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Acid-Base Equilibria You have just completed a chapter on equilibrium. That chapter focused primarily on gas phase reactions (with a few exceptions). This section on Acid-Base equilibria (along with the following section on solubility equilibrium) will deal exclusively with aqueous solution systems. You should be aware that the coverage of this topic in your text is a bit weak in a few areas (many freshman chemistry texts contain two chapters on acid-base equilibrum. This should not bother you . . . after all, you have already gained the necessary skills to handle the mathematics of equilibrium systems.

Objectives

1. Know and be capable of applying the Bronsted-Lowery model of acids and bases (inculdig the concepts related to conjugate acid-base pairs.

2. Understand and be able to apply the terms conjugate acid and conjugate base.

3. Understand the importance of the self-ionization of water and be able to perform calculations involving Kw or pKw.

4. Know how the pH scale works and be capable of perforiming calculations involving pH, pOH, [OH-], [H+] as needed.

5. Memorize the list of strong acids and strong bases. Be capable of performing calculations of pH, pOH, [OH-], [H+] for these solutions.

6. Be capable of performing equilibrium calculations for weak acids including the use of Ka, pKa, pH, pOH, [OH-], [H+].

7. Understand the ionization behavior of polyprotic acids and polyprotic bases. Particular care should be taken to understand the significance of the relative magnitude of the successive ionization constants and how to solve for equilibrium concentrations in solutions of polyprotic acids or bases.

8. Understand the nature of distribution curves and how they can depict the relative concentrations of conjugate acid-base pairs as a function of pH. Note: these curves can be useful for buffers, acid-base indicators and polyprotic acids or bases.

9. Be capable of performing equilibrium calculations for weak bases including the use of Kb, pKb, pH, pOH, [OH-], [H+].

10. Recognize and be able to use the relationship between Ka and Kb (or pKa and pKb) for conjugate pairs.

11. Understand the concept of and be able to find or use the numerical value of a percent ionization.

12. Understand how to determine if a salt solution will be acidic, basic, or neutral.

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13. Understand and be capable of applying the definitions related to buffers including the terms buffer range and buffer capacity.

14. Know the Henderson-Hasselbach equation and be able to solve buffer problems including determination of buffer pH, buffer range, conjugate base/acid ratios, and the affect of added base or acid on the pH of any given buffer.

15. Use the Henderson-Hasselbach equation to understand the pH dependence of the color of acid-base indicators.

16. Recognize that the equilibrium constants for acid-base neutralization reactions are large (this means that these reactions can be regarded as occurring to 100 % completion).

17. Understand the important features of titration curves including the significance of equivalence points (vs. endpoints). Pay particular attention to how the shape of the titration curve depends upon the type of titration being done (strong acid/strong base, weak acid/strong base, etc.).

18. Become proficient at performing titration curve calculations: recognize that the method used depends upon both the type of neutralization reaction that is being done and upon what species are present at a given point in a titration.

Bronsted-Lowry Acid-Base Model

Definitions and Examples

acid: a proton (H+) donor base: a proton (H+) acceptor acid-base reaction: a proton (H+) is transferred from an acid to a base

acid: HCl, H3PO4, H2O, etc. base: NaOH, PO4

3-, H2O, etc.

You will notice that water is listed as an example of both acid and base . . . a species that can act as either a proton donor (acid) OR a proton acceptor (base) is referred to as amphiprotic

Conjugate pairs: a pair of acid and base species that are related to one another structurally (they differ by the presence or absence of one proton).

HCl is an acid . . . when it loses a proton, it becomes Cl- (which is a base). HCl and Cl- are a conjugate acid-base pair.

PO43- is a base . . . when it gains a proton, it becomes HPO4

2- (which can be an acid). HPO42- and

PO43- are a conjugate acid-base pair.

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Water Ionization Constant

As mentioned above, water is amphiprotic. Since water can behave as either an acid or a base, it undergoes a reaction with itself. This process is referred to as autoprotolysis or self-ionization and can be represented as . . .

H2O (l) + H2O (l) <===> H3O+ (aq) + OH-

(aq) However, it is more simply represented as . . .

