Example 6

Electrical conductivity measurements indicate that 0.42% of the acetic acid molecules in a 1.00 M solution are ionized at 25°C. Calculate K_a and pK_a for acetic acid at this temperature.

Given: analytical concentration and percent ionization

Asked for: K_a and pK_a

Strategy:

A Write the balanced equilibrium equation for the reaction and derive the equilibrium constant expression.

B Use the data given and the stoichiometry of the reaction to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations for all species in the equilibrium constant expression.

C Substitute the final concentrations into the equilibrium constant expression and calculate the K_a. Take the negative logarithm of K_a to obtain the pK_a.

Solution:

A The balanced equilibrium equation for the dissociation of acetic acid is as follows:

$${\text{CH}}_3{\text{CO}}_2\text{H(aq)}\rightleftharpoons {\text{H}}^{+}\text{(aq)}+{\text{CH}}_3{\text{CO}}_2{}^{-}\text{(aq)}$$

and the equilibrium constant expression is as follows:

$$K_{\text{a}}=\frac{[{\text{H}}^{+}][{\text{CH}}_3{\text{CO}}_2{}^{-}]}{[{\text{CH}}_3{\text{CO}}_2\text{H}]}$$

B To calculate the K_a, we need to know the equilibrium concentrations of CH₃CO₂H, CH₃CO₂⁻, and H⁺. The most direct way to do this is to construct a table that lists the initial concentrations and the changes in concentrations that occur during the reaction to give the final concentrations, using the procedure introduced in Chapter 15 "Chemical Equilibrium". The initial concentration of unionized acetic acid ([CH₃CO₂H]_i) is the analytical concentration, 1.00 M, and the initial acetate concentration ([CH₃CO₂⁻]_i) is zero. The initial concentration of H⁺ is not zero, however; [H⁺]_i is 1.00 × 10⁻⁷ M due to the autoionization of water. The measured percent ionization tells us that 0.42% of the acetic acid molecules are ionized at equilibrium. Consequently, the change in the concentration of acetic acid is Δ[CH₃CO₂H] = −(4.2 × 10⁻³)(1.00 M) = −0.0042 M. Conversely, the change in the acetate concentration is Δ[CH₃CO₂⁻] = +0.0042 M because every 1 mol of acetic acid that ionizes gives 1 mol of acetate. Because one proton is produced for each acetate ion formed, Δ[H⁺] = +0.0042 M as well. These results are summarized in the following table.

The final concentrations of all species are therefore as follows:

$$\begin{array}{l}{[{\text{CH}}_3{\text{CO}}_2\text{H}]}_{\text{f}}={[{\text{CH}}_3{\text{CO}}_2\text{H}]}_{\text{i}}+\Delta [{\text{CH}}_3{\text{CO}}_2\text{H}]=1.00\text{ M}+(-0.0042\text{ M})=1.00\text{ M}\hfill \\ {[{\text{CH}}_3{\text{CO}}_2{}^{-}]}_{\text{f}}={[{\text{CH}}_3{\text{CO}}_2{}^{-}]}_{\text{i}}+\Delta [{\text{CH}}_3{\text{CO}}_2{}^{-}]=0\text{ M}+(+0.0042\text{ M})=0.0042\text{ M}\hfill \\ {[{\text{H}}^{+}]}_{\text{f}}={[{\text{H}}^{+}]}_{\text{i}}+\Delta [{\text{H}}^{+}]=1.00\times {10}^{-7}\text{ M}+(+0.0042\text{ M})=0.0042\text{ M}\hfill \end{array}$$

C We can now calculate K_a by inserting the final concentrations into the equilibrium constant expression:

$$K_{\text{a}}=\frac{[{\text{H}}^{+}][{\text{CH}}_3{\text{CO}}_2{}^{-}]}{[{\text{CH}}_3{\text{CO}}_2\text{H}]}=\frac{(0.0042)(0.0042)}{1.00}=1.8\times {10}^{-5}$$

The pK_a is the negative logarithm of K_a: pK_a = −log K_a = −log(1.8 × 10⁻⁵) = 4.74.

