Chapter 7 – Mixtures of Acids and Bases

Introduction

In Chapter 6, we examined the equilibrium concentrations in solutions of acids and solutions of bases. In this chapter, we continue our discussion of acids and bases by focusing on the equilibrium concentrations of solutions formed by mixing acids and bases.

7.1 The Common-Ion Effect

Introduction

Predicting the effect on an equilibrium mixture caused by the presence of an ion that appears in the equilibrium but is introduced from a separate source is something we will need to do frequently. Qualitative predictions of this type are frequently made with the use of the common ion effect, the topic of this section.

Objectives

• •
  Explain the common-ion effect and predict the effect of a common ion on an equilibrium mixture.

7.1-1. Definition

When solutions are prepared in such a way that there are two or more separate sources of an ion in an equilibrium, then that ion is referred to as a common ion. In this and the following chapters, we will have the occasion to deal with the effect on an equilibrium mixture caused by the addition of a common ion. The effect of a common ion can readily be predicted with Le Châtelier's principle, which predicts that the composition of the mixture shifts to counteract the addition. Consequently, the common ion effect is summarized as the following:

• •
  Common Ion Effect: The addition of a common ion to an equilibrium mixture shifts the equilibrium away from the side of the common ion.

Thus, the concentrations of those substances on the same side of the common ion decrease while those on the opposite side increase.

7.1-2. Mixture of a Weak Acid and a Strong Acid

We examine the effect that the presence of a strong acid has on the dissociation of a weak acid by considering the dissociation of 0.10 M HClO in water and then in a 0.05 M solution of a strong acid. We use Equation 6.6 to determine that [ClO^1–] = {(0.10)(3.5e–8)}^1/2 = 5.9e–5 M, which is a measure of the extent of the dissociation of HClO in water. We now consider the dissociation of HClO in 0.05 M HCl. H₃O¹⁺ is a common ion because it appears in the HClO equilibrium, but it has a separate source, the HCl. The common ion effect predicts that the presence of the common ion would reduce the concentrations of those substances on the same side of the equilibrium, so [ClO^1–] should drop, while it increases the concentrations of those substances on the opposite side, so [HClO] should increase. The solution is 0.10 M in HClO and 0.05 M in H₃O¹⁺ initially, so the reaction table takes the following form:

	HClO	+	H2O	⇌	ClO1–	+	H3O1+
initial	0.10				0		0.05
Δ	–x				+x		+x
equilibrium	0.10 – x				x		0.05 + x

The equilibrium constant expression is [ClO^1–] was 5.9e–5 M in water, and it will be less than that in the presence of acid, so we can assume that x is negligible in both the addition and the subtraction, and the equilibrium constant expression can be written as follows. Solving the above leads to x = [ClO^1–] = 7.0e–8 M, which is almost 1000 times less than in pure water. Furthermore, [H₃O¹⁺] = 0.05 + x = 0.05 M, i.e., the hydronium ion concentration in a solution of a strong acid and a weak acid equals the concentration of the strong acid. We conclude the following.

7.1-3. Mixture of a Weak Base and a Strong Base

We examine the effect that the presence of a strong base has on the dissociation of a weak base by determining the concentration of HClO in 0.10 M KClO that has been dissolved in water and then in a 0.05 M solution of a strong base. We use Equation 6.9 to determine that [HClO] = {(0.10)(2.9e–7)}^1/2 = 1.7e–4 M, which is a measure of the extent of the reaction in water. We now look at the reaction in 0.05 M NaOH. The common ion effect predicts that the presence of additional OH^1–, the common ion, will reduce [HClO] and increase [ClO^1–]. The solution is 0.10 M in ClO^1– and 0.05 M in OH^1– initially, so the reaction table takes the following form.

	ClO1–	+	H2O	⇌	HClO	+	OH1–
initial	0.10				0		0.05
Δ	–x				+x		+x
equilibrium	0.10 – x				x		0.05 + x

The equilibrium constant expression is [HClO] was 1.7e–4 M in water, and it will be less than that in the presence of base, so we can assume that x is negligible in both the addition and the subtraction. The equilibrium constant expression is then the following. Solving the above leads to x = [HClO] = 5.8e–7 M, which is almost 1000 times less than in pure water. Furthermore [OH^1–] = 0.05 + x = 0.05 M, i.e., the hydroxide ion concentration in a solution of a strong base and a weak base equals the concentration of the strong base. We conclude the following.