H2O (l) <===> H+ (aq) + OH-

(aq) You will notice that both of these representations are chemical equilibria. This reaction is of such great importance that it is given a special symbol for its equilibrium constant (Kw). The equilibrium expressions can be written in terms of the concentrations of the species present . . . either as

Kw = [H3O+][ OH-]

or more simply as

Kw = [H+][OH-]

You will note that the concentration of water itself does not appear in these expressions . . . why?

At 25oC Kw = 1.0 x 10-14. What does this tell you about the position of this equilibrium?

NOTE: as the concentration of OH- decreases, the concentration of H+ increases and visa versa (in other words these two concentrations are inversely related).

pH and pOH

In dealing with acid-base chemistry it is not uncommon to here the acidity or basicity of solutions described with the pH scale. This is merely a log10 expression of the proton concentration [H+] of a solution. The necessary relationships are shown below and it is worth noting that ANY TIME you see a “p function” you should recognize that it is a log10 expression of the parameter described (pH, pOH, pKw, pKa, pKb, and others).

Handling pH calculations (or similar calculations for other “p functions”) should not be viewed as “solving a problem”. These are routine transforms from a non-log scale to a log scale. You must master these conversions, becoming proficient at converting as needed.

Important Equations

pH = -log [H+] pOH = -log [OH-] pKw = -log Kw

At 25oC Kw = 1.0 x 10-14 So pKw = 14.00 and

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Since Kw = [H+][ OH-] (as discussed abvove) Then pKw = pH + pOH so . . . at 25oC 14.00 = pH + pOH

Summary of Relationships Between pH, [H+], [OH-], and pOH at 25oC

PH [H+], M [OH-], M pOH 14.00 1.0 x 10-14 1.0 0.00 13.00 1.0 x 10-13 1.0 x 10-1 1.00 12.00 1.0 x 10-12 1.0 x 10-2 2.00 11.00 1.0 x 10-11 1.0 x 10-3 3.00 10.00 1.0 x 10-10 1.0 x 10-4 4.00 9.00 1.0 x 10-9 1.0 x 10-5 5.00 8.00 1.0 x 10-8 1.0 x 10-6 6.00 7.00 1.0 x 10-7 1.0 x 10-7 7.00 6.00 1.0 x 10-6 1.0 x 10-8 8.00 5.00 1.0 x 10-5 1.0 x 10-9 9.00 4.00 1.0 x 10-4 1.0 x 10-10 10.00 3.00 1.0 x 10-3 1.0 x 10-11 11.00 2.00 1.0 x 10-2 1.0 x 10-12 12.00 1.00 1.0 x 10-1 1.0 x 10-13 13.00 0.00 1.0 1.0 x 10-14 14.00 PH [H+], M [OH-], M pOH

pH, pOH, etc. for Strong Acids and Strong Bases

The pH and other related terms are very easily calculated for most solutions of strong acids and strong bases. By definition strong acids dissociate essentially 100 % in water. The same can be said for strong bases. Therefore, we can quickly find pH, [H+], [OH-], and pOH for these solutions as shown in the following examples.

Strong Acid: let's examine a 0.015 M solution of HCl (aq)

[H+] = 0.015 M since it is a strong acid [OH-] = Kw/[H+] = 6.7 x 10-13 M pH = - log [H+] = 1.82 pOH = 14.00 - pH = 12.18

NOTE: after finding [H+] we could have found (in this order) pH, pOH, and then [OH-] since [OH-] = 10-pOH

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Strong Base: let's examine a 0.0223 M solution of Ca(OH)2 (aq)

[OH-] = 0.0446 M since it is a strong base that yields two hydroxides pOH = - log [OH-] = 1.351 pH = 14.00 - pOH = 12.649 [H+] = 10-pH = 2.24 x 10-13 M

NOTE: after finding [OH-] we could have found (in this order) [H+], pH, and pOH since [H+] = Kw/[OH-]

Weak Acids and Their Ionization Constants

Above you examined a strong acid and a strong base solution and solved for pH, pOH, etc. worrying only about one equilibrium (the water autoprotolysis, Kw, equilibrium). When dealing with weak acids or bases, we cannot assume 100% dissociation (or ionization) of the weak electrolyte (the weak acid or base). As such, there will be a minimum of two equilibria to consider . . . the weak acid (or base) equilibrium with water AND the equilibrium of water with itself (the water autoprotolysis, Kw, equilibrium).

