Chapter 8 revised

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Example A Weak-Base Problem
Find the pH of 0.10 M ammonia.
Solution When ammonia is dissolved in water, its reaction is
In Appendix G, we find ammonium ion, NH4+, listed next to ammonia. pKa for ammonium ion is 9.245. Therefore, Kb for NH3 is
14
Weak Is Conjugate to Weak
1. The conjugate base of a weak acid is a weak base. The conjugate acid of a weak base is a weak acid.

Consider a weak acid, HA, with Ka = 10 4. The conjugate base, A , has Kb = Kw/Ka = 10 10. That is, if HA is a weak acid, A is a weak base. If Ka were 10 5, then Kb would be 10 9.

2. As HA becomes a weaker acid, A becomes a stronger base (but never a strong base). Conversely, the greater the acid strength of HA, the less the base strength of A . However, if either A or HA is weak, so is its conjugate. If HA is strong (such as HCl), its conjugate base (Cl ) is so weak that it is not a base at all in water.
15
Using Appendix G
Appendix G lists acid dissociation constants. Each compound is shown in its fully protonated form.

Diethylamine, for example, is shown as , which is really the diethylammonium ion. The value of Ka(1.0 × 10 11) given for diethylamine is actually Ka for the diethylammonium ion. To find Kb for diethylamine, we write Kb = Kw/Ka = 1.0 × 10 14/1.0 × 10 11 = 1.0 × 10 3.
16
For polyprotic acids and bases, several Ka values are given. Pyridoxal phosphate is given in its fully protonated form as follows:
Polyprotic acids and bases
pK1 (1.4) is for dissociation of one of the phosphate protons
pK2 (3.44) is for the hydroxyl proton.
The third most acidic proton is the other phosphate proton, for which pK3 = 6.01, and the NH+ group is the least acidic (pK4 = 8.45).
17
Species drawn in Appendix G are fully protonated. If a structure in Appendix G has a charge other than 0, it is not the structure that belongs with the name in the appendix. Names refer to neutral molecules. The neutral molecule pyridoxal phosphate is not the species drawn above, which has a + 1 charge. The neutral molecule pyridoxal phosphate is
13
The acid HA and its corresponding base, A , are said to be a conjugate acid-base pair, because they are related by the gain or loss of a proton. Similarly, B and BH+ are a conjugate pair. The important relation between Ka and Kb for a conjugate acid-base pair, derived in Equation 6-35, is
(9-5)
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A weak base is one for which Reaction 8-4 does not go to completion.
pK is the negative logarithm of an equilibrium constant:
As K increases, pK decreases, and vice versa. Comparing formic and benzoic acids, we see that formic acid is stronger, with a larger Ka and smaller pKa, than benzoic acid.
11
8-2 Weak Acids and Bases
Let's review the meaning of the acid dissociation constant, Ka, for the acid HA:
(8-3)
A weak acid is one that is not completely dissociated. That is, Reaction 8-3 does not go to completion.