7.2 Buffers

Introduction

In Chapter 6, we discussed four types of acid-base solutions in some detail: strong acid, strong base, weak acid, and weak base. We now discuss buffers, the last type of acid-base solution to be considered. A buffer is defined in the dictionary as a "device that softens the shock of a blow." A chemical buffer serves much the same purpose: it softens the pH change resulting from the addition of a strong acid or base. For example, aspirin is an acid that can upset the stomach, but Bufferin^® contains buffers to reduce the pH changes that ordinarily accompany the addition of an acid. Much of the maintenance involved in the upkeep of an aquarium involves maintaining proper buffer levels in the water to assure that the pH stays in a range that is safe for the fish. Blood is also buffered, which is why it maintains a pH of about 7.4 even though many acid-base reactions take place in it. In this section, we discuss the action of buffers.

Objectives

• •
  Define a buffer and explain how it works.
• •
  Choose an appropriate conjugate acid-base pair to use in the preparation of a buffer of a known pH.
• •
  Determine the base to acid ratio required to produce a buffer of given pH.
• •
  Determine the pH of a buffer given the amounts of acid and base present.
• •
  Calculate the change in pH expected when a strong acid or base is added to a buffered solution.

7.2-1. Introduction to Buffers Video

7.2-2. Definition

A buffer is a solution of a weak acid and its conjugate base in comparable and appreciable amounts. They are able to minimize changes in pH because they contain both a weak acid and a weak base. Thus, if a strong acid is added to a buffered solution, it reacts with the weak base component of the buffer, and if a strong base is added to a buffered solution, it reacts with the weak acid component of the buffer. Either way, the effect of the addition of hydroxide or hydronium ion on the solution is dramatically reduced because these strong acids and bases are converted to much weaker acids and bases by the action of the buffer.

7.2-3. Hydronium Ion Concentrations in Buffer Solutions

A buffer contains both the weak acid and the weak base, so both K_a and K_b must be satisfied, and either can be used to determine the composition of the equilibrium mixture. However, because pH is more common than pOH, we use the acid dissociation reaction to discuss buffered solutions because it contains the hydronium ion directly. Consider a hypochlorous acid/hypochlorite ion buffer in which the initial concentrations of the acid (c_a) and the base (c_b) are comparable. The reaction table for the acid dissociation of acid in the presence of the base has the following form:

	HClO	+	H2O	⇌	ClO1–	+	H3O1+
initial	ca				cb		0
Δ	–x				+x		+x
equilibrium	ca – x				cb + x		x

The equilibrium constant expression for the reaction is: The presence of the conjugate base suppresses the dissociation of the acid, so x is negligible in both the addition and the subtraction. Consequently, the equilibrium constant expression can be written as the following: Solving for the above for hydronium ion concentration (x), we obtain Equation 7.1. c_a = n_a/V and c_b = n_b/V, and the volumes cancel in the ratio c_a/c_b. Thus, c_a/c_b = n_a/n_b, i.e., the ratio of the concentrations equals the ratio of the numbers of moles. Taking the negative logarithms of both sides of Equation 7.1

[H₃O¹⁺] = x = K_a

ca

cb

= K_a

na

nb

we obtain the equation that is known as the Henderson-Hasselbalch equation, where the identity log(x/y) = –log(y/x) has been used. Equation 7.2 is the most common method for determining the pH of a buffer.

7.2-4. Buffer pH Exercise

7.2-5. Buffer Action Video

7.2-6. Buffer Action Against Addition of Base

Buffers act to protect the solution from drastic pH changes resulting from the addition of hydronium or hydroxide ions. They do so by reacting with the added acid or base. However, in order for the buffer to function properly, it must be the excess reactant. Consider the case where x mmol OH^1– ion are added to a buffer solution that contains n_a mmol HClO and n_b mmol ClO^1–. The strong base reacts extensively with the weak acid in the buffer, so, as long as HClO is in excess, all of the hydroxide reacts. The reaction table has the following form:

	HClO	+	OH1–	⇌	ClO1–	+	H2O
initial	na		x		nb		0
Δ	–x		–x		+x		+x
equilibrium	na – x		~0		nb + x		x

Of course [OH^1–] cannot be zero, which is why the approximate sign is used, but it will be very small compared to the weak base concentration. The above reaction table is valid, and the solution is a buffer only so long as there is sufficient acid to react with the base. In other words, an appreciable amount of acid is required. The resulting solution is a solution of a weak acid and its conjugate base, so it is still a buffer solution, and Equation 7.2

pH = pK_a + log

cb

ca

= pK_a + log

nb

na

can still be used to determine its pH.