Exercise

Picric acid is the common name for 2,4,6-trinitrophenol, a derivative of phenol (C₆H₅OH) in which three H atoms are replaced by nitro (–NO₂) groups. The presence of the nitro groups removes electron density from the phenyl ring, making picric acid a much stronger acid than phenol (pK_a = 9.99). The nitro groups also make picric acid potentially explosive, as you might expect based on its chemical similarity to 2,4,6-trinitrotoluene, better known as TNT. A 0.20 M solution of picric acid is 73% ionized at 25°C. Calculate K_a and pK_a for picric acid.

Answer: K_a = 0.39; pK_a = 0.41

Example 7

A 1.0 M aqueous solution of ammonia has a pH of 11.63 at 25°C. Calculate K_b and pK_b for ammonia.

Given: analytical concentration and pH

Asked for: K_b and pK_b

Strategy:

A Write the balanced equilibrium equation for the reaction and derive the equilibrium constant expression.

B Use the data given and the stoichiometry of the reaction to construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations for all species in the equilibrium constant expression.

C Substitute the final concentrations into the equilibrium constant expression and calculate the K_b. Take the negative logarithm of K_b to obtain the pK_b.

Solution:

A The balanced equilibrium equation for the reaction of ammonia with water is as follows:

$${\text{NH}}_3\text{(aq)}+{\text{H}}_2\text{O(l)}\rightleftharpoons {\text{NH}}_4{}^{+}\text{(aq)}+{\text{OH}}^{-}\text{(aq)}$$

and the equilibrium constant expression is as follows:

$$K_{\text{b}}=\frac{[{\text{NH}}_4{}^{+}][{\text{OH}}^{-}]}{[{\text{NH}}_3]}$$

Remember that water does not appear in the equilibrium constant expression for K_b.

B To calculate K_b, we need to know the equilibrium concentrations of NH₃, NH₄⁺, and OH⁻. The initial concentration of NH₃ is the analytical concentration, 1.0 M, and the initial concentrations of NH₄⁺ and OH⁻ are 0 M and 1.00 × 10⁻⁷ M, respectively. In this case, we are given the pH of the solution, which allows us to calculate the final concentration of one species (OH⁻) directly, rather than the change in concentration. Recall that pK_w = pH + pOH = 14.00 at 25°C. Thus pOH = 14.00 − pH = 14.00 − 11.63 = 2.37, and [OH⁻]_f = 10^−2.37 = 4.3 × 10⁻³ M. Our data thus far are listed in the following table.

The final [OH⁻] is much greater than the initial [H⁺], so the change in [OH⁻] is as follows:

Δ[OH⁻] = (4.3 × 10⁻³ M) − (1.00 × 10⁻⁷ M) ≈ 4.3 × 10⁻³ M

The stoichiometry of the reaction tells us that 1 mol of NH₃ is converted to NH₄⁺ for each 1 mol of OH⁻ formed, so

Δ[NH₄⁺] = +4.3 × 10⁻³ M and Δ[NH₃] = −4.3 ×10⁻³ M

We can now insert these values for the changes in concentrations into the table, which enables us to complete the table.

C Inserting the final concentrations into the equilibrium constant expression gives K_b:

$$K_{\text{b}}=\frac{[{\text{NH}}_4{}^{+}][{\text{OH}}^{-}]}{[{\text{NH}}_3]}=\frac{{(4.3\times {10}^{-3})}^2}{1.0}=1.8\times {10}^{-5}$$

and pK_b = −log K_b = 4.74.

The K_b and the pK_b for ammonia are almost exactly the same as the K_a and the pK_a for acetic acid at 25°C. In other words, ammonia is almost exactly as strong a base as acetic acid is an acid. Consequently, the extent of the ionization reaction in an aqueous solution of ammonia at a given concentration is the same as in an aqueous solution of acetic acid at the same concentration.

Exercise

The pH of a 0.050 M solution of pyridine (C₆H₅N) is 8.96 at 25°C. Calculate K_b and pK_b for pyridine.

