w is calculated:
Kw = [H3O1+] [OH1-] = (1 X 10-7)2 = 1 X 10-14
K w = 1 X 10-14
Since water contains some hydrogen ions and some hydroxide ions, all water solutions must also
contain at least a small concentration of each of these ions. This includes all of the solutions of acids and
bases that we discussed in Chapter 21. Water solutions of acids contain both H1+ and OH 1- ions, and
water solutions of bases contain both H 1+ and OH 1- ions. Pure water is considered neutral because the
concentrations of H1+ and OH1- are equal at 1 X 10-7 M. Acid solutions have hydrogen ion concentrations
that are greater than their hydroxide ion concentrations.
Basic solutions have hydroxide ion
concentrations that are greater than their hydrogen ion concentrations. However, the product
[H1+][OH1-] for all water solutions must always equal 1 X 10-14.
Water: [H1+] = [OH1-] = 1 X 10-7 M
Acids: [H1+] > [OH1-]
Bases: [H1+] < [OH1-]
For Acids, Bases, and All Other Water Solutions: [H1+ ] [OH1-] = 1 X 10-14
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©1997, A.J. Girondi
Examine Table 22.2 below. It shows the relationship between [H1+], [OH1-], and Kw.
Table 22.2
The Relationship Between [H 1+], [OH 1-], and K w in Water Solutions
Strong Acids ----->
Weak Acids ---->
Neutral ----->
Weak Bases ---->
Strong Bases ---->
[H 3O 1+]
[OH 1-]
Kw
1 X 100
1 X 10-1
1 X 10-2
1 X 10-3
1 X 10-4
1 X 10-5
1 X 10-6
1 X 10-7
1 X 10-8
1 X 10-9
1 X 10-10
1 X 10-11
1 X 10-12
1 X 10-13
1 X 10-14
1 X 10-14
1 X 10-13
1 X 10-12
1 X 10-11
1 X 10-10
1 X 10-9
1 X 10-8
1 X 10-7
1 X 10-6
1 X 10-5
1 X 10-4
1 X 10-3
1 X 10-2
1 X 10-1
1 X 100
1 X 10-14
1 X 10-14
1 X 10-14
1 X 10-14
1 X 10-14
1 X 10-14
1 X 10-14
1 X 10-14
1 X 10-14
1 X 10-14
1 X 10-14
1 X 10-14
1 X 10-14
1 X 10-14
1 X 10-14
Now let's use this information and the concept of Kw to solve some problems. We'll begin with an
example:
Sample Problem: For a particular solution, [OH1-] = 1.0 X 10-5. Calculate [H1+] for this solution, and
identify it as acidic, basic, or neutral.
You are given [OH1-] and asked to find [H1+]. An equation that includes both of these quantities is the
expression for Kw: Kw = [H1+] [OH 1-]. Substitute the given value for [OH 1-] and for K w into the equation,
and solve for [H1+]:
K w = [H 1+][OH1- ]
[H 1+ ] =
Kw
[OH 1- ]
[H 1+ ] =
1 X 10 -14
1 X 10 -5
thus, [H1+ ] = 1 X 10 -9
We have found that [H1+] = 1 X 10-9, while [OH1-] = 1 X 10-5 for this solution. Since [OH1-] > [H 1+], the
solution is basic.
Table 22.2 reveals that as [H1+] gets larger, [OH 1-] gets smaller. K w remains constant. Now try the
problems below. Be sure to enter exponential numbers into your calculator in the correct way, using the
exponent key, either [EXP] or [EE]. Show your work.
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©1997, A.J. Girondi
Problem 8. A solution has [H 1+] = 1.0 X 10 -4. Calculate [OH1-] for this solution, and identify it as acidic,
basic, or neutral. Show work.
[OH1-] = _____________ Solution is _____________
Problem 9. A solution has [OH 1-] = 4.8 X 10 -6. Calculate [H1+] for this solution, and identify it as acidic,
basic, or neutral. Show work.
[H1+] = _____________ Solution is ______________
Problem 10. A solution has [H1+] = 8.3 X 10-2. Calculate [OH1-] for this solution, and identify it as acidic,
basic, or neutral. Show work.