This is important and a general point that will apply in this chapter and beyond . . . whenever we are dealing with more than one equilibrium the general rule is . . . perform all calculations involving the equilibrium with the large value of K first.

A generic weak acid can be represented as HB. Its conjugate base and hydronium ion are formed in the acid dissociation reaction shown below.

HB (aq) + H2O (l) <===> H3O+ (aq) + B-

(aq) The equilibrium constant for this process is called the acid ionization constant (Ka). For the strong acid case discussed above Ka is so large we can assume that essentially 100 % of the acid undergoes ionization. However, for a weak acid, only some fraction of the acid undergoes ionization. Let's see how we might handle a weak acid equilibrium . . .

Weak Acid: consider a 0.10 M solution of a weak acid, HB, that has an acid ionization constant, Ka = 1.0 x 10-7.

The reaction and starting information is summarized as . . .

HB (aq) + H2O (l) <===> H3O+ (aq) + B-

(aq) initial 0.10 M ---- ~ 0 M 0 M

The molarity of the conjugate base is zero before the start of the reaction. However, the molarity of hydronium ion is only approximately zero because water reacts with itself to produce some hydronium and hydroxide. Also, note that we are not going to concern ourselves with the water

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because it is a pure liquid. Based on what we know about reaction stoichiometry, we can express the changes in concentrations and the final equilibrium concentrations in terms of a variable . . .

HB (aq) + H2O (l) <===> H3O+ (aq) + B-

(aq) initial 0.10 M ---- ~ 0 M 0 M change - x M ---- + x M + x M

equilibrium 0.10 - x M ---- x M x M As you can see, all the concentrations are expressed in terms of a single variable. We do not concern ourselves with water since it is a pure liquid. The variable expressions of the concentrations can be plugged into the equilibrium constant expression and this expression can be solved. This should all look very familiar . . . do not forget what you learned in the previous chapter!

[ ][ ][ ]

( )( )( )

10.01000.1

10.02

7

3

xx

xxx

HBBOH

Ka

≈

−==

−

−+

NOTE: to solve the above expression, make the simplifying assumption that x is small compared to 0.10 M, cross multiply by 0.10 and take the square root. This assumption is a good idea since Ka has a small numerical value indicating that little forward reaction is required to reach equilibrium. After solving for x, check to make sure the assumption is valid. x = 1.0 x 10-4 M which is small compared to 0.10 M (verify by finding that 0.10 -x ≈ 0.10). This gives the following equilibrium conditions

HB (aq) + H2O (l) <===> H3O+ (aq) + B-

(aq) initial 0.10 M ---- ~ 0 M 0 M change - 1.0 x 10-4 M ---- + 1.0 x 10-4 M + 1.0 x 10-4 M equilibrium 0.10 M ---- 1.0 x 10-4 M 1.0 x 10-4 M

As you can plainly see the reaction only proceeded to a small extent in the forward direction in order to reach equilibrium. This extent can be expressed in terms of a parameter called percent ionization that is defined as follows:

[ ][ ]initial

equilHB

OHxionization

+

=3

100%

Where [H3O+]eq is the hydronium ion concentration at equilibrium and [HB]init is the initial concentration of the weak acid. For our example above, the percent ionization is 0.10 %.

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Polyprotic Weak Acids

Polyprotic acids are acids that are capable of donating more than one proton. For example: phosphoric acid, H3PO4, can donate three protons, H+ (as shown in the reactions below).

The reactions below point out several key features of polyprotic acids:

1. - the anion (ex. H2PO4-1) formed in one step produces another H+ in the next step,

2. - the acid dissociation constants generally become smaller with each successive step,

3. - and as a result, most of the H+ is produced by the first step. Solving polyprotic acid dissociation problems can seem very daunting at first. The table below illustrates the process for determining the equilibrium concentrations of the species found in a 3.0 M solution of H3PO4. You should take care to understand thoroughly the methods used to solve these problems. If you can master these problems, you will be able to handle the material in the subsequent chapters more easily.

Reaction Equilibrium Constant

first proton donation H3PO4 (aq) + H2O(l) <===> H2PO4

-1(aq) + H3O+

(aq) Ka1 = 7.1x10-3

initial 3.0 M --- 0 ~0

change - x --- +x +x

equilibrium 3.0 - x --- x x Assuming x is small (see prior example in last section), the equlibrium expression can be solved easily to find x = 0.15 M (the approximation is good to within 5%, so we do not need to solve the quadratic). This value of x will be used for the starting concentrations in the second ionization step.