For a base, B, the base hydrolysis constant, Kb, is defined by the reaction
(8-4)
7
The Dilemma
"What is the pH of 1.0 × 10 8 M KOH?" Applying our usual reasoning, we calculate
But how can the base KOH produce an acidic solution (pH < 7) when dissolved in pure water? It's impossible.
The Cure
Clearly, there is something wrong with our calculation. In particular, we have not considered the contribution of OH from the ionization of water. In pure water, [OH ] = 1.0 × 10 7 M, which is greater than the amount of KOH added to the solution. To handle this problem, we resort to the systematic treatment of equilibrium.
8
Step 2 Charge balance. The species in solution are K+, OH , and H+. So,
(8-2)
Step 3 Mass balance. All K+ comes from the KOH, so [K+] = 1.0 × 10 8 M.
Step 4 Equilibrium constant expression. Kw = [H+][OH ] = 1.0 × 10 14.
Step 5 Count. There are three equations and three unknowns ([H+], [OH ], [K+]), so we have enough information to solve the problem.
Step 6 Solve. Because we are seeking the pH, let's set [H+] = x. Writing [K+] = 1.0 × 10 8 M in Equation 8-2, we get
Step 1 Pertinent reactions. The only one is
9
Using this expression for [OH ] in the Kw equilibrium enables us to solve the problem:
Rejecting the negative concentration, we conclude that
This pH is eminently reasonable, because 10 8 M KOH should be very slightly basic.
10
Figure 8-1 shows the pH calculated for different concentrations of strong base or strong acid in water. There are three regions:
1. When the concentration is "high" ( 10 6 M), pH is calculated by just considering the added H+ or OH . That is, the pH of 10 5.00 M KOH is 9.00.
2. When the concentration is "low" ( 10 8 M), the pH is 7.00. We have not added enough acid or base to change the pH of the water itself.
3. At intermediate concentrations of 10 6 to 10 8 M, the effects of water ionization and the added acid or base are comparable. Only in this region is a systematic equilibrium calculation necessary.
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As another example ---
consider the molecule piperazine
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A Typical Weak-Acid Problem
The problem is to find the pH of a solution of the weak acid HA, given the formal concentration of HA and the value of Ka. Let's call the formal concentration F and use the systematic treatment of equilibrium:
(8-6)
(8-7)
(8-8)
8-3 Weak- Acid Equilibria
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There are four equations and four unknowns ([A ], [HA], [H+], [OH ]), so the problem is solved if we can just do the algebra.
We can make an excellent, simplifying approximation.
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Equation 8-11 can always be solved with the quadratic formula. However, an easier method worth trying first is to neglect x in the denominator. If x comes out much smaller than F, then your approximation was good and you need not use the quadratic formula.

For Equation 8-12, the approximation works like this:
The approximate solution (x 2.8 × 10 6) is much smaller than the term 0.050 in the denominator of Equation 8-12. Therefore, the approximate solution is fine. A reasonable rule of thumb is to accept the approximation if x comes out to be less than 1% of F.
A handy tip
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8-4 Weak-Base Equilibria
The treatment of weak bases is almost the same as that of weak acids.
We suppose that nearly all OH comes from the reaction of B + H2O, and little comes from dissociation of H2O. Setting [OH ] = x, we must also set [BH+] = x, because one BH+ is produced for each OH . Calling the formal concentration of base F (= [B] + [BH+]), we write
30
Plugging these values into the Kb equilibrium expression, we get
which looks a lot like a weak-acid problem, except that now x = [OH ].
31
If the formal concentration is 0.037 2 M, the problem is formulated as follows:
Because x = [OH ], we can write
This is a reasonable pH for a weak base.
A Typical Weak-Base Problem
Consider the commonly occurring weak base cocaine.
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Example A Weak-Acid Problem
Find the pH of 0.050 M trimethylammonium chloride.
Solution We assume that ammonium halide salts are completely dissociated to give (CH3)3NH+ and Cl .* We then recognize that trimethylammonium ion is a weak acid, being the conjugate acid of trimethylamine, (CH3)3N, a weak base. Cl has no basic or acidic properties and should be ignored. In Appendix G, we find trimethylammonium ion listed as trimethylamine, but drawn as trimethylammonium ion, with pKa = 9.799. So,
25
The Essence of a Weak-Acid Problem
When faced with finding the pH of a weak acid, you should immediately realize that [H+] = [A ] = x and proceed to set up and solve the equation
where F is the formal concentration of HA.

