Example 14

In Section 16.4 "Quantitative Aspects of Acid–Base Equilibriums", we calculated that a 0.150 M solution of formic acid at 25°C (pK_a = 3.75) has a pH of 2.28 and is 3.5% ionized.

1. Is there a change to the pH of the solution if enough solid sodium formate is added to make the final formate concentration 0.100 M (assume that the formic acid concentration does not change)?
2. What percentage of the formic acid is ionized if 0.200 M HCl is added to the system?

Given: solution concentration and pH, pK_a, and percent ionization of acid; final concentration of conjugate base or strong acid added

Asked for: pH and percent ionization of formic acid

Strategy:

A Write a balanced equilibrium equation for the ionization equilibrium of formic acid. Tabulate the initial concentrations, the changes, and the final concentrations.

B Substitute the expressions for the final concentrations into the expression for K_a. Calculate [H⁺] and the pH of the solution.

C Construct a table of concentrations for the dissociation of formic acid. To determine the percent ionization, determine the anion concentration, divide it by the initial concentration of formic acid, and multiply the result by 100.

Solution:

1. A Because sodium formate is a strong electrolyte, it ionizes completely in solution to give formate and sodium ions. The Na⁺ ions are spectator ions, so they can be ignored in the equilibrium equation. Because water is both a much weaker acid than formic acid and a much weaker base than formate, the acid–base properties of the solution are determined solely by the formic acid ionization equilibrium:

   $${\text{HCO}}_2\text{H(aq)}\rightleftharpoons {\text{HCO}}_2{}^{-}\text{(aq)}+{\text{H}}^{+}\text{(aq)}$$

   The initial concentrations, the changes in concentration that occur as equilibrium is reached, and the final concentrations can be tabulated.

   ${\text{HCO}}_2\text{H(aq)}\rightleftharpoons {\text{H}}^{+}\text{(aq)}+{\text{HCO}}_2{}^{-}\text{(aq)}$
   	[HCO2H]	[H+]	[HCO2−]
   initial	0.150	1.00 × 10−7	0.100
   change	−x	+x	+x
   final	(0.150 − x)	x	(0.100 + x)

   B We substitute the expressions for the final concentrations into the equilibrium constant expression and make our usual simplifying assumptions, so

   $$K_{\text{a}}=\frac{[{\text{H}}^{+}][{\text{HCO}}_2{}^{-}]}{[{\text{HCO}}_2\text{H}]}=\frac{(x)(0.100+x)}{0.150-x}=\frac{x(0.100)}{0.150}={10}^{-3.75}=1.8\times {10}^{-4}$$

   Rearranging and solving for x,

   $$x=(1.8\times {10}^{-4})\times \frac{0.150\text{ M}}{0.100\text{ M}}=2.7\times {10}^{-4}=[{\text{H}}^{+}]$$

   The value of x is small compared with 0.150 or 0.100 M, so our assumption about the extent of ionization is justified. Moreover, K_aC_HA = (1.8 × 10⁻⁴)(0.150) = 2.7 × 10⁻⁵, which is greater than 1.0 × 10⁻⁶, so again, our assumption is justified. The final pH is −log(2.7 × 10⁻⁴) = 3.57, compared with the initial value of 2.29. Thus adding a salt containing the conjugate base of the acid has increased the pH of the solution, as we expect based on Le Châtelier’s principle; the stress on the system has been relieved by the consumption of H⁺ ions, driving the equilibrium to the left.

2. C Because HCl is a strong acid, it ionizes completely, and chloride is a spectator ion that can be neglected. Thus the only relevant acid–base equilibrium is again the dissociation of formic acid, and initially the concentration of formate is zero. We can construct a table of initial concentrations, changes in concentration, and final concentrations.

