Chapter 6 – Acids and Bases

Introduction

Brønsted acid-base reactions are proton transfer reactions. Acids donate protons to bases. In the process, the acid is converted into its conjugate base and the base into its conjugate acid. A conjugate acid-base pair differ by one and only one proton. The relative strength of an acid is given by its acid dissociation constant, K_a, which is the equilibrium constant for the reaction of the acid with the weak base water. In this chapter, we show how to use the K_a expression to determine equilibrium concentrations in solutions of acids and bases.

6.1 Autoionization of Water

Introduction

All of the acid-base reactions we will discuss occur in water, so it is important to understand the acid-base properties of water. Water is said to be amphiprotic because it can behave as either an acid or a base. In fact, the acidity of an aqueous solution is the extent to which water reacts with an acid to produce its conjugate acid H₃O¹⁺, the hydronium ion. Similarly, the basicity of a solution is the extent to which water reacts with a base to produce its conjugate base, OH^1–. In this section, we examine the relationship between the hydronium and hydroxide ion concentrations in aqueous solutions.

Objectives

• •
  Convert between [H₃O¹⁺] and [OH^1–] in an aqueous solution.

6.1-1. Review of Acids and Bases

The following is a brief review of the more thorough treatment of acid-base chemistry which can be found in CAMS Chapter 12. You should refer to that chapter for a more in-depth discussion. The material in this and the following chapter assumes a knowledge of how to write chemical equations for acid-base reactions, so viewing the video in CAMS Section 12.8-1. would be especially helpful. A Lewis base is a substance that contains a lone pair that can be used in a coordinate covalent bond, and a Lewis acid is a substance that has an empty orbital that can be used to share the lone pair in the bond. A Lewis acid-base reaction is the formation of the bond between the acid and the base. The Lewis acid-base reaction between ammonia and acetic acid is represented in Figure 6.1a. In it, the lone pair on ammonia is used to form a covalent bond to a hydrogen atom on the acetic acid. Ammonia contains the lone pair, so it is the base, and acetic acid accepts the lone pair, so it is the acid. In the reverse reaction a lone pair on the acetate ion attacks a proton of the ammonium ion in another Lewis acid-base reaction. This very broad classification allows us to treat many reactions as acid-base reactions. However, the reaction in the figure can also be viewed has a proton transfer from the acid to the base. Although proton transfer reactions can be viewed as Lewis acid-base reactions, a different acid-base theory was developed for this very important branch of chemistry. In Brønsted-Lowery or simply Brønsted theory, an acid is a proton donor and a base is a proton acceptor. Acetic acid has a proton that it can transfer, so it is an acid, while ammonia can accept a proton, so it is a base. The loss of a proton converts the acid into its conjugate base, and the gain of the proton converts the base into its conjugate acid. An acid and a base differ by one proton only and are said to be a conjugate acid-base pair. The only reactants and products present in a Brønsted acid-base reaction are an acid, a base, and their conjugate base and acid. The brackets in Figure 6.1b identify the conjugate acid-base pairs in the reaction of acetic acid and ammonia.

6.1-2. Ion Product Constant

Since water is both an acid and a base, it can react with itself in a process called autoionization. The equilibrium constant expression for the autoionization, which is given in Equation 6.1, is called the ion product constant of water and given the symbol K_w. Water is the solvent, so it is treated as a pure liquid with an activity of unity, so it does not appear in the equilibrium constant expression. The subscript 'w' simply indicates 'water'. It is called an ion product constant because it involves only the product of the concentrations of two ions. K_w must be satisfied in all aqueous solutions, regardless of what other substances may be dissolved. Thus, the hydronium and hydroxide ion concentrations in ALL aqueous solutions are related by Equation 6.1.