Reaction Equilibrium Constant

second proton donation

H2PO4-1

(aq) + H2O(l) <===> HPO4-2

(aq) + H3O+(aq) Ka2 = 6.2 x 10-8

initial x M --- 0 x

change - y --- + y + y

equilibrium x - y --- y x + y

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Assuming y is small, the equlibrium expression can be solved easily to find y = Ka2 = 6.2 x 10-8 (the approximation basically assumes that (x - y) ≈ (x + y) ). This value of y will be used for a starting concentration in the third ionization step.

Reaction Equilibrium Constant

third proton donation HPO4

-2(aq) + H2O(l) <===> PO4

-3(aq) + H3O+

(aq) Ka3 = 4.5 x 10-13

initial y M --- 0 x

change - z --- + z + z

equilibrium y - z --- z x + z Assuming z is small (in comparison to both x and y), the equlibrium expression can be solved easily to find z = (Ka3)(y)/(x) = 1.9 x 10-19 (the approximation is valid ). By the way, the pH of this solution is 0.82, pOH = 13.18, and the [OH-] = 6.7 x 10-14 M.

You should also note that the method employed above only works well when the equilibrium constant is much smaller in each successive ionization step (normally, at least three orders of magnitude difference).

Weak Bases and Their Ionization Constants

We will find that working with weak bases is very similar to working with weak acids. In the following example we will consider how we can use experimental pH measurements to help determine equilibrium concentrations, and equilibrium constants.

A generic weak base can be represented as B. Its conjugate base and hydroxide ion are formed in the base dissociation reaction shown below.

B (aq) + H2O (l) <===> OH- (aq) + HB+

(aq) The equilibrium constant for this process is called the base ionization constant (Kb). For the strong base case discussed above Kb is so large we can assume that essentially 100 % of the base undergoes ionization. However, for a weak base, only some fraction of the base undergoes ionization. Let's see how we might determine a weak base ionization constant . . .

Weak Base: consider a 0.15 M solution of a weak base ammonia, NH3. Determine the ionization constant Kb if the solution has a pH = 11.20 at equilibrium.

B (aq) + H2O (l) <===> OH- (aq) + HB+

(aq) initial 0.15 M ---- ~ 0 M 0 M

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The molarity of the conjugate acid is zero before the start of the reaction. However, the molarity of hydroxide ion is only approximately zero because water reacts with itself to produce some hydronium and hydroxide. Also, note that we are not going to concern ourselves with the water because it is a pure liquid. Based on what we know about reaction stoichiometry and the pH of the solution at equilibrium we can complete the final equilibrium concentrations and the changes in concentrations for this reaction . . .

pH = 11.20, therefore, pOH = 2.80 which leads us to the conclusion that [OH-] = 10-pOH = 1.6 x 10-3 M

B (aq) + H2O (l) <===> OH- (aq) + HB+

(aq) initial 0.15 M ---- ~ 0 M 0 M change - 1.6 x 10-3 M ---- + 1.6 x 10-3 M + 1.6 x 10-3 M equilibrium 0.148* M ---- 1.6 x 10-3 M 1.6 x 10-3 M

* NOTE: [B] = 0.148 M has one extra significant figure, it really rounds to 0.15 M

As you can see, since we know the equilibrium concentration of one of the species we are able to determine all the equilibrium concentrations through the stoichiometry of the reaction. Thus, we are able to solve for Kb using these concentrations . . .

[ ][ ][ ]

( )( )( )

533

107.1148.0

106.1106.1 −+−

=−−

== xxxBBHOHKb

Ka and Kb for Conjugate Pairs

For a conjugate acid-base pair we can write the acid ionization reaction, the base ionization reaction, and add them to one another.