[OH1-] = ____________ Solution is ______________
SECTION 22.4 The pH Scale
As you can see in the problems above, in chemistry we often work with small concentrations of
H1+ and OH1- ions. The quantities in the measurements are usually negative exponential numbers.
These are not fun to work with! So, in order to characterize a solution as acidic, basic, or neutral without
having to use negative exponential numbers, scientists devised what is known as the pH scale. The small
letter "p" in chemistry is frequently used as an abbreviation meaning "negative logarithm of." The H stands
for hydrogen ion concentration: [H1+]. Thus, pH is a shorthand way of saying "the negative logarithm of
the hydrogen ion concentration of a solution." Or, pH = - log [H1+]. Now you must be wondering why this
concept was ever devised! Well, follow along with this explanation.
Logarithm is simply another word for exponent. Common logarithms are powers of 10. The
logarithm of 100 is 2. Why? Because 100 = 10 2. If someone asks you, "What is the log of 100?" What
they are really asking is, "To what power must you raise 10 to get 100?" The answer, of course, is 2. It is
easy to find "logs" using a scientific calculator. Simply enter the number and press the log key.
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©1997, A.J. Girondi
Problem 11. Find the logs of the numbers listed below.
a. log 1000 =
______________
d. log 450 =
______________
b. log 10 =
________ _____
e. log 8932 =
__ ___________
f. log 4.5 X 103 = ______________
c. log 10000 = ______________
To calculate the negative logarithm of a number, follow the procedure for finding the log on your
calculator, and then change the sign of your answer using the change sign key (+/-).
Problem 12. Do the problems below.
a. - log 679 =
________________
c. - log 0.034 =
_________________
b. - log (4.60 X 102)=
________________
d. - log (8.2 X 10-3) =
_________________
Let's find the negative log of [H 1+] for water: - log(1 X 10-7) = 7. You can see that by finding the negative
log of a negative exponential number, you end up with a positive integer. It is much easier to work with
positive integers than with negative exponents. So, scientists decided that it would be better to describe
the acidic or basic nature of solutions using positive integers than by using the actual values of [H1+] or
[OH1-] for the solution. Instead of talking about the hydrogen ion concentration of a solution, we can talk
about its pH. If we take the values in Table 22.2, and find the negative logs of them, we can express them
all as positive integers. [H1+] becomes pH, [OH1-] becomes pOH, and Kw becomes pK w. This has
been done for you in Table 22.3.
Table 22.3
The Relationship Between pH, pOH,
and pK w in Water Solutions
Strong Acids --->
Weak Acids --->
Neutral Solutions --->
Weak Bases --->
Strong Bases --->
pH
pOH
pK w
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
14
14
14
14
14
14
14
14
14
14
14
14
14
14
14
Notice that the sum of pH and pOH of a solution is always equal to 14
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©1997, A.J. Girondi
Problem 13. Find the pH of solutions with the following [H1+]:
a. 4.60 X 10-3 __________________
c. 0.00680
b. 8.80 X 10-9 __________________
d. 3.40 X 10-7 __________________
__________________
To do the next two problems, you will need to use the formulas for Kw and for pH:
Problem 14. What is the pH of a solution with [OH1-] = 5.20 X 10-5?
pH = __________
Problem 15. What is the pOH of a solution with [H1+] = 1.20 X 10-10?
pOH = __________
As pH {3}
{4}
creases, the strength of an acid solution decreases. Acids get stronger as their [H1+]
creases. A neutral solution has a pH of {5}___ _ . The stronger the acid, the
{6}____________
the pH. The stronger the base, the {7}_______________ the pH.
<--- stronger acids
0 1 2 3 4 5
6
7
8
9
10
stronger bases --->
11 12 13 14
Figure 22.1 The pH Scale
The pH scale was first devised by the Danish biochemist, Sven Sorenson in 1909. It is an openended scale which means that it is possible for solutions to have pH values which are less than 0 or greater
than 14. However, the vast majority of solutions fall into this range. The pH scale of 0 to 14 includes an
extremely large range of [H 1+] concentrations. Remember that pH values are logs, which means they are
powers of ten. When pH = 1, [H1+] = 1 X 10-1. When pH = 7, [H1+] = 1 X 10-7. This difference between pH
= 1 and pH = 7 represents a difference of 6 powers of 10 which is 1,000,000! Let's try the problem below.