   ${\text{HCO}}_2\text{H(aq)}\rightleftharpoons {\text{H}}^{+}\text{(aq)}+{\text{HCO}}_2{}^{-}\text{(aq)}$
   	[HCO2H]	[H+]	[HCO2−]
   initial	0.150	0.200	0
   change	−x	+x	+x
   final	(0.150 − x)	(0.200 + x)	x

   To calculate the percentage of formic acid that is ionized under these conditions, we have to determine the final [HCO₂⁻]. We substitute final concentrations into the equilibrium constant expression and make the usual simplifying assumptions, so

   $$K_{\text{a}}=\frac{[{\text{H}}^{+}][{\text{HCO}}_2{}^{-}]}{[{\text{HCO}}_2\text{H}]}=\frac{(0.200+x)(x)}{0.150-x}=\frac{x(0.200)}{0.150}=1.80\times {10}^{-4}$$

   Rearranging and solving for x,

   $$x=(1.80\times {10}^{-4})\times \frac{0.150\text{ M}}{0.200\text{ M}}=1.35\times {10}^{-4}=[{\text{HCO}}_2{}^{-}]$$

   Once again, our simplifying assumptions are justified. The percent ionization of formic acid is as follows:

   $$\text{percent ionization}=\frac{1.35\times {10}^{-4}\text{ M}}{0.150\text{ M}}\times 100=0.0900\%$$

   Adding the strong acid to the solution, as shown in the table, decreased the percent ionization of formic acid by a factor of approximately 38 (3.45%/0.0900%). Again, this is consistent with Le Châtelier’s principle: adding H⁺ ions drives the dissociation equlibrium to the left.

Exercise

As you learned in Example 8, a 0.225 M solution of ethylamine (CH₃CH₂NH₂, pK_b = 3.19) has a pH of 12.08 and a percent ionization of 5.4% at 20°C. Calculate the following:

1. the pH of the solution if enough solid ethylamine hydrochloride (EtNH₃Cl) is added to make the solution 0.100 M in EtNH₃⁺
2. the percentage of ethylamine that is ionized if enough solid NaOH is added to the original solution to give a final concentration of 0.050 M NaOH

Answer:

1. 11.16
2. 1.3%

Example 15

What is the pH of a solution that contains

1. 0.135 M HCO₂H and 0.215 M HCO₂Na? (The pK_a of formic acid is 3.75.)
2. 0.0135 M HCO₂H and 0.0215 M HCO₂Na?
3. 0.119 M pyridine and 0.234 M pyridine hydrochloride? (The pK_b of pyridine is 8.77.)

Given: concentration of acid, conjugate base, and pK_a; concentration of base, conjugate acid, and pK_b

Asked for: pH

Strategy:

Substitute values into either form of the Henderson-Hasselbalch equation (Equation 16.61 or Equation 16.62) to calculate the pH.

Solution:

1. According to the Henderson-Hasselbalch equation, the pH of a solution that contains both a weak acid and its conjugate base is pH = pK_a + log([A⁻]/[HA]). Inserting the given values into the equation,

   $$\text{pH}=3.75+\log \left(\frac{0.215}{0.135}\right)=3.75+\log 1.593=3.95$$

   This result makes sense because the [A⁻]/[HA] ratio is between 1 and 10, so the pH of the buffer must be between the pK_a (3.75) and pK_a + 1, or 4.75.

2. This is identical to part (a), except for the concentrations of the acid and the conjugate base, which are 10 times lower. Inserting the concentrations into the Henderson-Hasselbalch equation,

   $$\text{pH}=3.75+\log \left(\frac{0.0215}{0.0135}\right)=3.75+\log 1.593=3.95$$

   This result is identical to the result in part (a), which emphasizes the point that the pH of a buffer depends only on the ratio of the concentrations of the conjugate base and the acid, not on the magnitude of the concentrations. Because the [A⁻]/[HA] ratio is the same as in part (a), the pH of the buffer must also be the same (3.95).