6.1-3. Acidity and Basicity

If water is the only source of H₃O¹⁺ and OH^1– ions, then their concentrations must be equal because they are produced in a 1:1 ratio. However, addition of an acid to an aqueous solution increases [H₃O¹⁺] and decreases [OH^1–] so that the resulting concentrations obey Equation 6.1. Addition of a base has the opposite effect of increasing [OH^1–] and decreasing [H₃O¹⁺]. Consequently, aqueous solutions can be classified as one of the following types.

• •
  Neutral Solution: [H₃O¹⁺] = [OH^1–]
• •
  Acidic Solutions: [H₃O¹⁺] > [OH^1–]
• •
  Basic Solutions: [H₃O¹⁺] < [OH^1–]

However, Equation 6.1 is obeyed in all three types.

6.1-4. Exercise

6.2 The p-Scale

Introduction

Many of the numbers used in this chapter are very small, so they are often expressed on the p-scale to avoid the use of negative exponents or preceding zeroes.

Objective

• •
  Determine [H₃O¹⁺], [OH^1–], pH, and pOH of a solution of a strong acid or strong base of known concentration.

6.2-1. Method for Converting to the p-Scale

The p-scale is defined in Equation 6.2. A 'p' placed before a concentration or a constant means that the number is the negative log of the concentration or constant. For example, pH = –log [H₃O¹⁺] and pK_a = –log K_a. Taking the antilogarithm of both sides of Equation 6.2 gives the expression for converting the p-scale to the concentration or constant. The hydronium ion concentration in a solution with pH = 3.20 is [H₃O¹⁺] = 10^–pH = 10^–3.20 = 6.3e–04 and an acid with a pK_a = 5.62 has a K_a = 10^–5.62 = 2.4e–06. It might appear that the [H₃O¹⁺] and K_a do not have the correct number of significant figures. After all, the pH has three, while the [H₃O¹⁺] and K_a each have only two. The difference arises because the numbers to the left of the decimal in a logarithm indicate the exponent, not the significant figures. This is due to the way negative exponents are handled. An exponent of –3.20 is expressed as 0.80 – 4, so 10^–3.20 = (10^0.80)(10^–4) = (6.3)(10^–4). Thus, the significant figure, 6.3, is determined from 0.80, which comes from the 0.20 in the pH. The exponent, –4, is dictated by the number to the left of the decimal. We conclude that significant figures can be lost when converting from pX to X, while they can be gained in going from X to pX. This is the reason that pK_a values are usually reported to three significant figures, while K_a values have only two. The following are common uses of the p-scale.

pH = –log [H3O1+]	[H3O1+] = 10–pH
pOH = –log [OH1– ]	[OH1– ] = 10–pOH
pKw = –log(Kw) = –log(1.0 × 1014) = 14.00	Kw = 10–pKw = 10–14
pKa = –log(Ka)	Ka = 10–pKa

6.2-2. pK_w, pH, and pOH

We now express the ion product for water (Equation 6.1

K_w = [H₃O¹⁺][OH¹⁻] = 1.0 × 10⁻¹⁴ at 25 °C

) in terms of pH and pOH. Taking the negative logarithm of both sides, we obtain the following. Substitution of the p-scale definitions results in the relationship between pH, pOH, and pK_w. Equation 6.4 is an important relationship that we will use to convert between the pH and pOH of aqueous solutions. It is important to realize that pK_w = 14.00 only at 25 °C, but values at some selected temperatures are given in the following table.

T (°C)	pKw	Kw
0	14.94	1.15e–15
25	14.00	1.00e–14
50	13.28	5.25e–15
75	12.71	1.95e–13
100	12.26	5.50e–13