A solution with pH = 2 has how many times more H1+ ions per liter than one with pH = 7? {8}___
How many times more hydrogen ions are in a solution of pH 1 compared to a solution of pH 12?
Answer: {9}______________________________________
Another type of problem which is a little trickier than what you have done involves finding [H1+] for
a solution when you know its pH. For example, if the pH of a solution is 8.5, what is its [H1+]? The correct
answer is 3.2 X 10-9 M. This kind of problem involves the use of inverse log which is also known as antilog.
Problems of this type are discussed in Appendix F of your ALICE materials. Ask your teacher if he/she
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©1997, A.J. Girondi
wishes to include this material in this chapter. If so, read Appendix F and complete the problems there.
Otherwise, move on to Activity 22.3.
GO TO APPENDIX F -----> ???
(CHECK WITH YOUR INSTRUCTOR.)
ACTIVITY 22.5 Acid-Base Indicator Solutions
There are several ways to determine the pH of solutions. One method involves the use of
chemical indicators. These are substances which change color when added to acidic or basic solutions.
Bromthymol blue is an example. Many of these indicators are actually very, very weak acids which are plant
pigments. Let's abbreviate the chemical formula of bromthymol blue as "HIn." Since it is a very weak acid,
it exists in water in equilibrium:
HIn (aq) + H2O(l) <===> H3O1+(aq) + In 1-(aq)
yellow
blue
If we add few drops of this indicator solution to an acid (which has an excess of H3O1+), the H3O1+ ions in
the acid would push the equilibrium to the (left / right)
shift more to (yellow / blue) {11}
{10}_____________.
This would cause the color to
. If we add some bromthymol blue to a basic solution (which
has an excess of OH1- ions), the OH 1- ions in the base will react with the H 3O1+ in the indicator, producing
water. The result is a lower [H3O1+], which will shift the bromthymol blue equilibrium to the (left/right)
{12}_____________.
The color would then shift more to (yellow/blue)
. Notice that to
{13}
serve as an indicator, the molecule (HIn) form of the weak acid must have a different color than the anion
form (In 1-).
Bromthymol blue has a pH interval of 6.0 to 8.0. This means that it is yellow in a solution with a pH
of less than 6.0, while it is blue in a solution with a pH above 8.0. Between these two pH values the
indicator is changing its color. So, if you add bromthymol blue to a solution and the result is a yellow color,
you can predict that the pH is less than 6.0. If it turns blue, you would conclude that the pH is above 8.0. If
it turns green, then the pH is somewhere in the middle of the pH interval, around 7. There are many acidbase indicators, and they have different pH intervals. A few are listed in Table 22.3.
Table 22.4
A Selection of Acid-Base Indicators
Indicator
Acid Color
Base Color
bromthymol blue
litmus
methyl orange
methyl violet
phenolphthalein
alizarin yellow
yellow
red
red
yellow
colorless
yellow
blue
blue
yellow
blue
pink
red
pH Interval
6.0 - 8.0
5.5 - 8.0
3.1 - 4.4
0.0 - 1.6
8.2 - 10.0
10.1 - 12.0
In this activity you will use several acid-base indicators in an attempt to estimate the pH of some
selected solutions.
1. Obtain a dropper bottle of any solution with a pH < 3.1 and a bottle of any solution with a pH > 4.4. (The
exact pH values are not important.)
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©1997, A.J. Girondi
2. Place ten drops of these solutions into separate wells of a dropping plate. Add a drop of methyl orange
indicator to each solution. Note the colors. Do not discard these mixtures.
3. Using two clean wells of the same dropping plate, repeat the procedure above using any two solutions
with pH < 6.0 and pH > 8.0, and adding a drop of bromthymol blue to each. Note the colors. Do not
discard the mixtures.
4. Repeat the procedure once more using any two solutions with pH < 8.2 and pH > 10.0, and adding
phenolphthalein indicator solution. Note the colors. Do not discard the mixtures. Compare your results to
the colors and pH intervals listed in Table 22.4.
5. Add ten drops of white vinegar to each of three clean wells of a second dropping plate. Add a drop of
methyl orange to one of the wells of vinegar, a drop of bromthymol blue to the second, and a drop of
phenolphthalein to the third. Compare the colors to those of the indicators in the solutions of known pH.