7.2-7. Strong Base–Buffer Worked Example

7.2-8. Buffer Action with Strong Base Exercise

7.2-9. Buffer Action Against Addition of Acid

Consider the case where x mmol of H₃O¹⁺ ions are added to a buffer solution that contains n_a mmol HClO and n_b mmol ClO^1–. The strong acid reacts extensively with the weak base in the buffer, so, as long as ClO^1– is in excess, all of the hydronium reacts. The reaction table has the following form:

	ClO1–	+	H3O1+	⇌	HClO	+	H2O
initial	nb		x		na		0
Δ	–x				+x		+x
equilibrium	nb – x		~0		na + x		x

[H₃O¹⁺] cannot be zero, which is why the approximate sign is used, but it will be very small compared to the weak acid concentration. The above reaction table is valid, and the solution is a buffer only so long as there is sufficient base to react with the acid. In other words, an appreciable amount of base is required. The resulting solution is a solution of a weak acid and its conjugate base, so it is still a buffer solution, and Equation 7.2

pH = pK_a + log

cb

ca

= pK_a + log

nb

na

can still be used to determine its pH.

7.2-10. Buffer Action with Strong Acid Exercise

7.2-11. Buffer Exercise

7-2.12. Buffer Capacity

There must be appreciable amounts of the acid and its conjugate base in a buffer to assure that it has acceptable buffer capacity, which is the amount of acid or base that can be added without destroying the effectiveness of the buffer. Thus, a buffer solution has a large buffer capacity if it contains a large number of moles of the weak acid and of the weak base.

7.2-13. Buffer Range

A buffer functions well when the addition of large amounts of acid or base result in only small pH changes. The pH range over which the buffer is effective is called the buffer range. Figure 7.1 shows the relationship between pH of an acetic acid/acetate ion buffer and the mole fractions of the acid and the base. The buffer range is the pH range over which the mole fractions change most dramatically with pH (highlighted by the box in the figure). The center of the buffer range is where the two mole fractions (or concentrations) are equal (c_A/c_B = 1). Application of Equation 7.2

pH = pK_a + log

cb

ca

= pK_a + log

nb

na

and log 1 = 0 to the midpoint indicates that pH = pK_a. Thus, buffers are most effective at pH values near their pK_a. As shown in Figure 7.1, buffers operate acceptably in the range where the mole fraction of the acid or base is in the range of 0.1 to 0.9, which means that 0.1 < c_a/c_b < 10.

7.2-14. Preparing a pH = 7 Buffer Exercise

7.2-15. Preparing a pH = 9.8 Buffer Exercise

7.3 Acid-Base Composition from Reactant Amounts

Introduction

We have now treated five different types of acid-base solutions and shown how to determine [H₃O¹⁺], [OH^1–], and pH of each:

• 1
  Strong acid: Strong acids dissociate completely, so [H₃O¹⁺] = c_o.
• 2
  Strong base: Strong bases (M(OH)_n) dissociate completely, so [OH^1–] = nc_o.
• 3
  Weak acid: Solve the K_a expression for [H₃O¹⁺] using Equation 6.6.
• 4
  Weak base: Solve the K_b expression for [OH^1–] using Equation 6.9.
• 5
  Buffers: Solve the K_a expression for [H₃O¹⁺] using Equation 7.1
  [H₃O¹⁺] = x = K_a

  ca

  cb

  = K_a

  na

  nb

  or solve directly for pH using Equation 7.2
  pH = pK_a + log

  cb

  ca

  = pK_a + log

  nb

  na

  .

Thus, there are four types of acid-base reactions. (When buffers react, they do so as the weak acid or the weak base.)

• 1
  weak acid–weak base
• 2
  strong acid–strong base
• 3
  weak acid–strong base
• 4
  strong acid–weak base

In this section, we examine acid-base properties of the solutions that result from these reactions.