6.2-3. pH, Acidity, Basicity, and Neutrality

Acidic solutions: [H₃O¹⁺] > 1.0 × 10^–7. Taking the negative logarithm of both sides of this inequality produces
pH < 7.00 for acidic solutions. A solution in which [H₃O¹⁺] = 1.0 M (pH = 0.0) is fairly acidic solution, so most pH values for acidic solutions lie between 0 and 7. However, H₃O¹⁺ concentrations greater than 1.0 M are possible. We conclude that the pH of a solution decreases as the acidity increases, and the pH of most acidic solutions lies between 0 and 7, but negative values are possible. Basic Solutions: [OH^1–] > 1.0 × 10^–7, so [H₃O¹⁺] < 1.0 × 10^–7. Taking the negative logarithm of both sides yields pOH < 7 or pH > 7 for basic solutions. A solution in which [OH^1–] = 1.0 M has a pOH = –log(1.0) = 0.0, which gives it a pH = 14.0 – 0.0 = 14.0, which is a very basic solution. However, solutions in which [OH^1–] > 1 M are not uncommon. We conclude that the pH of a solution increases as the basicity increases, and the pH of most basic solutions lie between 7 and 14, but values higher than 14 are possible. Neutral Solutions: [H₃O¹⁺] = [OH^1–] = 1.0 × 10^–7, so pH = pOH = 7.00 (at 25 °C). Thus, a neutral solution at
25 °C is one with pH = 7.00. These conclusions are summarized in the following figure.

6.2-4. Converting to the p-Scale Exercises

6.2-5. Exercise

6.2-6. Exercise

6.2-7. Exercise

6.3 Strong Acids

Introduction

The common strong acids are HClO₄, HCl, HBr, HI, HNO₃, and H₂SO₄. In this section, we discuss how the pH of a strong acid solution is determined.

Objective

• •
  Calculate the pH of the solutions of strong acids and of strong bases.

6.3-1. Strong Acid Video

6.3-2. Strong Acid Solutions

Hydrochloric acid is a strong acid, so essentially all of the HCl reacts. The reaction is so extensive that it is usually represented with a single arrow.

	HCl	+	H2O	→	H3O1+	+	Cl1–
initial	co				0		0
Δ	–co				+co		co
eq	0				co		co

In the reaction table above, c_o is the makeup concentration of the acid, which is the concentration given on the acid bottle. For strong, monoprotic acids, the hydronium ion concentration is given by the makeup concentration because all of the acid reacts. The case of H₂SO₄ is a little more complicated because only the first proton is strongly acidic. We will consider sulfuric acid in more detail in the section on polyprotic acids.

6.3-3. Exercises

6.4 Weak Acids

Introduction

Weak acids react only slightly with water, so the hydronium ion concentration in a solution of a weak acid does not equal the makeup concentration. In this section, we show how to determine the concentrations of all of the species in a solution of a weak acid.

Objectives

• •
  Write the K_a expression for a weak acid.
• •
  Determine the equilibrium concentrations of all species present in and the pH of solutions of weak acids.
• •
  Determine the pH, K_a, or initial concentration of a weak acid solution given the other two.
• •
  Calculate the percent ionization of a weak acid.
• •
  Convert between pK_a and K_a.

6.4-1. Weak Acid Video

6.4-2. Weak Acid-Water Reaction Table

The equilibrium constant for the reaction of an acid and water is called the acid dissociation constant, K_a. K_a is so large for strong acids that all of the acid reacts to produce hydronium ion and the conjugate base of the acid. K_a is small for weak acids, so only a portion of a weak acid reacts with water. Thus, the Δ line entries of the reaction table are unknown. Consider the reaction of a generic weak acid HA with water.

	HA	+	H2O	⇌	H3O1+	+	A1–
initial	co				0		0
Δ	–x				x		x
initial	co – x				x		x

Water is the solvent, so it is treated as a pure liquid (activity = 1), so the K_a expression is the following. The values of K_a and pK_a for several acids are given in the Acid-Base Table with pK_a.