Record your observations in Table 22.5 below, and estimate the pH of the vinegar. Your estimates might
be something like these: < 4.4; > 8.2; or maybe, < 2.3.
6. Repeat the procedure in step 5 with the following solutions: household ammonia, a solution of
laboratory detergent, a colorless soft drink solution (like 7-up), and tap water. Record all observations and
pH estimates in Table 22.5 below.
Table 22.5
pH of Common Household Substances
Substance
Color in
Methyl Orange
Color in
Bromthymol Blue
Color in
Phenolphthalein
pH
vinegar
___________
_____________
_____________
______
ammonia
___________
detergent
___________
soft drink
___________
A more accurate method for determining pH involves the use of an instrument called a pH meter.
It is a sensitive device that must be used with care, and the proper procedure must be followed. Your
teacher will assist you in the use of the pH meter as you measure the pH of each of the substances in
Table 22.5 above. Record your results below:
Vinegar: __________ Ammonia: __________ Detergent: __________ Soft Drink: __________
Do all of the estimates of pH which you made using the indicators agree with the more specific results
obtained using the pH meter?
If not, which ones did not agree?_____________________
. They all should be in agreement, unless there was
some error in the procedure or equipment.
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©1997, A.J. Girondi
SECTION 22.6
Acid–Base Titrations
At this point we are ready to determine the outcome when acidic and basic solutions are added
together. Earlier in this chapter you completed an exercise in which reactions between acids and bases
produced salts and water. These were examples of neutralization reactions. Some degree of
neutralization occurs whenever an acid is mixed with a base. Water is always one of the products formed in
such a reaction. Consider the neutralization reaction that occurs when HCl and NaOH solutions are mixed.
One way that we wrote this reaction was:
HCl(aq) + NaOH(aq) ----> HOH(l) + NaCl(aq)
According to this
equation, 1 mole of HCl is required to neutralize 1 mole of NaOH. Laboratory experiments often involve
neutralization reactions. Suppose you knew the molar concentration (M) of one of the solutions (the acid
or the base) and you wanted to determine the concentration of the other solution. You could determine
the unknown concentration by mixing the acid and base together until the neutralization was complete.
Then, the unknown concentration can be calculated using the volumes of the acid and base solutions
which were consumed. Such an experimental procedure is called a titration. Let's try some examples:
Sample Problem: What is the molarity of a solution of HCl if 48 mL of 0.25 M NaOH solution are
required to neutralize 35 mL of HCl?
HCl(aq) + NaOH(aq) ----> HOH(l) + NaCl(aq)
We will solve this problem by setting up a "fencepost" and by using dimensional (unit) analysis. Since we
are looking for the molarity of HCl, we will want to be left with units of moles HCl divided by liters of HCl
solution: moles HCl / L sol'n. Although we could start the fencepost with any of the information given,
let's go with 48 mL NaOH:
48 mL NaOH X
1 L NaOH
0.25 mole NaOH
1 mole HCl
1
1000 mL HCl
X
X
X
X
1000 mL NaOH
1 L NaOH
1 mole NaOH 35 mL HCl
1 L HCl
=
0.34 mole HCl
or, 0.34 M HCl
1 L HCl
Sample Problem: What is the molarity (M) of a solution of KOH if 45 mL of 0.20 M H2SO 4 are required to
neutralize 34 mL of KOH solution?
H2SO 4 + 2 KOH ----> 2 HOH + K2SO 4
45mL H 2 SO4 X
1 L H 2SO 4
0.20 mole H 2 SO 4
2 moles KOH
1
X
X
X
...
1000 mL H 2 SO 4
1 L H 2 SO 4
1 mole H 2 SO 4
34 mL KOH
X
1000 mL KOH
0.53 mole KOH
=
or, 0.53 M KOH
1 L KOH
1 L KOH
Sample Problem: A beaker contains 0.11 L of 0.35 M Ca(OH)2 solution. What volume (in L) of 0.50 M
H3PO 4 solution will be required to neutralize it?