Objectives

• •
  Write acid-base reactions and determine their equilibrium constants from the K_a's of the reacting and produced acids.
• •
  Determine the concentrations of all species in a solution prepared by mixing a strong acid (or base) with a strong base (or acid) or a weak base (or acid).

Weak Acid–Weak Base Reactions

7.3-1. The Equilibrium Constant

Consider the following reaction. The equilibrium constant expression for the reaction is Compare that with the expression obtained by dividing the K_a expression of the reacting acid (HF) by that of the produced acid (HOCl). Note that the hydronium ion concentrations cancel to produce the K_a expression, which is identical to the equilibrium expression for the reaction of HF with OCl^1– ion. Thus, the equilibrium constant for the reaction can be determined as shown below. We conclude that the equilibrium constant for an acid-base reaction can be determined from the acid dissociation constants of the reacting and produced acids as follows. The produced acid is the conjugate acid of the reacting base. The strength of an acid is inversely proportional to the strength of its conjugate base, so the above expression indicates that the value of K depends upon the relative strengths of the reacting acid and base. If both are relatively strong, then both K_a (reacting acid) and K_b (reacting base) are large. However, if K_b (reacting base) is large, then K_a (produced acid) must be small, which makes the equilibrium constant of the reaction large.

7.3-2. Determining K Exercise

7.3-3. Mixing Exercise

7.3-4. Overview of Extensive Reactions

The remainder of the acid-base reaction types all involve a strong acid and/or a strong base, so they are all extensive. Consequently, the limiting reactant is assumed to disappear completely, which means that there are no unknowns in the reaction table. We will use the following method in each case:

• 1
  Write the chemical equation and construct the reaction table. We will use millimoles in the reaction table because the volumes are all given in milliliters, but molarities can also be used.
• 2
  Identify the type of solution that results from the mixing based upon the equilibrium entries.
  • •
    Strong acid if the entry under [H₃O¹⁺] is not zero. There can be no base present if there is some H₃O¹⁺ because they would react, and a weak acid can be ignored in the presence of a strong acid.
  • •
    Strong base if the entry under OH^1– is not zero. There can be no acid present if there is some OH^1– because they would react, and a weak base can be ignored in the presence of a strong one.
  • •
    Weak acid if the only nonzero entry lies under a weak acid.
  • •
    Weak base if the only nonzero entry lies under a weak base.
  • •
    Buffer if the entries under both the weak acid and its conjugate base are nonzero.
• 3
  Determine [H₃O¹⁺], [OH^1–], pH, or pOH using the equation appropriate to the solution type above.

We will apply the above method to each of the reaction types in the following sections.

7.3-5. Titration Curves

Recall from Section 2.4-5. that the equivalence point in an acid-base titration is that point where the number of moles of analyte in the initial flask equals the number of moles of titrant that are added, which is summarized as follows.

• •
  Equivalence Point: moles acid = moles base, so M_acidV_acid = M_baseV_base

The above expression was then used to determine M_acid from a known M_base and the known volumes. We now examine the pH of the acid solution as a function of the volume of base that is added to generate the titration curve for the titration. A titration curve is a plot of the pH of the solution as a function of the volume of a titrant. The shape of the curve depends on the nature of both the acid and base, and we start by examining the titration of a strong acid with a strong base in the following video.

Strong Acid–Strong Base Reactions

7.3-6. Strong Acid–Strong Base Titration Video

7.3-7. Method

Our first titration curve is for the titration of a strong acid with a strong base. We start out with n_a moles of acid and determine the pH after n_b moles of base have been added. The reaction is extensive, so there is essentially no limiting reactant present at equilibrium and the solution type is dictated solely by the reactant in excess. The solution type depends upon the relative values of n_a and n_b, and the following reaction table shows the case where the base is the limiting reactant:

	H3O1+	+	OH1–	→	H2O	+	H2O
initial	na		nb
Δ	–nb		–nb
equilibrium	na – nb		~0

The equilibrium solution is a strong acid that contains n_a – n_b mol of acid in a total volume V_total = V_acid + V_base, so [H₃O¹⁺] = (n_a – n_b)/V_total. At the equivalence point, [H₃O¹⁺] = [OH^1–], so pH = 7.0 at 25 °C. Beyond the equivalence point, OH^1– is in excess and the equilibrium solution is one of a strong base. The following table summarizes these conclusions:

