Titrations of Strong Acids and Bases

Part (a) of Figure 16.17 "Solution pH as a Function of the Volume of a Strong Acid or a Strong Base Added to Distilled Water" shows a plot of the pH as 0.20 M HCl is gradually added to 50.00 mL of pure water. The pH of the sample in the flask is initially 7.00 (as expected for pure water), but it drops very rapidly as HCl is added. Eventually the pH becomes constant at 0.70—a point well beyond its value of 1.00 with the addition of 50.0 mL of HCl (0.70 is the pH of 0.20 M HCl). In contrast, when 0.20 M NaOH is added to 50.00 mL of distilled water, the pH (initially 7.00) climbs very rapidly at first but then more gradually, eventually approaching a limit of 13.30 (the pH of 0.20 M NaOH), again well beyond its value of 13.00 with the addition of 50.0 mL of NaOH as shown in part (b) in Figure 16.17 "Solution pH as a Function of the Volume of a Strong Acid or a Strong Base Added to Distilled Water". As you can see from these plots, the titration curve for adding a base is the mirror image of the curve for adding an acid.

Figure 16.17 Solution pH as a Function of the Volume of a Strong Acid or a Strong Base Added to Distilled Water

(a) When 0.20 M HCl is added to 50.0 mL of distilled water, the pH rapidly decreases until it reaches a minimum at the pH of 0.20 M HCl. (b) Conversely, when 0.20 M NaOH is added to 50.0 mL of distilled water, the pH rapidly increases until it reaches a maximum at the pH of 0.20 M NaOH.

Suppose that we now add 0.20 M NaOH to 50.0 mL of a 0.10 M solution of HCl. Because HCl is a strong acid that is completely ionized in water, the initial [H⁺] is 0.10 M, and the initial pH is 1.00. Adding NaOH decreases the concentration of H⁺ because of the neutralization reaction: $({\text{OH}}^{-}+{\text{H}}^{+}\rightleftharpoons {\text{H}}_2\text{O})$ (in part (a) in Figure 16.18 "The Titration of (a) a Strong Acid with a Strong Base and (b) a Strong Base with a Strong Acid"). Thus the pH of the solution increases gradually. Near the equivalence pointThe point in a titration where a stoichiometric amount of the titrant has been added., however, the point at which the number of moles of base (or acid) added equals the number of moles of acid (or base) originally present in the solution, the pH increases much more rapidly because most of the H⁺ ions originally present have been consumed. (For more information on titrations and the equivalence point, see Chapter 4 "Reactions in Aqueous Solution", Section 4.9 "Quantitative Analysis Using Titrations".) For the titration of a monoprotic strong acid (HCl) with a monobasic strong base (NaOH), we can calculate the volume of base needed to reach the equivalence point from the following relationship:

If 0.20 M NaOH is added to 50.0 mL of a 0.10 M solution of HCl, we solve for V_b:

Figure 16.18 The Titration of (a) a Strong Acid with a Strong Base and (b) a Strong Base with a Strong Acid

(a) As 0.20 M NaOH is slowly added to 50.0 mL of 0.10 M HCl, the pH increases slowly at first, then increases very rapidly as the equivalence point is approached, and finally increases slowly once more. (b) Conversely, as 0.20 M HCl is slowly added to 50.0 mL of 0.10 M NaOH, the pH decreases slowly at first, then decreases very rapidly as the equivalence point is approached, and finally decreases slowly once more.

At the equivalence point (when 25.0 mL of NaOH solution has been added), the neutralization is complete: only a salt remains in solution (NaCl), and the pH of the solution is 7.00. Adding more NaOH produces a rapid increase in pH, but eventually the pH levels off at a value of about 13.30, the pH of 0.20 M NaOH.

As shown in part (b) in Figure 16.18 "The Titration of (a) a Strong Acid with a Strong Base and (b) a Strong Base with a Strong Acid", the titration of 50.0 mL of a 0.10 M solution of NaOH with 0.20 M HCl produces a titration curve that is nearly the mirror image of the titration curve in part (a) in Figure 16.18 "The Titration of (a) a Strong Acid with a Strong Base and (b) a Strong Base with a Strong Acid". The pH is initially 13.00, and it slowly decreases as HCl is added. As the equivalence point is approached, the pH drops rapidly before leveling off at a value of about 0.70, the pH of 0.20 M HCl.