2 H3PO 4 + 3 Ca(OH)2 ----> 6 HOH + Ca3(PO4)2
0.11 L Ca(OH) 2 X
0.35 mole Ca(OH)2
2 moles H 3 PO4
1 L H 3 PO4
X
X
= 0.051 L H 3PO 4
1 L Ca(OH)2
3 moles Ca(OH)2
0.50 mole H 3 PO4
Study the three sample problems above carefully, and then do the problems which follow. Show your setup in each case. All measurements should have units. Complete a balanced equation before starting
each problem.
22-14
©1997, A.J. Girondi
Problem
16.
If 24.60 mL of 0.18 M HNO 3 is titrated with 22 mL of KOH solution.
concentration (molarity) of the base?
HNO3 + KOH ---> _________
What is the
_______
__________ M KOH
Problem
17. If 40.0 mL of H3PO 4 are neutralized by 22 mL of 0.60 M NaOH, What is the concentration
of the acid?
H3PO 4 + 3 NaOH ---> __________________________
_________ M H3PO 4
Problem
18. How many mL of 2.0 M H 2SO 4 will be needed to neutralize 40.0 mL of 0.80 M Ca(OH)2?
H2SO 4 + Ca(OH)2 ---> CaSO4 + ______________________
__________ mL H2SO 4
Problem
19. If 16 mL of 0.15 M H 3PO 4 are needed to neutralize a 0.10 M solution of Mg(OH)2, what is
the volume (in mL) of the base solution?
2 H3PO 4 + 3 Mg(OH)2 ---> ________________________
__________ mL Mg(OH)2
22-15
©1997, A.J. Girondi
ACTIVITY 22.7
A Titration of Vinegar
You are now going to put what you have learned to
practical use. You are going to determine the percentage of
acetic acid in vinegar. You will perform a titration by adding a
base to the acetic acid until neutralization occurs. One of the
most important parts of a titration, is knowing when the acid and
base have been completely neutralized. We will get this
information by using phenolphthalein indicator solution.
When doing titrations, volumes of solutions must be
very carefully measured. A graduated cylinder is normally used
to measure volume. But, in titrations, we use a piece of
glassware that is even more accurate. It is called a buret (also
spelled burette). The volume of solution in the buret should
be read the same way that you read a graduated cylinder.
Read the bottom of the meniscus.
Figure 22.2
Reading a Meniscus
Get the materials labeled 22.7 from the materials shelf.
Follow the procedure below. Wear safety glasses and an
apron.
Procedure:
1. Rinse two burets with some distilled water. Mount the clean burets on a ring
stand with a double buret holder as shown in Figure 22.3.
2. Place funnels in the mouths of the burets, and fill one of the burets with
water. Practice using the burets by measuring out volumes of 5 and 10 mL.
Drain any remaining water.
3. Pour a few mL of 0.50 M NaOH into one buret to rinse it. Drain the solution
out, and then fill that buret with the 0.5 M NaOH solution. Remove the air in the
tip of the buret below the stopcock by draining some of the solution. Record
the starting level and the concentration of NaOH (0.50 M) in Table 22.6.
4. For the other buret, follow the procedure in step 3 but use white vinegar
instead of NaOH solution.
5. Carefully measure three 10 mL volumes of vinegar from the buret into three
125 mL Erlenmeyer flasks. Label your flasks 1, 2, and 3. Add 5 to 10 drops of
phenolphthalein indicator to each flask.
Figure 22.3
Burets and Holder
6. Place one flask on a small piece of white paper under the buret containing NaOH, and add a few mL of
the NaOH solution from the buret while you swirl the solution in the flask. The pink color of the indicator
may appear briefly, but swirl the solution until the color disappears. (Use a magnetic stirrer if one is
available in your lab. It will make stirring easier.)
7. Continue to add NaOH solution to the flask, dropwise, with thorough mixing. We will use this first
titration to get a rough idea of about how much NaOH is required to neutralize the vinegar. Therefore, you
can allow the drop rate to go rather rapidly. When you reach the point where a slight pink color persists
(does not disappear), you have reached the end point of the titration. Record the level of base left in the
buret in Table 22.6.