Ka rxn HB (aq) + H2O (l) <===> H3O+ (aq) + B-

(aq) Kb rxn B-

(aq) + H2O (l) <===> OH- (aq) + HB (aq)

Kw rxn H2O (l) + H2O (l) <===> H3O+ (aq) + OH-

(aq) When this is done the resulting reaction is the autoionization reaction for water. Therefore, for acid-base conjugate pairs we can assert that the following is true:

Kw = (Ka)(Kb)

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In other words, there is an inverse relationship between Ka and Kb for conjugate pairs. If a weak acid has a small value of Ka, its conjugate base will have a relatively large Kb. . If a weak acid has a large value of Ka, its conjugate base will have a relatively small Kb.

Acid-Base Properties of Salt Solutions

A salt is an ionic solid that contains a cation other than H+ and an anion other than OH- or O2-. In order to determine the effect a salt has on pH of a solution, follow the rules below.

1. Decide what effect, if any, the cation has on the pH of water. 2. Decide what effect, if any, the anion has on the pH of water. 3. Combine the two effects to decide upon the behavior of the salt.

When an ion reacts with water to form either H3O+ or OH- the reaction is referred to as a hydrolysis reaction.

Cations: typically cations are capable of creating H3O+ in water . . .

ex. NH4+

(aq) + H2O (l) <===> H3O+ (aq) + NH3 (aq)

The cations that are NOT capable of doing so are the ones derived from strong bases, e.g. the alkali metal cations - (Li+, Na+, K+, ...) and the heavier alkaline earth metal cations - (Ca2+, Sr2+, Ba2+)

Since these cations do not react with water they are spectator ions.

Anions: typically anions are capable of creating OH- in water . . .

ex. F- (aq) + H2O (l) <===> OH-

(aq) + HF (aq)

The anions that are NOT capable of doing so are the ones derived from strong acids, e.g. (Cl-, Br-, I-, NO3

-, ClO4-, and SO4

2-)

Since these anions do not react with water they are spectator ions.

Applying the Rules:

1. If the salt contains a cation that hydrolyzes and an anion that does not, the salt will create an acidic solution in water (example: NH4Cl) 2. If the salt contains an anion that hydrolyzes and a cation that does not, the salt will create a basic solution in water (example: Na3CO3) 3. If the salt contains a cation that does not hydrolyzes and an anion that does not, the salt will not alter the pH of water (example: NaCl) 4. If salt contains a cation that hydrolyzes and an anion that also hydrolyzses, the salt will create an acidic or basic solution in water depending upon the relative acid and base strength of the ions (example 1: NH4ClO will be basic because Kb for ClO- > Ka for NH4

+) (example 2: NH4F will be acidic because Kb for F- < Ka for NH4

+)

99

Solving equilibrium problems for salts that fall into categories 1 and 2 above is the same as solving problems for either weak acid or weak bases since that is what they are!

Buffers

Buffers are mixtures of a weak acid-base conjugate pair that resist changes in pH when either acid or base are added to them. In the previous chapter, you learned how to perform equilibrium calculations involving weak acid and weak bases in separate solutions. In those solutions, the concentration of the conjugate species that was formed through the ionization process was generally small in comparison to the initial concentration of the weak acid (or base). Stated differently we might say that since the percent ionization is low for weak acids or bases, the amount of the conjugate found in the solutions of weak acids or bases is small.

The types of calculations used for buffer solutions are no different . . . except that the solutions are prepared with significant quantities of both forms of the conjugate pair present in the buffer solution. As usual, it is probably easiest to demonstrate the calculations by example.

The following examples will be based on a solution that is prepared from ammonia, NH3, and ammonium chloride, NH4Cl. The first thing to recognize is that this is a solution of a weak base, ammonia (NH3), and its conjugate acid (NH4

+). You should also recall that the chloride ion, Cl-, is a spectator ion (i.e. it does not significantly hydrolize water).

In order to understand how a buffer works, it may be useful to calculate the pH of several solutions. First let's look at a case in which the soltuion is prepared with an initial concentration of 1.0 M in both ammonia, NH3, and ammonium chloride, NH4Cl. As usual, it is a good idea to write a reaction that includes all the species present in the equilibrium mixture. Generally speaking, it is most common to represent buffer solutions by writing the acid dissociation equilibrium reaction (Note: the buffer calculations could be done using the base equilibria . . . but it is not a convenient for reasons that should become clear shorlty). In our case the pertinent reaction is:

NH4+

(aq) + H2O (l) <===> H3O+ (aq) + NH3 (aq)

Initial 1.0 M ---- ~ 0 M 1.0 M

Change - x M ---- + x M + x M

Equilibrium 1.0 - x M ---- x M 1.0 + x M

Note that each of the conjugate pair species has an inital concentration of 1.0 M. The changes are defined in the same way as was done for any acid-base equilibrium problem. The equilibrium expression and solution are shown below.