The approximation [H+] = [A ] would be poor only if the acid were too dilute or too weak, neither of which constitutes a practical problem.
21
For any respectable weak acid, [H+] from HA will be much greater than [H+] from H2O. When HA dissociates, it produces A . When H2O dissociates, it produces OH . If dissociation of HA is much greater than H2O dissociation, then [A ] >> [OH ], and Equation 8-6 reduces to
To solve the problem, first set [H+] = x. Equation 8-9 says that [A ] also is equal to x. Equation 8-7 says that [HA] = F [A ] = F x. Putting these expressions into Equation 8-8 gives
22
Was the approximation [H+] [A ] justified? The calculated pH is 2.17, which means that [OH ] = Kw/[H+] = 1.5 × 10 12 M.
Setting F = 0.050 0 M and Ka = 1.07 × 10 3 for o-hydroxybenzoic acid, we can solve the equation, because it is just a quadratic equation.
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The assumption that H+ is derived mainly from HA is excellent.
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Fraction of Dissociation
The fraction of dissociation, α, is defined as the fraction of the acid in the form A :
For 0.050 0 M o-hydroxybenzoic acid, we find
6
Finding the pH of other concentrations of KOH is pretty trivial:
A generally useful relation is
(9-1)
5
How do we calculate the pH of 0.10 M KOH? KOH is a strong base (completely dissociated), so [OH ] = 0.10 M. Using Kw = [H+][OH ], we write
4
Example Activity Coefficient in a Strong-Acid Calculation
Calculate the pH of 0.10 M HBr, using activity coefficients.
Solution The ionic strength of 0.10 M HBr is 0.10 M, at which the activity coefficient of H+ is 0.83 (Table 7-1). Remember that pH is
, not log[H+]:
Example Calculating How to Prepare a Buffer Solution
How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride to give a pH of 7.60 in a final volume of 250 mL?
Solution The moles of tris hydrochloride in 10.0 g are (10.0 g)/(157.596 g/mol) = 0.063 5. We can make a table to help solve the problem:

The Henderson-Hasselbalch equation allows us to find x, because we know pH and pKa.
We see that the pH of a buffer does not change very much when a limited amount of strong acid or base is added. Addition of 12.0 mL of 1.00 M HCl changed the pH from 8.61 to 8.41. Addition of 12.0 mL of 1.00 M HCl to 1.00 L of unbuffered solution would have lowered the pH to 1.93.

A buffer resists changes in pH

But why does a buffer resist changes in pH? It does so because the strong acid or base is consumed by B or BH+. If you add HCl to tris, B is converted into BH+. If you add NaOH, BH+ is converted into B. As long as you don't use up the B or BH+ by adding too much HCl or NaOH, the log term of the Henderson-Hasselbalch equation does not change very much and the pH does not change very much.

The buffer has its maximum capacity to resist changes of pH when pH = pKa. We will return to this point later.

Example Effect of adding acid to buffer
If we add 12.0 mL of 1.00 M HCl to the solution used in the previous example, what will be the new pH?
Solution The key to this problem is to realize that, when a strong acid is added to a weak base, they react completely to give BH+ (Box 9-3). We are adding 12.0 mL of 1.00 M HCl, which contains (0.012 0 L)(1.00 mol/L) = 0.012 0 mol of H+. This much H+ will consume 0.012 0 mol of B to create 0.012 0 mol of BH+, which is shown conveniently in a little table:

The information in the table allows us to calculate the pH:
The volume of the solution is irrelevant.
Example A Buffer Solution

Find the pH of a solution prepared by dissolving 12.43 g of tris (FM 121.135) plus 4.67 g of tris hydrochloride (FM 157.596) in 1.00 L of water.
Solution The concentrations of B and BH+ added to the solution are

Assuming that what we mixed stays in the same form, we plug these concentrations into the Henderson-Hasselbalch equation to find the pH:

Notice that the volume of solution is irrelevant, because volume cancels in the numerator and denominator of the log term:

Preparing a Buffer in Real Life!
If you really wanted to prepare a tris buffer of pH 7.60, you would not do it by calculating what to mix. Suppose that you wish to prepare 1.00 L of buffer containing 0.100 M tris at a pH of 7.60. You have available solid tris hydrochloride and approximately 1 M NaOH. Here's how I would do it:

1. Weigh out 0.100 mol of tris hydrochloride and dissolve it in a beaker containing about 800 mL of water.
2. Place a calibrated pH electrode in the solution and monitor the pH.
3. Add NaOH solution until the pH is exactly 7.60.
4. Transfer the solution to a volumetric flask and wash the beaker a few times. Add the washings to the volumetric flask.
5. Dilute to the mark and mix.
You do not simply add the calculated quantity of NaOH, because it would not give exactly the desired pH. The reason for using 800 mL of water in the first step is so that the volume will be reasonably close to the final volume during pH adjustment. Otherwise, the pH will change slightly when the sample is diluted to its final volume and the ionic strength changes.