 

Answer: K_b = 1.7 × 10⁻⁹; pK_b = 8.77

Example 8

Calculate the pH and percent ionization of a 0.225 M solution of ethylamine (CH₃CH₂NH₂), which is used in the synthesis of some dyes and medicines. The pK_b of ethylamine is 3.19 at 20°C.

Given: concentration and pK_b

Asked for: pH and percent ionization

Strategy:

A Write the balanced equilibrium equation for the reaction and the equilibrium constant expression. Calculate K_b from pK_b.

B Use Equation 16.45 to see whether you can ignore [H⁺] due to the autoionization of water. Then use a tabular format to write expressions for the final concentrations of all species in solution. Substitute these values into the equilibrium equation and solve for [OH⁻]. Use Equation 16.42 to calculate the percent ionization.

C Use the relationship K_w = [OH⁻][H⁺] to obtain [H⁺]. Then calculate the pH of the solution.

Solution:

A We begin by writing the balanced equilibrium equation for the reaction:

$${\text{CH}}_3{\text{CH}}_2{\text{NH}}_2\text{(aq)}+{\text{H}}_2\text{O(l)}\rightleftharpoons {\text{CH}}_3{\text{CH}}_2{\text{NH}}_3{}^{+}\text{(aq)}+{\text{OH}}^{-}\text{(aq)}$$

The corresponding equilibrium constant expression is as follows:

$$K_{\text{b}}=\frac{[{\text{CH}}_3{\text{CH}}_2{\text{NH}}_3{}^{+}][{\text{OH}}^{-}]}{[{\text{CH}}_3{\text{CH}}_2{\text{NH}}_2]}$$

From the pK_b, we have K_b = 10^−3.19 = 6.5 × 10⁻⁴.

B To calculate the pH, we need to determine the H⁺ concentration. Unfortunately, H⁺ does not appear in either the chemical equation or the equilibrium constant expression. However, [H⁺] and [OH⁻] in an aqueous solution are related by K_w = [H⁺][OH⁻]. Hence if we can determine [OH⁻], we can calculate [H⁺] and then the pH. The initial concentration of CH₃CH₂NH₂ is 0.225 M, and the initial [OH⁻] is 1.00 × 10⁻⁷ M. Because ethylamine is a weak base, the extent of the reaction will be small, and it makes sense to let x equal the amount of CH₃CH₂NH₂ that reacts with water. The change in [CH₃CH₂NH₂] is therefore −x, and the change in both [CH₃CH₂NH₃⁺] and [OH⁻] is +x. To see whether the autoionization of water can safely be ignored, we substitute K_b and C_B into Equation 16.46:

K_bC_B = (6.5 × 10⁻⁴)(0.225) = 1.5 × 10⁻⁴ > 1.0 × 10⁻⁶

Thus the simplifying assumption is valid, and we will not include [OH⁻] due to the autoionization of water in our calculations.

Substituting the quantities from the last line of the table into the equilibrium constant expression,

$$K_{\text{b}}=\frac{[{\text{CH}}_3{\text{CH}}_2{\text{NH}}_3{}^{+}][{\text{OH}}^{-}]}{[{\text{CH}}_3{\text{CH}}_2{\text{NH}}_2]}=\frac{(x)(x)}{0.225-x}=6.5\times {10}^{-4}$$

As before, we assume the amount of CH₃CH₂NH₂ that ionizes is small compared with the initial concentration, so [CH₃CH₂NH₂]_f = 0.225 − x ≈ 0.225. With this assumption, we can simplify the equilibrium equation and solve for x:

$$\begin{array}{ccc}\hfill K_{\text{b}}& =\frac{x^2}{0.225}\hfill & =6.5\times {10}^{-4}\hfill \\ \hfill & x& =0.012={[{\text{CH}}_3{\text{CH}}_2{\text{NH}}_3{}^{+}]}_{\text{f}}={[{\text{OH}}^{-}]}_{\text{f}}\hfill \end{array}$$

The percent ionization is therefore

$$\text{percent ionization}=\frac{[{\text{OH}}^{-}]}{C_{\text{B}}}\times 100=\frac{0.012\text{ M}}{0.225\text{ M}}\times 100=5.4\%$$

which is at the upper limit of the approximately 5% range that can be ignored. The final hydroxide concentration is thus 0.012 M.