6.4-3. Solving the K_a Expression

We must solve Equation 6.5 for x to obtain the equilibrium concentrations in an aqueous solution of the weak acid HA. where x = [H₃O¹⁺] = [A^1–] and c_o – x = [HA]. The equation is a quadratic that could be solved by gathering terms and using the quadratic formula. However, it can be solved more simply if x is negligible compared to c_o, i.e., if c_o – x ~ c_o. In other words, if so little of the acid reacts that the acid concentration is essentially unchanged. Making this substitution, we obtain the following expression for a weak acid which can be solved for x, the concentration of the hydronium ion or the conjugate base (A^1–), with Equation 6.6. Although neglecting x in the c_o – x term is acceptable most of the time, there are times (K_a is large or c_o is small) when x is not negligible. Consequently, you should use the above expression only when solving for the squared term, and you should always check your result to see if it is indeed negligible. An easy way to determine whether the approximation is valid is given next.

6.4-4. Five Percent Rule

Equation 6.6 is an easy way to determine the concentrations in a solution of a weak acid, but the approximation used to derive it is not always valid, so the answer should always be checked. There are two ways to check whether x is negligible:

• •
  Inspection: The value of x calculated with Equation 6.6
  [H₃O¹⁺] = [A¹⁻] =

  Kaco

  is often so small that it is obviously negligible compared to c_o. For example, if c_o = 0.100 M and x = 0.0000001 M then c_o – x = 0.100 M and the assumption is good.
• •
  5% Rule: If x is less than 5% of c_o, then approximating c_o for c_o – x will produce the same answer as the quadratic formula. For example, substitution of c_o = 0.100 and K_a = 1.6e–04 into Equation 6.6
  [H₃O¹⁺] = [A¹⁻] =

  Kaco

  gives x = 0.004 M. In this case, c_o – x = 0.100 – 0.004 = 0.096, which is not equal to c_o. However, (x/c_o) = 0.04, which is 4%. Thus, x is less than 5%, so using the quadratic formula would result in the same answer for x.

6.4-5. Solving Without Approximation

If the answer produced by Equation 6.6

[H₃O¹⁺] = [A¹⁻] =

Kaco

is greater than 5% of c_o, the terms in Equation 6.5

K_a =

[H3O1+][A1−]

[HA]

=

(x)(x)

co − x

=

x2

co − x

must be gathered in the form of a quadratic equation

(ax² + bx + c = 0)

and solved with the quadratic formula given below. However, Equation 6.6

[H₃O¹⁺] = [A¹⁻] =

Kaco

is so easy to use, that it is usually simpler to use it and check to be certain that x is within the 5% limit. This is especially true since most of the acids and concentrations we deal with in this course are so weak that the 5% rule is usually obeyed. However, for the stronger of the weak acids or for very dilute solutions, more than 5% of the acid may react, so always check.

6.4-6. Method for Weak Acid Problems

Equations 6.5

K_a =

[H3O1+][A1−]

[HA]

=

(x)(x)

co − x

=

x2

co − x

and 6.6

[H₃O¹⁺] = [A¹⁻] =

Kaco

indicate that three terms (c_o, x, and K_a) are required to establish a weak acid equilibrium. Consequently, there are three different types of problems that are solved using these equations. The three differ in the identity of the unknown.

• •
  x: What is the pH, [H₃O¹⁺], or [A^1–]? Given: K_a (or the identity of the acid so you can look it up) and the makeup concentration. Method: Use Equation 6.6
  [H₃O¹⁺] = [A¹⁻] =

  Kaco

  to solve for the unknown then check that x is indeed small enough to neglect. If x is not small enough, rearrange the expression so that it is in the form of a quadratic equation and use the quadratic formula to solve for the unknown concentration.
• •
  K_a: What is the value of K_a? Given: x as the pH, [H₃O¹⁺], or [A^1–] and the makeup concentration. Method: Substitute the known values into Equation 6.5
  K_a =

  [H3O1+][A1−]

  [HA]

  =

  (x)(x)

  co − x

  =

  x2

  co − x

  . Since x and c_o are known, there is no reason to make the neglect-x approximation, so make the subtraction.
• •
  c_o: How much acid must be used to make a solution? Given: x as pH, [H₃O¹⁺], or [A^1–], and K_a (or the identity of the acid so you can look it up). Method: Solve Equation 6.5
  K_a =

  [H3O1+][A1−]

  [HA]

  =

  (x)(x)

  co − x

  =

  x2

  co − x

  for the equilibrium concentration of the weak acid (c_o – x), then solve for c_o using the given value of x.