Moles	Solution Type	Moles of Excess Reactant
na > nb	strong acid	na – nb mol H3O1+	pH = –log [H3O1+]
na = nb	neutral	no excess reactant	pH = 7.00
na < nb	strong base	nb – na mol OH1–	pOH = –log [OH1–]

7.3-8. SA–SB Titration Curve Exercise

Weak Acid–Strong Base Reactions

7.3-9. Weak Acid–Strong Base Titration Video

7.3-10. Weak Acid–Strong Base Titration Calculations

We now examine the titration curve for the titration of n_a moles of a weak acid HA with hydroxide ion. Initially the solution is one of the weak acid HA, so if less than 5% of the acid reacts, we can write where c is the concentration of HA. OH^1– is limiting reactant. The reaction table has the following form:

	HA	+	OH1–	→	A1–	+	H2O
initial	na		nb		0
Δ	–nb		–nb		+nb
equilibrium	na – nb		~0		nb

The equilibrium line contains both a weak acid and its conjugate base, so it is a buffer solution, and we can use the Henderson-Hasselbalch equation to determine the pH. A special case in this region occurs at the midpoint of the titration; i.e., when n_b = n_a/2. In this special case the table would be

	HA	+	OH1–	→	A1–	+	H2O
initial	na		na/2		0
Δ	–na/2		–na/2		+na/2
equilibrium	na/2		~0		na/2

Thus, the resulting solution is still a buffer, but at the midpoint of the titration the number of moles of HA and A^1– are equal. The ratio (n_b/n_a) = 1 and log(1) = 0, so the Henderson-Hasselbalch equation becomes, At the equivalence point, n_a = n_b = n, so the reaction table would be

	HA	+	OH1–	→	A1–	+	H2O
initial	n		n		0
Δ	–n		–n		+n
equilibrium	~0		~0		n

The only equilibrium entry that is not zero is under the weak base, so the solution is a weak base solution that contains n mol of the weak base. The [OH^1–] of this weak acid solution would be determined as follows: Note that in the titration of a weak acid with a strong base, the solution at the equivalence point is that of a weak base, so pH > 7. When HA is the limiting reactant, the reaction table is given as the following:

	HA	+	OH1–	→	A1–	+	H2O
initial	na		nb		0
Δ	–na		–na		+na
equilibrium	~0		nb – na		na

The equilibrium entry under OH^1– is not zero, so the solution is a strong base that contains n_b – n_a mol of base. The presence of the weak base does not impact the pH of the solution due to the common ion effect, so we can write

7.3-11. WA–SB Titration Curve Exercise

7.3-12. Using a Titration Curve Exercise

7.3-13. The Shape of an Acid-Strong Base Titration Curve

The general shape of a titration curve consists of a sharp rise in pH in the vicinity of the equivalence point, but how steep the slope of the rise is depends upon the strength of the acid. Compare the titration curves for the titrations of 0.1 M HCl, 0.1 M HC₂H₃O₂, and 0.1 M HCN. In each case, the final pH approaches the pH of the strong base, which would be 13.0 in the case of 0.1 M NaOH. However, the pH at the midpoint of the titration varies with the acid strength. Thus, the pH is about 1.5 at the midpoint in the titration of 0.10 M HCl, but it equals the pK_a of the acid for a weak acid, which is 4.74 for HC₂H₃O₂ and 9.40 for HCN. Thus, the rise of the curve in the case of a strong acid is about 12 pH units, but it is only about 8 pH units for HC₂H₃O₂ and about 3 pH units for HCN. The less dramatic change in pH associated with very weak acids can make it difficult to get as a precise determination of the equivalence point.

7.3-14. Mixing Exercise

7.3-15. Polyprotic Acid–Strong Base Titration Video

7.3-16. Polyprotic Acid–Strong Base

Polyprotic acids dissociate one proton at a time, and they are deprotonated by OH^1– ion one proton at a time. The loss of each proton gives rise to one equivalence point. The equilibrium constants for the individual dissociations are usually very different, so only one or two species usually have non-negligible concentrations in polyprotic acid solutions. Indeed, their equilibria are identical to those of monoprotic acids except that the base may be amphiprotic. Consider the titration of H₂A with a strong base. The following reaction table covers the reaction up to the first equivalence point.