3. In this case, we have a weak base, pyridine (Py), and its conjugate acid, the pyridinium ion (HPy⁺). We will therefore use Equation 16.62, the more general form of the Henderson-Hasselbalch equation, in which “base” and “acid” refer to the appropriate species of the conjugate acid–base pair. We are given [base] = [Py] = 0.119 M and [acid] = [HPy⁺] = 0.234 M. We also are given pK_b = 8.77 for pyridine, but we need pK_a for the pyridinium ion. Recall from Equation 16.23 that the pK_b of a weak base and the pK_a of its conjugate acid are related: pK_a + pK_b = pK_w. Thus pK_a for the pyridinium ion is pK_w − pK_b = 14.00 − 8.77 = 5.23. Substituting this pK_a value into the Henderson-Hasselbalch equation,

   $$\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{[\text{base}]}{\text{[acid]}}\right)=5.23+\log \left(\frac{0.119}{0.234}\right)=5.23-0.294=4.94$$

   Once again, this result makes sense: the [B]/[BH⁺] ratio is about 1/2, which is between 1 and 0.1, so the final pH must be between the pK_a (5.23) and pK_a − 1, or 4.23.

Exercise

What is the pH of a solution that contains

1. 0.333 M benzoic acid and 0.252 M sodium benzoate?
2. 0.050 M trimethylamine and 0.066 M trimethylamine hydrochloride?

The pK_a of benzoic acid is 4.20, and the pK_b of trimethylamine is also 4.20.

Answer:

1. 4.08
2. 9.68

Example 16

The buffer solution in Example 15 contained 0.135 M HCO₂H and 0.215 M HCO₂Na and had a pH of 3.95.

1. What is the final pH if 5.00 mL of 1.00 M HCl are added to 100 mL of this solution?
2. What is the final pH if 5.00 mL of 1.00 M NaOH are added?

Given: composition and pH of buffer; concentration and volume of added acid or base

Asked for: final pH

Strategy:

A Calculate the amounts of formic acid and formate present in the buffer solution using the procedure from Example 14. Then calculate the amount of acid or base added.

B Construct a table showing the amounts of all species after the neutralization reaction. Use the final volume of the solution to calculate the concentrations of all species. Finally, substitute the appropriate values into the Henderson-Hasselbalch equation (Equation 16.62) to obtain the pH.

Solution:

The added HCl (a strong acid) or NaOH (a strong base) will react completely with formate (a weak base) or formic acid (a weak acid), respectively, to give formic acid or formate and water. We must therefore calculate the amounts of formic acid and formate present after the neutralization reaction.

1. A We begin by calculating the millimoles of formic acid and formate present in 100 mL of the initial pH 3.95 buffer:

   $$100\cancel{\text{mL}}\left(\frac{0.135{\text{ mmol HCO}}_2\text{H}}{\cancel{\text{mL}}}\right)=13.5{\text{ mmol HCO}}_2\text{H}$$ $$100\cancel{\text{mL}}\left(\frac{0.215{\text{ mmol HCO}}_2{}^{-}}{\cancel{\text{mL}}}\right)=21.5{\text{ mmol HCO}}_2{}^{-}$$

   The millimoles of H⁺ in 5.00 mL of 1.00 M HCl is as follows:

   $$5.00\cancel{\text{mL}}\left(\frac{1.00{\text{ mmol H}}^{+}}{\cancel{\text{mL}}}\right)=5.00{\text{ mmol H}}^{+}$$

   B Next, we construct a table of initial amounts, changes in amounts, and final amounts:

   HCO2−(aq) + H+(aq) → HCO2H(aq)
   	[HCO2−]	[H+]	[HCO2H]
   initial	21.5 mmol	5.00 mmol	13.5 mmol
   change	−5.00 mmol	−5.00 mmol	+5.00 mmol
   final	16.5 mmol	∼0 mmol	18.5 mmol