Reasons why a calculation would be wrong:
1. You might have ignored activity coefficients.
2. The temperature might not be just right.
3. The approximations that [HA] = FHA and [A ] = FA– could be in error.
4. The pKa reported for tris in your favorite table is probably not what you would measure in your lab.
5. You will probably make an arithmetic error anyway.

Buffer Capacity
Buffer capacity, β, is a measure of how well a solution resists changes in pH when strong acid or base is added. Buffer capacity is defined as

where Ca and Cb are the number of moles of strong acid and strong base per liter needed to produce a unit change in pH. The greater the buffer capacity, the more resistant the solution is to pH change.

Figure 8-4 (a) Cb versus pH for a solution containing 0.100 F HA with pKa = 5.00. (b) Buffer capacity versus pH for the same system reaches a maximum when pH = pKa. The lower curve is the derivative of the upper curve.

Figure 8-4a shows Cb versus pH for a solution containing 0.100 F HA with pKa = 5.00. The ordinate (Cb) is the formal concentration of strong base needed to be mixed with 0.100 F HA to give the indicated pH. For example, a solution containing 0.050 F OH plus 0.100 F HA would have a pH of 5.00 (neglecting activities).

Figure 8-4b, which is the derivative of the upper curve, shows buffer capacity for the same system. Buffer capacity reaches a maximum when pH = pKa. That is, a buffer is most effective in resisting changes in pH when pH = pKa (that is, when [HA] = [A ]).

In choosing a buffer, seek one whose pKa is as close as possible to the desired pH. The useful pH range of a buffer is usually considered to be pKa ± 1 pH unit. Outside this range, there is not enough of either the weak acid or the weak base to react with added base or acid. Buffer capacity can be increased by increasing the concentration of the buffer.

Buffer pH Depends on Ionic Strength and Temperature
The correct Henderson-Hasselbalch equation, 9-18, includes activity coefficients. Failure to include activity coefficients is the principal reason why calculated pH values do not agree with measured values. The H2PO4-/ HPO42- buffer has a pKa of 7.20 at 0 ionic strength. At 0.1 M ionic strength, the pH of a 1:1 mole mixture of H2PO4-/ HPO42- is 6.86 (see Problem 8.43). Molecular biology lab manuals list pKa for phosphoric acid as 6.86, which is representative for ionic strengths employed in the lab. As another example of ionic strength effects, when a 0.5 M stock solution of phosphate buffer at pH 6.6 is diluted to 0.05 M, the pH rises to 6.9—a rather significant effect.

Changing ionic strength changes pH.

Buffer pKa depends on temperature. Tris has an exceptionally large dependence, 0.028 pKa units per degree, near room temperature. A solution of tris with pH 8.07 at 25°C will have pH 8.7 at 4°C and pH 7.7 at 37°C.