[ ][ ][ ]

( )( )( )

( )( )( )0.10.1

0.10.1106.5

4

3310 xxxx

NH

NHOHxKa ≈

−

+===

+

+−

Solving the expression is very easy because the change, x, is very small. It is found that:

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x = [H3O+] = 5.6 x 10-10 which gives a pH = -log [H3O+] = - log (5.6 x 10-10) = 9.25

Note that this is a basic buffer. The reason that this buffer is basic is because the conjugate base (NH3 with Kb = 1.8 x 10-5) is better at being a base than the conjugate acid (NH4

+ with Ka = 5.6 x 10-10) is at being an acid. Remember the discussion about the pH of salts above!

In the calculations above, we used the equilibrium expression for acid dissociation. This equation has the general form:

[ ][ ][ ]HB

BOHKa

−+= 3

This expression can be rewritten in a very useful form for dealing with buffers if we take the negative log of both sides of the equation and use the following definitions:

• pH = - log [H3O+]

• pKa = - log Ka

These are not new definitions (remember we have quite a few "p" functions . . . pH, pOH, pKw, pKa, pKb). The resulting equation is called the Henderson-Hasselbalch equation:

[ ][ ]HBBpKpH a−

+= log

Note: this is not a "new equation". It is merely the -log transform of an equation you are already familiar with. The new form of the equation is nice to use because it includes pH.

At this point, it may be wise to introduce two parameters that are used to characterize a buffer:

• buffer range - the pH range over which a buffer is useful (extends 2 pH units from pKa - 1 to pKa + 1)

• buffer capacity - a measure of how effectively a buffer neutralizes added acid or base

The useful range for a buffer is determined by how well it resists pH changes. The buffer is most resistant to changes in pH at the middle of its buffer range (i.e. when its pH is equal to pKa). If we look at the example we did above, we find that the ammonium ion has pKa = 9.25 (the pH of the buffer in the problem). By looking at the Henderson-Hasselbalch equation, you can see that whenever the concentration of the conjugate acid-base species are equal, the ratio of their concentration will be equal to one and the log term in the equation becomes zero. The result is pH = pKa for any buffer that has equal amounts of the conjugate acid-base present.

However, there are times when we might want to create a buffer with a specific pH (for example pH = 9.00) and we may not have an acid available with pKa = 9.00. In order to create a buffer in which pH does not equal pKa we simply alter the ratio of conjugate base to conjugate acid. Again, let's look at the system we were studying above. If we want to create a buffer with a pH = 9.00 from the NH3 / NH4

+ system we should first recognize that the desired pH (9.00) is lower

101

than pKa. It should be obvious that in order to make the buffer more acidic, the amount of conjugate acid will have to be greater than the amount of conjugate base present. We can find out what the ratio must be by using the Henderson-Hasselbalch equation:

[ ][ ][ ][ ][ ][ ]

[ ][ ]

[ ][ ]+

+−

+

+

−

=

=

+=

+=

+=

4

3

4

325.0

4

3

4

3

56.0

10

log25.900.9

log

log

NH

NHNH

NHNH

NHNH

NHpKpH

HBBpKpH

a

a

This ratio could be created by preparing a buffer with [NH4+] = 1.0 M and [NH3] = 0.56 M (or

any other concentrations with a ratio of 0.56, ex. 1.12 M to 2.00 M). As predicted above, the amount of conjugate acid is larger than the conjugate base resulting in a pH lower than pKa.

Generally speaking, it is best to create a buffer with a conjugate acid-base pair in which pKA is close to the pH desired for the buffer. No let's turn our attention to the meaning of buffer capacity. Let us work with the pH = 9.00 buffer that we just discussed.