6.4-7. Determining the pH Exercise

6.4-8. Determining the pH Exercise

6.4-9. Determining K_a Exercise

6.4-10. Determining the Makeup Concentration Exercise

6.4-11. Percent Ionization

We now show that the fraction of an acid that reacts depends upon both its concentration and its acid dissociation constant. We start by reproducing the equilibrium line in the reaction table for a weak acid reacting with water.

	HA	+	H2O	⇌	H3O1+		A1–
eq	co – x				x		x

Of the initial concentration c_o, x mol/L react, so the fraction that reacts is x/c_o. If we assume that the amount that reacts is negligible (c_o – x = c_o), we can use Equation 6.6

[H₃O¹⁺] = [A¹⁻] =

Kaco

to determine x as (c_oK_a)^1/2. Substitution of this value of x into the expression for the fraction that reacts produces the following. Multiplication of the above by 100% gives us the percent ionization (reaction) of a weak acid. Note that the percent that reacts depends upon both K_a and c_o. Thus, the 5% rule is obeyed by acids with low K_a values and relatively high concentrations. Consider the following table that compares the percent ionization of nitrous (K_a = 4.0e–4), acetic (K_a = 1.8e–5), and hypochlorous (K_a = 3.5e–8) acids as a function of their concentrations.

Molarity	HNO2	CH3COOH	HOCl
1.0	2.0%	0.42%	0.019%
0.10	6.1%*	1.3%	0.059%
0.010	18%*	4.2%	0.19%
0.00010	83%*	34%*	1.9%

*The percent ionization exceeds 5% in these cases, so the quadratic formula was used to determine x.

6.4-12. Percent Ionization Exercise

6.5 Polyprotic Acids

Introduction

Polyprotic acids contain two or more protons. For example, H₂S is diprotic and H₃PO₄ is triprotic. There are some polyprotic acids that contain more than three protons, but we will not consider any of these acids. The conjugate base of a polyprotic acid is also an acid, and the solution of a polyprotic acid is really a solution of more than one acid. Although it might seem that this would make the treatment of these acids very difficult, determining the concentrations of the species present in a polyprotic acid is not much different than determining the concentrations in a solution of a monoprotic acid.

Objective

• •
  Determine the concentrations of all species in a solution of a polyprotic acid of known concentration.

6.5-1. Phosphoric Acid Overview

We examine a 0.10-M solution of phosphoric acid as an example because it contains three protons and is the most complicated. The three deprotonation steps are the following.

H3PO4 + H2O ⇌ H2PO41– + H3O1+	K1 = 7.5 × 10–5
H2PO41– + H2O ⇌ HPO42– + H3O1+	K2 = 6.2 × 10–8
HPO42– + H2O ⇌ PO43– + H3O1+	K3 = 4.8 × 10–13

Note that each K value is over 1,000 times smaller than the preceding value. This is the case for most polyprotic acids, and it is why polyprotic acids can be treated in a relatively simple manner. The large decrease in K means that a negligible amount of each acid reacts in each step, so the concentration of any species is the concentration determined in the first step in which it is produced. The method we will use to find the concentrations of all species is the following:

• 1
  Determine [H₃O¹⁺], [H₂PO₄^1–], and [H₃PO₄] from the first acid dissociation. These concentrations are not affected by subsequent steps because the subsequent equilibrium constants are so small.
• 2
  Set up the reaction table for the second reaction using the [H₃O¹⁺] and [H₂PO₄^1–] determined in step 1 as the initial conditions. Solve the equilibrium problem for [HPO₄^2–], which will be shown to equal K₂.
• 3
  Set up the reaction table for the third reaction using [H₃O¹⁺] and [HPO₄^2–] determined in steps 1 and 2 as initial concentrations. Solve the equilibrium problem for [PO₄^3–].