C We can now determine the [H⁺] using the expression for K_w:

$$\begin{array}{cc}\hfill K_{\text{w}}& =[{\text{OH}}^{-}][{\text{H}}^{+}]\hfill \\ \hfill 1.01\times {10}^{-14}& =(0.012\text{ M})[{\text{H}}^{+}]\hfill \\ \hfill 8.4\times {10}^{-13}\text{ M}& =[{\text{H}}^{+}]\hfill \end{array}$$

The pH of the solution is −log(8.4 × 10⁻¹³) = 12.08. Alternatively, we could have calculated pOH as −log(0.012) = 1.92 and determined the pH as follows:

$$\begin{array}{cc}\hfill \text{pH} + \text{pOH}& =\text{p}{K}_{\text{w}}=14.00\hfill \\ \hfill \text{pH}& =14.00-1.92=12.08\hfill \end{array}$$

The two methods are equivalent.

Exercise

Aromatic amines, in which the nitrogen atom is bonded directly to a phenyl ring (−C₆H₅) tend to be much weaker bases than simple alkylamines. For example, aniline (C₆H₅NH₂) has a pK_b of 9.13 at 25°C. What is the pH of a 0.050 M solution of aniline?

Answer: 8.78

Example 9

Benzoic acid (C₆H₅CO₂H) is used in the food industry as a preservative and medically as an antifungal agent. Its pK_a at 25°C is 4.20, making it a somewhat stronger acid than acetic acid. Calculate the percentage of benzoic acid molecules that are ionized in each solution.

1. a 0.0500 M solution
2. a 0.00500 M solution

Given: concentrations and pK_a

Asked for: percent ionization

Strategy:

A Write both the balanced equilibrium equation for the ionization reaction and the equilibrium equation (Equation 16.15). Use Equation 16.25 to calculate the K_a from the pK_a.

B For both the concentrated solution and the dilute solution, use a tabular format to write expressions for the final concentrations of all species in solution. Substitute these values into the equilibrium equation and solve for [C₆H₅CO₂⁻]_f for each solution.

C Use the values of [C₆H₅CO₂⁻]_f and Equation 16.41 to calculate the percent ionization.

Solution:

A If we abbreviate benzoic acid as PhCO₂H where Ph = –C₆H₅, the balanced equilibrium equation for the ionization reaction and the equilibrium equation can be written as follows:

$$\begin{array}{c}\hfill {\text{PhCO}}_2\text{H(aq)}\rightleftharpoons {\text{H}}^{+}\text{(aq)}+{\text{PhCO}}_2{}^{-}\text{(aq)}\\ \hfill K_{\text{a}}=\frac{[{\text{H}}^{+}][{\text{PhCO}}_2{}^{-}]}{[{\text{PhCO}}_2\text{H}]}\end{array}$$

From the pK_a, we have K_a = 10^−4.20 = 6.3 × 10⁻⁵.

1. B For the more concentrated solution, we set up our table of initial concentrations, changes in concentrations, and final concentrations:

   ${\text{PhCO}}_2\text{H(aq)}\rightleftharpoons {\text{H}}^{+}\text{(aq)}+{\text{PhCO}}_2{}^{-}\text{(aq)}$
   	[PhCO2H]	[H+]	[PhCO2−]
   initial	0.0500	1.00 × 10−7	0
   change	−x	+x	+x
   final	(0.0500 − x)	(1.00 × 10−7 + x)	x