	H2A	+	OH1–	→	HA1–	+	H2O
initial	na		nb		0			mmol
Δ	–nb		–nb		+nb			mmol
equilibrium	na – nb		~0		+nb			mmol

Thus, adding hydroxide to H₂A produces a solution that can be treated as an H₂A/HA^1– buffer. At the first equivalence point, the solution is that of an amphiprotic substance, so we can use Equation 6.13, which is shown below, to determine its pH. The following reaction table then covers the reaction between the first and second equivalence points. Note that the first titration creates n_a mmol of HA^1–, so we use that for the initial amount of weak acid. n_b is the number of mmoles of base added after the first equivalence point.

	HA1–	+	OH1–	→	A2–	+	H2O
intial	na		nb		0			mmol
Δ	–nb		–nb		+nb			mmol
equilibrium	na – nb		~0		+nb			mmol

Adding hydroxide to HA^1– produces a solution that can be treated as an HA^1–/A^2– buffer. At the second equivalence point, the solution is that of the weak base A^2–. After the second equivalence point the solution is a strong base. The resulting titration curve is shown in Figure 7.2.

7.3-17. Polyprotic Titration Curve Exercise

Strong Acid–Weak Base Reactions

7.3-18. Summary of Strong Acid–Weak Base Reactions

The reaction between a strong acid and a weak base is treated analogously with that between a weak acid and a strong base. The nature of the resulting solution depends upon the relative number of moles of acid (n_a) and base (n_b). n_a < n_b: acid is limiting reactant

	H3O1+	+	A1–	→	HA	+	H2O
equilibrium	~0		nb – na		na			Buffer

n_a = n_b = n: acid and base in stoichiometric amounts

	H3O1+	+	A1–	→	HA	+	H2O
equilibrium	~0		~0		n			Weak Acid

n_a > n_b: base is limiting reactant

	H3O1+	+	A1–	→	HA	+	H2O
equilibrium	na – nb		~0		na			Strong Acid

7.3-19. Strong Acid–Weak Base Titration

Thank you for your patience as this section undergoes re-construction.

7.3-20. Strong Acid–Weak Base Titration Curve

Making Buffers by Reaction

7.3-21. Three Ways to Make Buffers

We have now seen three ways in which buffers can be made:

• 1
  mixing the acid and its conjugate base directly in the appropriate ratio to achieve the required pH
• 2
  adding a strong acid to an excess of weak base until the required pH is achieved
• 3
  adding a strong base to an excess of weak acid until the required pH is achieved

Examples of the first method were presented in CAQS Section 7.2, and an example of the second method is presented in the following section.

7.3-22. Preparing a Buffer with a Strong Acid and a Weak Base

7.4 Acid-Base Composition from Equilibrium pH

Introduction

Solutions are frequently prepared by adjusting the solution pH to a desired level rather than adding a known volume of acid or base. For example, the easiest way to make a buffer of a desired pH is to decide on the proper acid-base pair, make a solution that contains an appreciable amount of the acid or base and then add a strong base or acid until the solution has the required pH. The question then becomes: Once you have attained the appropriate pH, what are the concentrations of the acid(s) and base(s) in the solution? We answer that question in this section.

7.4-1. Method

One equation is required for each unknown concentration in a solution, so determining [H₂X], [HX^1–], and [X^2–] in a solution of a diprotic acid involves three equations and three unknowns. The equilibrium constant expressions for K₁ and K₂ are two of the equations. The third equation is mass balance, which is based on the fact that even though reaction may take place in solution, the total amount of X does not change; i.e., c_o = [H₂X] + [HX^1–] + [X^2–]. Fortunately, chemistry can often be used to simplify the algebra because only one or two of the concentrations are usually appreciable at any pH. Once the one or two appreciable concentrations have been determined at a given pH, the other concentrations can be readily determined from the given K values, c_o, and pH. Thus, our first task is to identify these one or two species that dominate the concentration. We do so by assuming that the concentration of a substance is negligible if it is less than 1% of the concentration of either its conjugate acid or base, so

• •
  the acid concentration is negligible if [base]/[acid] > 100, and
• •
  the base concentration is negligible if [base]/[acid] < 0.01

We use the above conditions in the Henderson-Hasselbalch equation to conclude the following.