22-16
©1997, A.J. Girondi
Table 22.6
Titration of Vinegar
Sample 1
Sample 2
0.50
0.50
Sample 3
1. Initial buret reading (mL)
2. Final buret reading (mL)
3. mL of base added
4. Volume of vinegar (mL)
5. Concentration (M) of NaOH
0.50
8. Repeat the titration for the other two samples of vinegar. In each case, fill the NaOH buret, record the
initial level, run the titration, and record the final level of base. The second two titrations should go more
quickly, because you already know the approximate final level of base in the buret. Run an amount of
base into the second flask that is 1 or 2 mL less than that required in the first titration. Then complete the
titration by carefully adding the last amount of base dropwise with good mixing. A good end point will
leave the mixture in the flask with only a faint pink color.
9. Rinse both burets with water when you are finished. Record all data in Table 22.6.
Calculations:
1. Calculate the concentration (M) of acetic acid (HC2H3O2) in the vinegar. Follow the examples in section
22.6. (For the volume of NaOH, use an average of the two closest volumes from your 3 titrations.)
HC2H3O2 + NaOH ---> HOH + NaC2H3O2
__________ M
2. Using atomic masses from the periodic table, calculate the molecular mass of acetic acid, HC2H3O2.
__________ grams / mole
3. Calculate the mass in grams of acetic acid in a liter of vinegar. (Your answer to calculation one is in moles
per liter. Change that value to grams per liter using the molecular mass you found in calculation 2.)
__________ g HC2H3O2 / L
22-17
©1997, A.J. Girondi
4. Assume that a liter of vinegar has a mass of 1000. grams. Calculate the mass percent of acetic acid in
vinegar:
mass of acid per liter
X 100 = % acetic acid in vinegar
mass of 1 liter
__________% HC2H3O2
5. Check the bottle of vinegar from which your samples came. What is the percent of acid in the vinegar
according to the label? _____________ %
6. Comment on how your result compares to that on the label of the vinegar bottle.
7. Calculate your percent error. The formula is in the reference section of your ALICE materials.
__________ % Error
SECTION 22.8
Optional Review Problems
The following problems are optional, and can be done on a separate sheet of paper if you feel that
you need additional practice.
Problem 20. Write and balance the three forms of the equation for the reaction between HClO3(aq) and
Ca(OH)2(aq).
Problem 21. If the [H3O1+] of a solution is 8.6 X 10-12, find the [OH1-] of the same solution.
Problem 22. If the [H3O1+] of a solution is 5.2 X 10-9, calculate the pH and pOH and identify the solution
as an acid or a base.
Problem 23. If the [OH1-] of a solution is 3.5 X 10-4, calculate the pH and pOH and identify the solution
as an acid or a base.
Problem 24. What volume (in mL) of a 0.22M HCl solution are required to neutralize 64 mL of a 0.12M
solution of Ba(OH)2?
Problem 25. During a titration procedure, it is found that 38 mL of a 2.5M solution of KOH are needed
to neutralize 57 mL of a solution of H2SO 4. What is the molarity (M) of the H2SO 4 solution?
22-18
©1997, A.J. Girondi
SECTION 22.9
Learning Outcomes
Understanding acid-base chemistry is very important, because many chemical reactions involve acids
and bases. Check the learning outcomes below. When you feel you have mastered them, arrange to take
any quizzes or exams on Chapter 22. Then, move on to Chapter 23.
_____1. Given an acid and a base, properly complete and balance the equation for the reaction that
occurs between them in the formula, ionic, and net ionic forms.
_____2. Explain pH and classify a solution as acid, base, or neutral given its pH, pOH, [H1+], or [OH1-].
_____3. Describe the relationship between the conductivity of an acid solution, the size of its Ka, and its
pH.
_____4. Given [H1+], calculate [OH1-] and pH; Given [OH1-], calculate pH and [H1+].
_____5. Explain the proper procedure for conducting a titration.
_____6. Calculate concentrations or volumes in titration problems.
The following learning outcome pertains to Appendix F:
_____7. Determine the [H 1+] or [OH1-] of a solution, given its pH or pOH.
22-19
©1997, A.J. Girondi
SECTION 22.10
Answers to Questions and Problems
Questions:
{1} H1+(aq) + OH1-(aq) ----> HOH(l) ; {2} Br1- and K1+; {3} in; {4} de; {5} 7.0; {6} lower; {7} higher;
{8} 1 X 105; {9} 1 X 1011; {10} left; {11} yellow; {12} right; {13} blue
Problems:
1. a. HCl + LiOH ---> HOH + LiCl
b. HNO3 + KOH ---> HOH + KNO3
c. HBr + NaOH ---> HOH + NaBr
2. a. HBr + KOH ---> HOH + KBr
b. H1+(aq) + Br1-(aq) + K1+(aq) + OH1-(aq) ---> HOH(l) + Br1-(aq) + K1+(aq)
H1+(aq) + OH1-(aq) ---> HOH(l)
c.