The titration of either a strong acid with a strong base or a strong base with a strong acid produces an S-shaped curve. The curve is somewhat asymmetrical because the steady increase in the volume of the solution during the titration causes the solution to become more dilute. Due to the leveling effect, the shape of the curve for a titration involving a strong acid and a strong base depends on only the concentrations of the acid and base, not their identities.

Titrations of Weak Acids and Bases

In contrast to strong acids and bases, the shape of the titration curve for a weak acid or a weak base depends dramatically on the identity of the acid or the base and the corresponding K_a or K_b. As we shall see, the pH also changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. As you learned in Section 16.4 "Quantitative Aspects of Acid–Base Equilibriums", [H⁺] of a solution of a weak acid (HA) is not equal to the concentration of the acid but depends on both its pK_a and its concentration. Because only a fraction of a weak acid dissociates, [H⁺] is less than [HA]. Thus the pH of a solution of a weak acid is greater than the pH of a solution of a strong acid of the same concentration. Part (a) in Figure 16.19 "The Titration of (a) a Weak Acid with a Strong Base and (b) a Weak Base with a Strong Acid" shows the titration curve for 50.0 mL of a 0.100 M solution of acetic acid with 0.200 M NaOH superimposed on the curve for the titration of 0.100 M HCl shown in part (a) in Figure 16.18 "The Titration of (a) a Strong Acid with a Strong Base and (b) a Strong Base with a Strong Acid". Below the equivalence point, the two curves are very different. Before any base is added, the pH of the acetic acid solution is greater than the pH of the HCl solution, and the pH changes more rapidly during the first part of the titration. Note also that the pH of the acetic acid solution at the equivalence point is greater than 7.00. That is, at the equivalence point, the solution is basic. In addition, the change in pH around the equivalence point is only about half as large as for the HCl titration; the magnitude of the pH change at the equivalence point depends on the pK_a of the acid being titrated. Above the equivalence point, however, the two curves are identical. Once the acid has been neutralized, the pH of the solution is controlled only by the amount of excess NaOH present, regardless of whether the acid is weak or strong.

The titration curve in part (a) in Figure 16.19 "The Titration of (a) a Weak Acid with a Strong Base and (b) a Weak Base with a Strong Acid" was created by calculating the starting pH of the acetic acid solution before any NaOH is added and then calculating the pH of the solution after adding increasing volumes of NaOH. The procedure is illustrated in the following subsection and Example 12 for three points on the titration curve, using the pK_a of acetic acid (4.76 at 25°C; K_a = 1.7 × 10⁻⁵).

Figure 16.19 The Titration of (a) a Weak Acid with a Strong Base and (b) a Weak Base with a Strong Acid

(a) As 0.200 M NaOH is slowly added to 50.0 mL of 0.100 M acetic acid, the pH increases slowly at first, then increases rapidly as the equivalence point is approached, and then again increases more slowly. The corresponding curve for the titration of 50.0 mL of 0.100 M HCl with 0.200 M NaOH is shown as a dashed line. (b) As 0.200 M HCl is slowly added to 50.0 mL of 0.100 M NH₃, the pH decreases slowly at first, then decreases rapidly as the equivalence point is approached, and then again decreases more slowly. The corresponding curve for the titration of 50.0 mL of 0.100 M NaOH with 0.200 M HCl is shown as a dashed line.

Calculating the pH during the Titration of a Weak Acid or a Weak Base

Now consider what happens when we add 5.00 mL of 0.200 M NaOH to 50.00 mL of 0.100 M CH₃CO₂H (part (a) in Figure 16.19 "The Titration of (a) a Weak Acid with a Strong Base and (b) a Weak Base with a Strong Acid"). Because the neutralization reaction proceeds to completion, all of the OH⁻ ions added will react with the acetic acid to generate acetate ion and water:

Equation 16.51

CH₃CO₂H(aq) + OH⁻(aq) → CH₃CO₂⁻(aq) + H₂O(l)

All problems of this type must be solved in two steps: a stoichiometric calculation followed by an equilibrium calculation. In the first step, we use the stoichiometry of the neutralization reaction to calculate the amounts of acid and conjugate base present in solution after the neutralization reaction has occurred. In the second step, we use the equilibrium equation (Equation 16.15) to determine [H⁺] of the resulting solution.