• •
  the acid is negligible if pH > pK_a + log(100) = pK_a + 2
• •
  the base is negligible if pH < pK_a + log(0.01) = pK_a – 2

7.4-2. Concentrations vs. pH

Monoprotic Acid: The concentrations of CH₃COOH and CH₃COO^1– as a function of pH are shown in Figure 7.3. The pK_a of CH₃COOH is 4.7, so [CH₃COOH] = c_o if pH < 2.7 (pH = 4.7 – 2), but [CH₃COO^1–] = c_o if pH > 6.7 (4.7 + 2). At intermediate pH values (2.7 < pH < 6.7), neither concentration is negligible, so we use [CH₃COOH] + [CH₃COO^1–] = c_o. Thus, the solution is treated as a buffer in the yellow region of the figure, as an acetic acid solution at lower pH, and as an acetate ion solution at higher pH. Diprotic Acid: The case of the diprotic acid H₂S (pK₁ = 7.0, pK₂ = 12.9) is shown in Figure 7.4. The two yellow boxes represent the two pH ranges (pK_a ± 2) in which both the acid (H₂S or HS^1–) and its conjugate base (HS^1– or S^2–) are appreciable. Only one substance is appreciable at all other values of pH. Triprotic Acid: The example of a triprotic acid, H₃PO₄, is considered in the following exercise. Even though there are four concentrations to determine, no more than two are appreciable at any pH.

7.4-3. Triprotic Exercise

7.4-4. Combining Equilibria

While the protons of a polyprotic acid are removed individually, the equilibria can be combined to eliminate concentrations of intermediate ions. For example, consider the dissociation equilibria of H₂S. Adding the above chemical equations results in the cancellation of the HS^1– ion concentration and eliminates it from the equilibrium constant expression. The equilibrium constant for the reaction that results from the addition of two equilibria equals the product of the two equilibrium constants, so we can write the following for the loss of both protons in one equation. The above affords us a quick way to determine the concentration of sulfide ion in a solution where both the H₃O¹⁺ and H₂S concentrations are known. However, the H₂S concentration is its equilibrium concentration, not its makeup concentration. Referring to Figure 7.4, we conclude that

• •
  The above K₁₂ expression is best used for solutions in which pH < 5, because above this pH, the H₂S concentration is no longer equal to the makeup concentration and would have to be determined with K₁.

7.4-5. Exercise

7.5 Acid-Base Indicators

Introduction

Acid-base indicators change color in small ranges of pH change and indicate the endpoint of an acid-base titration, but they too are weak acids and bases.

7.5-1. Acid-Base Indicators

Acid-base indicators are organic dyes that are also weak acids. They function because the weak acid (HIn) and its conjugate base (In^1–) differ in color. Consider the K_a reaction of the indicator HIn. The K_a expression is Solving the above for the ratio of the concentrations of the base and the acid: The human eye perceives the color of In^1– when the ratio is greater than 10 and the color of HIn if the ratio is less than 0.1. In order to change the ratio from 0.1 to 10, the hydronium ion concentration must decrease by a factor of 100, which means that the pH must increase by 2 pH units. For this reason, most indicators have a useful range of about 2 pH units. Thus, a good indicator has a K_a that is close to the hydronium ion concentration at the equivalence point. Alternatively, a good indicator has a pK_a that is close to the pH at the equivalence point of the titration. The following table gives the useful range of some common acid-base indicators.

Indicator	pH Range	Color Change
Methyl violet	0.0–1.6	yellow → blue
Thymol blue	1.2–2.8	red → yellow
Methyl orange	3.2–4.4	red → yellow
Bromocresol green	3.8–5.4	yellow → blue
Alizarin	5.4–6.6	colorless → yellow
Bromothymol blue	6.0–7.6	yellow → blue
Phenolphthalein	8.2–10.0	colorless → pink
Alizarin yellow R	10.1–12.0	yellow → red

7.5-2. Titration Exercise

7.6 Exercises and Solutions

Select the links to view either the end-of-chapter exercises or the solutions to the odd exercises.

Copyright © 2014 Advanced Instructional Systems Inc. and NC State College of Sciences Foundation | Credits