We next examine each of these steps more closely.

6.5-2. Phosphoric Acid Step 1

The reaction table for the first step in the deprotonation of 0.10 M H₃PO₄ is as follows.

	H3PO4	+	H2O	⇌	H2PO41–	+	H3O1+	K1 = 7.5 × 10–3
initial	0.10				0		0	M
Δ	–x				+x		+x	M
eq	0.10 – x				x		x	M

H₃PO₄ is about five hundred times stronger than acetic acid so x is probably not negligible in a 0.10 M solution, but we try the approximation to be sure. The result represents a 27% dissociation, which is well above the 5% limit. The approximation cannot be used, so we must solve for x with the quadratic formula as follows: Next we use these concentrations in the second deprotonation.

6.5-3. Phosphoric Acid Step 2

The reaction table for the second deprotonation step is the same as that for the dissociation of 0.024 M H₂PO₄^1– except that the initial concentration of [H₃O¹⁺] is also 0.024 M as a result of the dissociation of H₃PO₄.

	H2PO41–	+	H2O	⇌	HPO42–	+	[H3O1+]	K2 = 6.2 × 10–8
initial	0.024				0		0.024	M
Δ	–x				0		+x	M
eq	0.024 – x				x		0.024 + x	M

Set up the equilibrium constant expression. Substitute the given K_a and the entries in the equilibrium line. If x is negligible then 0.024 + x = 0.024 – x and the two terms cancel in the K₂ expression to yield [HPO₄^2–] = K₂ = 6.2e–8, which is certainly negligible compared to 0.024. We conclude the following.

6.5-4. Phosphoric Acid Step 3

The reaction table for the third deprotonation step is the same as dissociation of HPO₄^2– except that the initial concentration of [H₃O¹⁺] is 0.024 M from the first step and that of H₂PO₄^1– is 6.2e–8 M as a result of the second step.

	HPO42–	+	H2O	⇌	PO43–	+	[H3O1+]	K3 = 4.8 × 10–13
initial	6.2e–8				0		0.024	M
Δ	–x				+x		+x	M
eq	6.2e–8				x		0.024	M

x has been assumed to be negligible compared to the initial concentrations. Set up the equilibrium constant expression. Substitute the given K_a and the entries in the equilibrium line. Solve for x = [PO₄^3–] to obtain which is negligible. Indeed, there is hardly any phosphate ion present in 0.1 M phosphoric acid. In summary, the concentrations of the phosphorus containing species are as follows.

• •
  [H₃PO₄] = 0.08 M
• •
  [H₂PO₄^1–] = 0.024 M
• •
  [HPO₄^2–] = 6.2e–8 M
• •
  [PO₄^3–] = 1.2e–18 M

Note that the sum of the concentrations should be 0.10 M, the make-up concentration of H₃PO₄. The fact that the sum, 0.08 + 0.024 = 0.104 M, does not equal the total added initially is due to the number of significant figures; i.e., the final digit in the sum is not significant.

6.5-5. Hydrogen Sulfide Exercise

6.5-6. Carbonic Acid Exercise

6.5-7. Sulfuric Acid Exercise

6.6 Strong Bases

Introduction

The solutions of strong bases are usually made by dissolving metal hydroxides in water. In this section, we discuss how the pH of a strong base solution is determined.

Objective

• •
  Calculate the pH of the solutions of strong bases.

6.6-1. Strong Base Solutions

Dissolving a metal hydroxide in water is represented by the following chemical equation. [OH^1–] = nc_o in such a solution. However, most metal hydroxides are only sparingly soluble, so the metal is normally a Group 1 metal ion (n = 1) or Ba²⁺ (n = 2) in solutions with appreciable hydroxide ion concentrations.