First we will stipulate that we are working with 1.00 L of the pH = 9.00 buffer in which [NH4+] =

1.0 M and [NH3] = 0.56 M. Let's calculate what would happen to the pH of this buffer if we added either 0.100 mole of HCl(g) or 0.100 mole of NaOH(s).

a. If 0.100 mole of HCl(g) is bubbled into 1.00 L of solution, it will create [H3O+] = 0.100 M that needs to be neutralized. Since this is added acid, it will be neutralized by the conjugate base present in the buffer, NH3. The following reaction represents that neutralization:

NH3 (aq) + H3O+ (aq) <===> H2O (l) + NH4

+(aq)

b. It should be clear to you that for each mole of H3O+ neutralized, one mole of NH3 is converted into one mole of NH4

+. The effect of this can be shown in the Henderson-Hasselbalch equation.

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[ ][ ][ ]

[ ]87.8

added is HClafter 10.000.1

]10.056.0[log25.9

added is HCl before 00.1

]56.0[log25.9

log4

3

=

+

−+=

+=

+=+

pH

pH

pH

NH

NHpKpH a

c. Adding acid made the buffer more acidic, however, the pH was only lowered from 9.00 to 8.87. This is not a large change. Consider that the pH of water is lowered from neutral to a pH of 1.00 when a concentration of 0.10 M strong acid is added.

d. If 0.100 mole of NaOH(s) is added into 1.00 L of the original pH 9.00 buffer solution, it will create [OH-] = 0.100 M that needs to be neutralized. Since this is added base, it will be neutralized by the conjugate acid present in the buffer, NH4

+. The following reaction represents that neutralization:

NH4+

(aq) + OH-(aq) <===> H2O (l) + NH3 (aq)

e. It should be clear to you that for each mole of OH- neutralized, one mole of NH4+ is

converted into one mole of NH3. The effect of this can be shown in the Henderson-Hasselbalch equation.

[ ][ ][ ]

[ ]12.9

added is NaOHafter 10.000.1

]10.056.0[log25.9

added is NaOH before 00.1

]56.0[log25.9

log4

3

=

−

++=

+=

+=+

pH

pH

pH

NH

NHpKpH a

f. Adding base made the buffer more basic, however, the pH was only raised from 9.00 to 9.12. This is not a large change. Consider that the pH of water is rasied from neutral to a pH of 13.00 when a concentration of 0.10 M strong base is added.

The amount of pH change experienced depends upon the actual concentrations of the conjugate acid-base used in the buffer system. For example, if 1.00 L of our pH 9.00 buffer had been made with [NH4

+] = 2.00 M and [NH3] = 1.12 M (double the concentrations described above), then the resulting pH values upon the addition of 0.100 mole of HCl(g) or 0.100 mole of NaOH(s) would have been:

• With the addition of 0.100 mole of HCl(g) to 1.00 L of the buffer, the pH goes from 9.00 to 8.94 (compared to 8.87 for the buffer described previously).

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• With the addition of 0.100 mole of NaOH(s)to 1.00 L of the buffer, the pH from 9.00 to 9.06 (compared to 9.12 for the buffer described previously).

In other words, the higher the concentration of the buffer components, the larger the buffer capacity (added acid or base changes the pH less).

Hopefully, you were able to find the answers listed above. If not, make sure you get some help with this kind of problem.

Acid-Base Indicators

Acid-base indicators are themselves weak acid molecules. However, they have the useful property that the acid form produces a different color in solution than the basic form. Therefore, the color of the acid-base indicator depends upon the pH of the solution it is present in. Consider the following example.

HIn(aq) + H2O (l) <===> H3O+(aq) + In-

(aq) KIn = 1.0 x 10-4

acid color yellow base color

blue

Where HIn is the acid for of the indicator, In- is the base form of the indicator, and KIn is the acid dissociation constant of the indicator.

There are a few other things to know about acid base indicators. First, the indicators intereact with light very strongly. Therefore, the concentration of indicator present is kept much lower than the other acid-base species in the solution. In other words, the concentration of the indicator is so low that its presence does not alter the pH of solutions. Second, the change from the acid color to the base color generally occurs over a range of about two pH units centered on pKIn. In the example above, the acid color will dominate at pH = 3.00 or lower. The base form will dominate at pH = 5.00 and higher. In the range from pH = 3.00 to pH = 5.00 the color of the indicator will pass through various shades of green (more from yellow-green near pH 3.00 and more blue-green near pH = 5.00.