   The final amount of H⁺ in solution is given as “∼0 mmol.” For the purposes of the stoichiometry calculation, this is essentially true, but remember that the point of the problem is to calculate the final [H⁺] and thus the pH. We now have all the information we need to calculate the pH. We can use either the lengthy procedure of Example 14 or the Henderson–Hasselbach equation. Because we have performed many equilibrium calculations in this chapter, we’ll take the latter approach. The Henderson-Hasselbalch equation requires the concentrations of HCO₂⁻ and HCO₂H, which can be calculated using the number of millimoles (n) of each and the total volume (V_T). Substituting these values into the Henderson-Hasselbalch equation,

   $$\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{[{\text{HCO}}_2{}^{-}]}{[{\text{HCO}}_2\text{H}]}\right)=\text{p}{K}_{\text{a}}+\log \left(\frac{n_{{\text{HCO}}_2{}^{-}}\text{/}{V}_{\text{f}}}{n_{{\text{HCO}}_2\text{H}}\text{/}{V}_{\text{f}}}\right)=\text{p}{K}_{\text{a}}+\log \left(\frac{n_{{\text{HCO}}_2{}^{-}}}{n_{{\text{HCO}}_2\text{H}}}\right)$$

   Because the total volume appears in both the numerator and denominator, it cancels. We therefore need to use only the ratio of the number of millimoles of the conjugate base to the number of millimoles of the weak acid. So

   $$\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{n_{{\text{HCO}}_2{}^{-}}}{n_{{\text{HCO}}_2\text{H}}}\right)=3.75+\log \left(\frac{16.5\text{ mmol}}{18.5\text{ mmol}}\right)=3.75-0.050=3.70$$

   Once again, this result makes sense on two levels. First, the addition of HCl has decreased the pH from 3.95, as expected. Second, the ratio of HCO₂⁻ to HCO₂H is slightly less than 1, so the pH should be between the pK_a and pK_a − 1.

2. A The procedure for solving this part of the problem is exactly the same as that used in part (a). We have already calculated the numbers of millimoles of formic acid and formate in 100 mL of the initial pH 3.95 buffer: 13.5 mmol of HCO₂H and 21.5 mmol of HCO₂⁻. The number of millimoles of OH⁻ in 5.00 mL of 1.00 M NaOH is as follows:

   $$5.00\cancel{\text{mL}}\left(\frac{1.00{\text{ mmol OH}}^{-}}{\cancel{\text{mL}}}\right)=5.00{\text{ mmol OH}}^{-}$$

   B With this information, we can construct a table of initial amounts, changes in amounts, and final amounts.

   HCO2H(aq) + OH−(aq) → HCO2−(aq) + H2O(l)
   	[HCO2H]	[OH−]	[HCO2−]
   initial	13.5 mmol	5.00 mmol	21.5 mmol
   change	−5.00 mmol	−5.00 mmol	+5.00 mmol
   final	8.5 mmol	∼0 mmol	26.5 mmol

   The final amount of OH⁻ in solution is not actually zero; this is only approximately true based on the stoichiometric calculation. We can calculate the final pH by inserting the numbers of millimoles of both HCO₂⁻ and HCO₂H into the simplified Henderson-Hasselbalch expression used in part (a) because the volume cancels:

   $$\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{n_{{\text{HCO}}_2{}^{-}}}{n_{{\text{HCO}}_2\text{H}}}\right)=3.75+\log \left(\frac{26.5\text{ mmol}}{8.5\text{ mmol}}\right)=3.75+0.494=4.24$$

   Once again, this result makes chemical sense: the pH has increased, as would be expected after adding a strong base, and the final pH is between the pK_a and pK_a + 1, as expected for a solution with a HCO₂⁻/HCO₂H ratio between 1 and 10.

Exercise

The buffer solution from Example 15 contained 0.119 M pyridine and 0.234 M pyridine hydrochloride and had a pH of 4.94.

1. What is the final pH if 12.0 mL of 1.5 M NaOH are added to 250 mL of this solution?
2. What is the final pH if 12.0 mL of 1.5 M HCl are added?

Answer:

1. 5.30
2. 4.42