6.6-2. Exercise

6.7 Weak Bases

Introduction

Weak bases are treated in a manner identical to that used for weak acids.

Objectives

• •
  Write the K_b expression for a weak base.
• •
  Determine the equilibrium concentrations of all species present in and the pH of solutions of weak bases.
• •
  Determine the pH, K_b, or initial concentration of a weak base solution given the other two.
• •
  Convert between pK_b and K_b.
• •
  Convert between the K_a and K_b of a conjugate acid-base pair.

6.7-1. Weak Base-Water Reaction

The equilibrium constant for the reaction of a base and water has the symbol, K_b. K_b is small for weak bases, so only a portion of a weak base reacts with water. Thus, the Δ line entries of the reaction table are unknown. Consider the reaction of a generic weak base A^1– with water.

	A1–	+	H2O	→	OH1–	+	HA
initial	co				0		0
Δ	–x				x		x
initial	co – x				x		x

Water is the solvent, so it is treated as a pure liquid (activity = 1), so the K_b expression is the following. If we assume that the extent of reaction of the weak base is negligible, i.e., c_o – x = c_o, we can express the K_b expression as follows. which can be solved for x, the concentration of the hydroxide ion or the conjugate acid (HA).

6.7-2. Relating K_a and K_b

Consider the product of the K_a of HA (a weak acid) and the K_b of A^1– (its conjugate base). Note that [A^1–] appears in the numerator of K_a and in the denominator of K_b, so the two terms cancel. Similarly, [HA] appears in the denominator of K_a and in the numerator of K_b, so these two terms cancel as well. Canceling these terms results in the following. Substitution of K_w for [H₃O¹⁺][OH^1–] produces Equation 6.10. The Acid-Base Table with pK_a gives only K_a and pK_a, so Equation 6.10 must be used to determine the K_b of the conjugate base. At 25 °C, K_w = 1.0e–14, so the following can be used to determine the K_b at 25 °C.

6.7-3. Weak Base pH Exercise

6.7-4. pK_a and pK_b

Acid and base constants are frequently given on the p-scale as follows. Due to the minus sign in the definition, a higher pK_a or pK_b indicates a weaker acid or base. Taking the negative logarithm of both sides of Equation 6.10

K_aK_b = K_w

produces the following relationship.

6.7-5. Determining K_b Exercise

6.7-6. pK_a, pK_b, K_a, and K_b Exercise

6.7-7. Bases of Polyprotic Acids

Bases that can accept more than one proton are treated much the same way as polyprotic acids. Consider the case of the sulfide ion, which can accept two protons.

• 1
  S^2– + H₂O ⇌ HS^1– + OH^1–    K₁ = 0.077
• 2
  HS^1– + H₂O ⇌ H₂S + OH^1–    K₂ = 1.0 × 10^–7

Note that the two K_b values differ by a factor of about a million. Thus, the hydroxide produced in the second step is negligible compared to that produced in the first step. Consequently, the hydroxide ion concentration in an aqueous solution of a base that can accept more than one proton can usually be determined by considering only the first step.

6.7-8. Base of a Polyprotic Acid Exercise

6.8 Salts of Weak Acids and Bases

Introduction

Salts are ionic compounds that are produced in acid-base reactions. The anion of the salt is the conjugate base of the reacting acid, and the cation is usually a metal ion that was associated with the reacting base. For example, the reaction between NaOH and HCl is the acid-base reaction between H₃O¹⁺ and OH^1–. Na¹⁺ ion, the metal ion associated with the reacting base and Cl^1–, the conjugate base of the reacting acid serve as spectator ions. Combination of the spectator ions produces a salt, NaCl. Salts can be acidic, basic, or neutral depending upon the relative acidity/basicity of the cation and anion. In this section, we discuss the acid-base properties of salts.

Objective

• •
  Define the term salt and predict whether a salt is neutral, basic, or acidic.