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What fraction of cocaine has reacted with water? We can formulate α for a base, called the fraction of association:
Only 0.83% of the base has reacted.
(9-14)
A Buffer in Action
In Appendix G, we find pKa for the conjugate acid of tris listed as 8.072. An example of a salt containing the BH+ cation is tris hydrochloride, which is BH+Cl . When BH+Cl is dissolved in water, it dissociates to BH+ and Cl .
For illustration, we choose a widely used buffer called "tris," which is short for tris(hydroxymethyl)aminomethane.
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To find the pH of 0.10 M NH3, we set up and solve the equation
8. Monoprotic Acid-Base Equilibria
1
2
Acids and bases are essential to virtually every application of chemistry and for the intelligent use of analytical procedures such as chromatography and electrophoresis. It would be difficult to have a meaningful discussion of, say, protein purification or the weathering of rocks without understanding acids and bases. This chapter covers acid-base equilibria and buffers.
3
8-1 Strong Acids and Bases
What could be easier than calculating the pH of 0.10 M HBr? HBr is a strong acid, so the reaction
goes to completion, and the concentration of H3O+ is 0.10 M. It will be our custom to write H+ instead of H3O+, and we can say
Example Using the Henderson-Hasselbalch Equation
Sodium hypochlorite (NaOCl, the active ingredient of almost all bleaches) was dissolved in a solution buffered to pH 6.20. Find the ratio [OCl ]/[HOCl] in this solution.
Solution In Appendix G, we find that pKa = 7.53 for hypochlorous acid, HOCl. The pH is known, so the ratio [OCl ]/[HOCl] can be calculated from the Henderson-Hasselbalch equation.
Finding the ratio [OCl ]/[HOCl] requires knowing only the pH and the pKa. We do not need to know how much NaOCl was added, or the volume.
9-5 Buffers

A buffered solution resists changes in pH when acids or bases are added or when dilution occurs. The buffer is a mixture of an acid and its conjugate base.

There must be comparable amounts of the conjugate acid and base (say, within a factor of 10) to exert significant buffering.

The acid is only 3.1% dissociated under these conditions.
In a solution containing 0.10 mol of A dissolved in 1.00 L, the extent of reaction of A with water is even smaller.

HA dissociates very little, and adding extra A to the solution will make the HA dissociate even less. Similarly, A does not react very much with water, and adding extra HA makes A react even less. If 0.050 mol of A plus 0.036 mol of HA are added to water, there will be close to 0.050 mol of A and close to 0.036 mol of HA in the solution at equilibrium.
Mixing a Weak Acid and Its Conjugate Base
If you mix A moles of a weak acid with B moles of its conjugate base, the moles of acid remain close to A and the moles of base remain close to B. Very little reaction occurs to change either concentration.

To understand why this should be so, look at the Ka and Kb reactions in terms of Le Châtelier's principle. Consider an acid with pKa = 4.00 and its conjugate base with pKb = 10.00. Let's calculate the fraction of acid that dissociates in a 0.10 M solution of HA.

Properties of the Henderson-Hasselbalch Equation
if [A ] = [HA], then pH = pKa.
Regardless of how complex a solution may be, whenever pH = pKa, [A ] must equal [HA]. This relation is true because all equilibria must be satisfied simultaneously in any solution at equilibrium. If there are 10 different acids and bases in the solution, the 10 forms of Equation 8-16 must all give the same pH, because there can be only one concentration of H+ in a solution.

Another feature of the Henderson-Hasselbalch equation is that, for every power-of-10 change in the ratio [A ]/[HA], the pH changes by one unit. As the base ( A ) increases, the pH goes up. As the acid (HA) increases, the pH goes down. For any conjugate acid-base pair, you can say, for example, that, if pH = pKa 1, there must be 10 times as much HA as A . Ten-elevenths is in the form HA and one-eleventh is in the form A .
Henderson-Hasselbalch Equation
The central equation for buffers is the Henderson-Hasselbalch equation, which is merely a rearranged form of the Ka equilibrium expression.

The Henderson-Hasselbalch equation tells us the pH of a solution, provided we know the ratio of the concentrations of conjugate acid and base, as well as pKa for the acid. If a solution is prepared from the weak base B and its conjugate acid, the analogous equation is
where pKa is the acid dissociation constant of the weak acid BH+. The important features of Equation 9-16 and Equation 9-17 are that the base (A or B) appears in the numerator of both equations, and the equilibrium constant is Ka of the acid in the denominator.

Equation 8-16 and Equation 8-17 are sensible only when the base (A or B) is in the numerator. When the concentration of base increases, the log term increases and the pH increases.
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Chapter 8 revised

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