This again can be understood by referring to the Henderson-Hasselbalch equation. When the pH is 1.00 unit lower than pKIn, the acid form of the indicator will be present at ten times higher concentration than the base form of the indicator. Likewise, at a pH 1.00 unit higher than pKIn, the base form of the indicator will be present at ten times the concentration of the acid form. When pH = pKIn, the concentrations of the two species will be equal and both will contribute to the color.

There are of course additional factors to keep in mind.

Some indicators exhibit only one color. For example, phenolphthalein is color less in acidci solution but turns bright pink in basic solution (the "in between" color is a very faint pink).

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Some indicators are diprotic weak acids and change color twice. For example, thymol blue changes from red (at about pH 1) to yellow (at about pH 3) then remains yellow up to about pH 7.5 changing over to blue by about pH 9.5. From these ranges we can estimate that pKIn1 = 2 (where the indicator would appear orange) and pKIn2 = 8.5 (where the indicator would appear green) for thymol blue.

It is quite fortunate that we have a variety of acid-base indicators with different pKIn values to choose from. It allows us to use them to indicate the endpoints in the different types of acid-base titrations discussed in the next section.

Acid-Base Titrations

Acid-base titrations are important methods for determining the unknown concnetrations of acids or bases and for determining various acid-base equilibrium constants. The calculations used to determine what an acid-base titration curve will look like will be discussed at some length in class. For now, you should refer to the following titration curves and make sure you understand why they exhibit the general features indicated in each curve.

Strong Acid-Strong Base Titrations

You can examine several titration curves for the reaction:

HCl(aq) + NaOH (aq) <===> H2O (aq) + NaCl (aq)

In each case, 25.00 mL of HCl is titrated by NaOH of the same concentration. Since the stoichiometry is 1:1 in this reaction and the acid and base solutions are equal in concentration, the volume of NaOH required to reach the equivalence point will be 25.00 mL in each case.

However, since the concentrations of both the acid and the base can be changed, the starting pH and pH past the equivalence point will differ from one curve to another. However, the pH at the equivalence point will be 7.00 in all the curves (because the salt produced, NaCl, is composed of spectator ions . . . refer to the previous lesson).

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Note that the starting pH is determined by the pH of the 1.00 M HCl and the final pH will approach (but not exceed) the value of the pH of the 1.00 M NaOH used in the titration. See how this compares with the two figures below.

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Weak Acid-Strong Base Titrations

Sample titration curves of this type are available at the on-line text site. For the reaction:

HB(aq) + NaOH (aq) <===> H2O (aq) + NaB (aq)

In each case, 25.00 mL of a 0.10 M solution of the weak acid HB is titrated by 0.10 M NaOH. The paramter that is altered in these curves is the Ka of the weak acid. Since the stoichiometry is 1:1 in this reaction and the acid and base solutions are equal in concentration, the volume of NaOH required to reach the equivalence point will still be 25.00 mL in each case (acid strength does not alter the stoichiometry.). However, since the acid dissociation constant of the acid is changed, the pH values on the curve will differ from one curve to another. However, the pH beyond the equivalence point will be the same in all the curves (because those pH values are determined by the excess NaOH present).

Strong Acid-Weak Base Titrations

Sample titration curves of this type are available at the on-line text site. For the reaction:

HCl(aq) + NaB (aq) <===> HB (aq) + NaCl (aq)

In each case, 25.00 mL of a 0.10 M solution of the weak base B- is titrated by 0.10 M HCl. The paramter that is altered in these curves is the Kb of the weak base. Since the stoichiometry is 1:1 in this reaction and the acid and base solutions are equal in concentration, the volume of NaOH required to reach the equivalence point will still be 25.00 mL in each case (base strength does not alter the stoichiometry.). However, since the base dissociation constant of the base is changed, the pH values on the curve will differ from one curve to another. However, the pH beyond the equivalence point will be the same in all the curves (because those pH values are determined by the excess HCl present).

Keywords acid, strong acid, weak acidic amphiprotic base, strong base, weak basic buffer conjugate acid conjugate base

distribution curve endpoint equivalence point indicator ionization constant, water (Kw) ionization constant, acid (Ka) ionization constant, base (Kb) molarity neutral pH pKa

pKb pKw pOH percent ionization polyprotic acid polyprotic base quadratic formula salt successive approximations titration