   Inserting the expressions for the final concentrations into the equilibrium equation and making our usual assumptions, that [PhCO₂⁻] and [H⁺] are negligible due to the autoionization of water,

   $$\begin{array}{cccc}\hfill K_{\text{a}}& =\frac{[{\text{H}}^{+}][{\text{PhCO}}_2{}^{-}]}{[{\text{PhCO}}_2\text{H}]}\hfill & =\frac{(x)(x)}{0.0500-x}\hfill & =\frac{x^2}{0.0500}=6.3\times {10}^{-5}\hfill \\ \hfill & 1.8\times {10}^{-3}& =x\hfill & \end{array}$$

   This value is less than 5% of 0.0500, so our simplifying assumption is justified, and [PhCO₂⁻] at equilibrium is 1.8 × 10⁻³ M. We reach the same conclusion using C_HA: K_aC_HA = (6.3 × 10⁻⁵)(0.0500) = 3.2 × 10⁻⁶ > 1.0 × 10⁻⁶.

   C The percent ionized is the ratio of the concentration of PhCO₂⁻ to the analytical concentration, multiplied by 100:

   $$\text{percent ionized}=\frac{[{\text{PhCO}}_2{}^{-}]}{C_{{\text{PhCO}}_2\text{H}}}\times 100=\frac{1.8\times {10}^{-3}}{0.0500}\times 100=3.6\%$$

   Because only 3.6% of the benzoic acid molecules are ionized in a 0.0500 M solution, our simplifying assumptions are confirmed.

2. B For the more dilute solution, we proceed in exactly the same manner. Our table of concentrations is therefore as follows:

   ${\text{PhCO}}_2\text{H(aq)}\rightleftharpoons {\text{H}}^{+}\text{(aq)}+{\text{PhCO}}_2{}^{-}\text{(aq)}$
   	[PhCO2H]	[H+]	[PhCO2−]
   initial	0.00500	1.00 × 10−7	0
   change	−x	+x	+x
   final	(0.00500 − x)	(1.00 × 10−7 + x)	x

   Inserting the expressions for the final concentrations into the equilibrium equation and making our usual simplifying assumptions,

   $$\begin{array}{cccc}\hfill K_{\text{a}}& =\frac{[{\text{H}}^{+}][{\text{PhCO}}_2{}^{-}]}{[{\text{PhCO}}_2\text{H}]}\hfill & =\frac{(x)(x)}{0.00500-x}\hfill & =\frac{x^2}{0.00500}=6.3\times {10}^{-5}\hfill \\ \hfill & 5.6\times {10}^{-4}& =x\hfill & \end{array}$$

   Unfortunately, this number is greater than 10% of 0.00500, so our assumption that the fraction of benzoic acid that is ionized in this solution could be neglected and that (0.00500 − x) ≈ x is not valid. Furthermore, we see that K_aC_HA = (6.3 × 10⁻⁵)(0.00500) = 3.2 × 10⁻⁷ < 1.0 × 10⁻⁶. Thus the relevant equation is as follows:

   $$\frac{x^2}{0.00500-x}=6.3\times {10}^{-5}$$

   which must be solved using the quadratic formula. Multiplying out the quantities,

   x² = (6.3 × 10⁻⁵)(0.00500 − x) = (3.2 × 10⁻⁷) − (6.3 × 10⁻⁵)x

   Rearranging the equation to fit the standard quadratic equation format,

   x² + (6.3 × 10⁻⁵)x − (3.2 × 10⁻⁷) = 0

   This equation can be solved by using the quadratic formula:

   $$\begin{array}{cc}\hfill x& =\frac{-b\pm \sqrt{b^2-4ac}}{2a}\hfill \\ \hfill & =\frac{-(6.3\times {10}^{-5})\pm \sqrt{{(6.3\times {10}^{-5})}^2-4(1)(-3.2\times {10}^{-7})}}{2\left(1\right)}\hfill \\ \hfill & =\frac{-(6.3\times {10}^{-5})\pm (1.1\times {10}^{-3})}{2}=5.3\times {10}^{-4}\text{ or}-5.9\times {10}^{-4}\hfill \end{array}$$

   Because a negative x value corresponds to a negative [PhCO₂⁻], which is not physically meaningful, we use the positive solution: x = 5.3 × 10⁻⁴. Thus [PhCO₂⁻] = 5.3 × 10⁻⁴ M.