Step 1: To determine the amount of acid and conjugate base in solution after the neutralization reaction, we calculate the amount of CH₃CO₂H in the original solution and the amount of OH⁻ in the NaOH solution that was added. The acetic acid solution contained

$$50.00\cancel{\text{mL}}\left(\frac{0.100{\text{ mmol CH}}_3{\text{CO}}_2\text{H}}{\cancel{\text{mL}}}\right)=5.00{\text{ mmol CH}}_3{\text{CO}}_2\text{H}$$

The NaOH solution contained

$$5.00\cancel{\text{mL}}\left(\frac{0.200\text{ mmol NaOH}}{\cancel{\text{mL}}}\right)=1.00\text{ mmol NaOH}$$

Comparing the amounts shows that CH₃CO₂H is in excess. Because OH⁻ reacts with CH₃CO₂H in a 1:1 stoichiometry, the amount of excess CH₃CO₂H is as follows:

5.00 mmol CH₃CO₂H − 1.00 mmol OH⁻ = 4.00 mmol CH₃CO₂H

Each 1 mmol of OH⁻ reacts to produce 1 mmol of acetate ion, so the final amount of CH₃CO₂⁻ is 1.00 mmol.

The stoichiometry of the reaction is summarized in the following table, which shows the numbers of moles of the various species, not their concentrations.

${\text{CH}}_3{\text{CO}}_2\text{H(aq)}+{\text{OH}}^{-}\text{(aq)}\to {\text{CH}}_3{\text{CO}}_2{}^{-}\text{(aq)}+{\text{H}}_2\text{O(l)}$
	[CH3CO2H]	[OH−]	[CH3CO2−]
initial	5.00 mmol	1.00 mmol	0 mmol
change	−1.00 mmol	−1.00 mmol	+1.00 mmol
final	4.00 mmol	0 mmol	1.00 mmol

This table gives the initial amount of acetate and the final amount of OH⁻ ions as 0. Because an aqueous solution of acetic acid always contains at least a small amount of acetate ion in equilibrium with acetic acid, however, the initial acetate concentration is not actually 0. The value can be ignored in this calculation because the amount of CH₃CO₂⁻ in equilibrium is insignificant compared to the amount of OH⁻ added. Moreover, due to the autoionization of water, no aqueous solution can contain 0 mmol of OH⁻, but the amount of OH⁻ due to the autoionization of water is insignificant compared to the amount of OH⁻ added. We use the initial amounts of the reactants to determine the stoichiometry of the reaction and defer a consideration of the equilibrium until the second half of the problem.

Step 2: To calculate [H⁺] at equilibrium following the addition of NaOH, we must first calculate [CH₃CO₂H] and [CH₃CO₂⁻] using the number of millimoles of each and the total volume of the solution at this point in the titration:

$$\begin{array}{cc}\hfill \text{final volume}& =50.00\text{ mL}+5.00\text{ mL}=55.00\text{ mL}\hfill \\ \hfill [{\text{CH}}_3{\text{CO}}_2\text{H}]& =\frac{4.00{\text{ mmol CH}}_3{\text{CO}}_2\text{H}}{55.00\text{ mL}}=7.27\times {10}^{-2}\text{ M}\hfill \\ \hfill [{\text{CH}}_3{\text{CO}}_2{}^{-}]& =\frac{1.00{\text{ mmol CH}}_3{\text{CO}}_2{}^{-}}{55.00\text{ mL}}=1.82\times {10}^{-2}\text{ M}\hfill \end{array}$$

Knowing the concentrations of acetic acid and acetate ion at equilibrium and K_a for acetic acid (1.74 × 10⁻⁵), we can use Equation 16.15 to calculate [H⁺] at equilibrium:

$$K_{\text{a}}=\frac{[{\text{CH}}_3{\text{CO}}_2{}^{-}][{\text{H}}^{+}]}{[{\text{CH}}_3{\text{CO}}_2\text{H}]}$$ $$[{\text{H}}^{+}]=\frac{K_{\text{a}}[{\text{CH}}_3{\text{CO}}_2\text{H}]}{[{\text{CH}}_3{\text{CO}}_2{}^{-}]}=\frac{(1.74\times {10}^{-5})(7.27\times {10}^{-2}\text{ M})}{1.82\times {10}^{-2}}=6.95\times {10}^{-5}\text{ M}$$

Calculating −log[H⁺] gives pH = −log(6.95 × 10⁻⁵) = 4.158.

Comparing the titration curves for HCl and acetic acid in part (a) in Figure 16.19 "The Titration of (a) a Weak Acid with a Strong Base and (b) a Weak Base with a Strong Acid", we see that adding the same amount (5.00 mL) of 0.200 M NaOH to 50 mL of a 0.100 M solution of both acids causes a much smaller pH change for HCl (from 1.00 to 1.14) than for acetic acid (2.88 to 4.16). This is consistent with the qualitative description of the shapes of the titration curves at the beginning of this section. In Example 12, we calculate another point for constructing the titration curve of acetic acid.