3. a. HI (aq) + LiOH(aq) ---> HOH(l) + LiI (aq)
b. H1+(aq) + I1-(aq) + Li1+(aq) + OH1-(aq) ---> HOH(l) + I1-(aq) + Li1+(aq)
c. H1+(aq) + OH1-(aq) ---> HOH(l)
4. a. H2SO 4(aq) + Ca(OH)2(aq) ----> 2 HOH(l) + CaSO4(aq)
b. 2 H3PO 4(aq) + 3 Ba(OH)2(aq) ----> 6 HOH(l) + Ba3(PO4)2(s)
c. H2CO3(aq) + 2 CsOH(aq) ----> 2 HOH(l) + Cs2CO3(aq)
d. 2 HClO4(aq) + Sr(OH)2(aq) ----> 2 HOH(l) + Sr(ClO4)2(aq)
5. a. H2SO 4(aq) + 2 LiOH(aq) ----> 2 HOH(l) + Li2SO 4(aq)
b. 2 H1+(aq) + SO42-(aq) + 2 Li1+(aq) + 2 OH1-(aq) ----> 2 HOH(l) + 2 Li1+(aq) + SO42-(aq)
c. 2 H1+(aq) + 2 OH1-(aq) ---> 2 HOH(l) which reduces to H1+(aq) + OH1-(aq) ---> HOH(l)
6. a. H3PO 4(aq) + 3 NaOH(aq) ----> 3 HOH(l) + Na3PO 4(aq)
b. 3 H1+(aq) + PO43-(aq) + 3 Na1+(aq) + 3 OH1-(aq) ----> 3 HOH(l) + PO43-(aq) + 3 Na1+(aq)
c. 3 H1+(aq) + 3 OH1-(aq) ---> 3 HOH(l) which reduces to H1+(aq) + OH1-(aq) ---> HOH(l)
7. a. 2 HBr(aq) + Ba(OH)2(aq) ----> 2 HOH(l) + BaBr2(aq)
b. 2 H1+(aq) + 2 Br1-(aq) + Ba2+(aq) + 2 OH1-(aq) ----> 2 HOH(l) + 2 Br1-(aq) + Ba2+(aq)
c. 2 H1+(aq) + 2 OH1-(aq) ---> 2 HOH(l) which reduces to H1+(aq) + OH1-(aq) ---> HOH(l)
8. 1.0 X 10-10; acid
9. 2.1 X 10-9; base
10. 1.2 X 10-13; acid
11. a. 3; b. 1; c. 4; d. 2.7; e. 3.951; f. 3.7
12. a. -2.83; b. -2.66; c. 1.5; d. 2.1
13. a. 2.34; b. 8.06; c. 2.17; d. 6.47
14. pH = 9.72
15. pOH = 4.08
16. 0.20 M KOH
17. 0.11 M H3PO 4
18. 16 mL H2SO 4
19. 36 mL Mg(OH)2
20. 2 HClO3(aq) + Ca(OH)2(aq) ----> 2 HOH(l) + Ca(ClO3)2(aq)
2 H1+(aq) + 2 ClO31-(aq) + Ca2+(aq) + 2 OH1-(aq) ----> 2 HOH(l) + Ca2+(aq) + 2 ClO31-(aq)
2 H1+(aq) + 2 OH1-(aq) ----> 2 HOH(l) which reduces to H1+(aq) + OH1-(aq) ----> HOH(l)
21. 1.16 X 10-3
22. pH = 8.28; pOH = 5.72; base
23. pH = 10.5; pOH = 3.46; base
24. 69.8 mL HCl
25. 0.83M H2SO 4
22-20
©1997, A.J. Girondi