   C The percent ionized is therefore

   $$\text{percent ionized}=\frac{[{\text{PhCO}}_2{}^{-}]}{C_{{\text{PhCO}}_2\text{H}}}\times 100=\frac{5.3\times {10}^{-4}}{0.00500}\times 100=11\%$$

   In the more dilute solution (C = 0.00500 M), 11% of the benzoic acid molecules are ionized versus only 3.6% in the more concentrated solution (C = 0.0500 M). Decreasing the analytical concentration by a factor of 10 results in an approximately threefold increase in the percentage of benzoic acid molecules that are ionized.

Exercise

Lactic acid (CH₃CH(OH)CO₂H) is a weak acid with a pK_a of 3.86 at 25°C. What percentage of the lactic acid is ionized in each solution?

1. a 0.10 M solution
2. a 0.0020 M solution

Answer:

1. 3.7%
2. 23%

Example 10

Fish tend to spoil rapidly, even when refrigerated. The cause of the resulting “fishy” odor is a mixture of amines, particularly methylamine (CH₃NH₂), a volatile weak base (pK_b = 3.34). Fish is often served with a wedge of lemon because lemon juice contains citric acid, a triprotic acid with pK_a values of 3.13, 4.76, and 6.40 that can neutralize amines. Calculate the equilibrium constant for the reaction of excess citric acid with methylamine, assuming that only the first dissociation constant of citric acid is important.

Given: pK_b for base and pK_a for acid

Asked for: K

Strategy:

A Write the balanced equilibrium equation and the equilibrium constant expression for the reaction.

B Convert pK_a and pK_b to K_a and K_b and then use Equation 16.49 to calculate K.

Solution:

A If we abbreviate citric acid as H₃citrate, the equilibrium equation for its reaction with methylamine is as follows:

$${\text{CH}}_3{\text{NH}}_2\text{(aq)}+{\text{H}}_3\text{citrate(aq)}\rightleftharpoons {\text{CH}}_3{\text{NH}}_3{}^{+}\text{(aq)}+{\text{H}}_2{\text{citrate}}^{-}\text{(aq)}$$

The equilibrium constant expression for this reaction is as follows:

$$K=\frac{[{\text{CH}}_3{\text{NH}}_3{}^{+}][{\text{H}}_2{\text{citrate}}^{-}]}{[{\text{CH}}_3{\text{NH}}_2][{\text{H}}_3\text{citrate}]}$$

B Equation 16.49 is K = (K_aK_b)/K_w. Converting pK_a and pK_b to K_a and K_b gives K_a = 10^−3.13 = 7.4 × 10⁻⁴ for citric acid and K_b = 10^−3.34 = 4.6 × 10⁻⁴ for methylamine. Substituting these values into the equilibrium equation,

$$K=\frac{K_{\text{a}}{K}_{\text{b}}}{K_{\text{w}}}=\frac{(7.4\times {10}^{-4})(4.6\times {10}^{-4})}{1.01\times {10}^{-14}}=3.4\times {10}^7$$

The value of pK can also be calculated directly by taking the negative logarithm of both sides of Equation 16.49, which gives

pK = pK_a + pK_b − pK_w = 3.13 + 3.34 − 14.00 = −7.53

Thus K = 10^−(−7.53) = 3.4 × 10⁷, in agreement with the earlier value. In either case, the K values show that the reaction of citric acid with the volatile, foul-smelling methylamine lies very far to the right, favoring the formation of a much less volatile salt with no odor. This is one reason a little lemon juice helps make less-than-fresh fish more appetizing.

Exercise

Dilute aqueous ammonia solution, often used as a cleaning agent, is also effective as a deodorizing agent. To see why, calculate the equilibrium constant for the reaction of aqueous ammonia with butyric acid (CH₃CH₂CH₂CO₂H), a particularly foul-smelling substance associated with the odor of rancid butter and smelly socks. The pK_b of ammonia is 4.75, and the pK_a of butyric acid is 4.83.

Answer: 2.6 × 10⁴