6.8-1. Acid-Base Properties of Cations and Anions

Small, highly-charged metal cations have orbitals available to accept electron pairs to form covalent bonds, so they are Lewis acidic. However, the available orbitals of the 1A metal ions are so high in energy that these ions are not acidic in aqueous solutions. Thus, 1A metal ions do not impact the acidity or basicity of a salt. For simplicity, we will limit our use of metal ions to 1A metal ions. The only other common cation is NH₄¹⁺, which is acidic. Most anions readily accept the positive charge of a proton, so anions are usually good bases. The exceptions are the conjugate bases of the strong acids (ClO₄^1–, Cl^1–, Br^1–, I^1–, and NO₃^1–), which are neutral, and the protonated anions (HSO₄^1–, HSO₃^1– and H₂PO₄^1–), which are acidic.

6.8-2. Acid-Base Properties of Salts

A salt can be acidic, basic, or neutral depending upon the relative acid and base strengths of the cation and anion.

• 1
  Neutral Salts: The reaction of a strong acid and a strong base produces a neutral salt. The cation in such reactions is a 1A metal and the anion is the conjugate base of a strong acid. Thus, both ions are neutral, which produces a neutral salt. The salt of a weak acid and a weak base can be neutral only when the K_a of the cation equals the K_b of the anion. Ammonium acetate is a neutral salt formed in the reaction of acetic acid and ammonia. It is neutral because the K_a of the ammonium ion equals the K_b of the acetate ion.
• 2
  Basic Salts: Basic Salts are produced in the reaction between a weak acid and a strong base. For example, NaCN is a basic salt that is produced by the reaction of HCN and NaOH. It is basic because sodium ion does not impact the acid-base properties, while the cyanide ion is a basic anion. Basic salts are produced in the reaction between a weak acid and a weak base if the K_b of the base is greater than the K_a of the acid. Ammonium sulfide is a basic salt because the K_b of the sulfide ion is greater than the K_a of the ammonium ion. Basic salts are the most common source of weak bases.
• 3
  Acidic salts: The reaction of a strong acid and a weak base produces acidic salts. NH₄Cl is an acidic salt produced by the reaction between HCl and NH₃. It is acidic because the ammonium ion is a weak acid, but chloride ion is the conjugate base of a strong acid, so it is not basic. Ammonium fluoride is acidic because the K_a of ammonium ion is greater than the K_b of the fluoride ion.

6.8-3. Predicting Salt Basicity and Acidity Exercise

6.8-4. Determining the pH of a Salt Exercise

6.9 Amphiprotic Salts

Introduction

Amphiprotic salts are both acids and bases, so their aqueous equilibria are slightly more complicated than those of weak acids or bases. In this section, we discuss the equilibria and give a simple expression for determining their pH.

Objective

• •
  Determine the pH of a solution of an amphiprotic salt.

6.9-1. Acid Equilibrium

As shown in the figure, HCO₃^1– is an amphiprotic substance because it can behave as both an acid and a base. It produces hydronium ion through its K_a reaction. For every mole of hydronium produced, a mole of carbonate ion is also produced; i.e., The concentration of hydronium ion produced in this step equals the equilibrium concentration of its conjugate base, carbonate ion. However, HCO₃^1– also produces hydroxide ion through its K_b reaction. For every mole of hydroxide ion produced, a mole of carbonic acid is also produced; i.e., Each mole of hydroxide ion that is produced consumes a mole of hydronium ion (H₃O¹⁺ + OH^1– → 2 H₂O). Thus, the following is true. The equilibrium hydronium ion concentration is the following. The concentrations of CO₃^2– and H₂CO₃ can be obtained from K_a2 and K_b1. After some algebra and the assumption that [HCO₃^1–] >> K_a1, we arrive at the result below. The pH of an amphiprotic substance is half-way between its pK_a (pK₂) and that of its conjugate acid (pK₁) so long as its concentration is much larger than the K_a of its conjugate acid, K_a1.

6.10 Exercises and Solutions

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

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