Example 12

What is the pH of the solution after 25.00 mL of 0.200 M NaOH is added to 50.00 mL of 0.100 M acetic acid?

Given: volume and molarity of base and acid

Asked for: pH

Strategy:

A Write the balanced chemical equation for the reaction. Then calculate the initial numbers of millimoles of OH⁻ and CH₃CO₂H. Determine which species, if either, is present in excess.

B Tabulate the results showing initial numbers, changes, and final numbers of millimoles.

C If excess acetate is present after the reaction with OH⁻, write the equation for the reaction of acetate with water. Use a tabular format to obtain the concentrations of all the species present.

D Calculate K_b using the relationship K_w = K_aK_b (Equation 16.23). Calculate [OH⁻] and use this to calculate the pH of the solution.

Solution:

A Ignoring the spectator ion (Na⁺), the equation for this reaction is as follows:

CH₃CO₂H(aq) + OH⁻(aq) → CH₃CO₂⁻(aq) + H₂O(l)

The initial numbers of millimoles of OH⁻ and CH₃CO₂H are as follows:

$$25.00\cancel{\text{mL}}\left(\frac{0.200{\text{ mmol OH}}^{-}}{\cancel{\text{mL}}}\right)=5.00{\text{ mmol OH}}^{-}$$ $$50.00\cancel{\text{mL}}\left(\frac{0.100{\text{ CH}}_3{\text{CO}}_2\text{H}}{\cancel{\text{mL}}}\right)=5.00{\text{ mmol CH}}_3{\text{CO}}_2\text{H}$$

The number of millimoles of OH⁻ equals the number of millimoles of CH₃CO₂H, so neither species is present in excess.

B Because the number of millimoles of OH⁻ added corresponds to the number of millimoles of acetic acid in solution, this is the equivalence point. The results of the neutralization reaction can be summarized in tabular form.

${\text{CH}}_3{\text{CO}}_2\text{H(aq)}+{\text{OH}}^{-}\text{(aq)}\to {\text{CH}}_3{\text{CO}}_2{}^{-}\text{(aq)}+{\text{H}}_2\text{O(l)}$
	[CH3CO2H]	[OH−]	[CH3CO2−]
initial	5.00 mmol	5.00 mmol	0 mmol
change	−5.00 mmol	−5.00 mmol	+5.00 mmol
final	0 mmol	0 mmol	5.00 mmol

C Because the product of the neutralization reaction is a weak base, we must consider the reaction of the weak base with water to calculate [H⁺] at equilibrium and thus the final pH of the solution. The initial concentration of acetate is obtained from the neutralization reaction:

$$[{\text{CH}}_3{\text{CO}}_2]=\frac{5.00{\text{ mmol CH}}_3{\text{CO}}_2}{(50.00+25.00)\text{ mL}}=6.67\times {10}^{-2}\text{ M}$$

The equilibrium reaction of acetate with water is as follows:

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

The equilibrium constant for this reaction is K_b = K_w/K_a, where K_a is the acid ionization constant of acetic acid. We therefore define x as [OH⁻] produced by the reaction of acetate with water. Here is the completed table of concentrations:

${\text{H}}_2\text{O(l)}+{\text{CH}}_3{\text{CO}}_2{}^{-}\text{(aq)}\rightleftharpoons {\text{CH}}_3{\text{CO}}_2\text{H(aq)}+{\text{OH}}^{-}\text{(aq)}$
	[CH3CO2−]	[CH3CO2H]	[OH−]
initial	0.0667	0	1.00 × 10−7
change	−x	+x	+x
final	(0.0667 − x)	x	x

D Substituting the expressions for the final values from this table into Equation 16.18,

$$K_{\text{b}}=\frac{[{\text{CH}}_3{\text{CO}}_2\text{H}][{\text{OH}}^{-}]}{[{\text{CH}}_3{\text{CO}}_2{}^{-}]}=\frac{(x)(x)}{0.0667-x}\approx \frac{x^2}{0.0667}$$

We can obtain K_b by rearranging Equation 16.23 and substituting the known values:

$$K_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a}}}=\frac{1.01\times {10}^{-14}}{1.74\times {10}^{-5}}=5.80\times {10}^{-10}=\frac{x^2}{0.0667}$$

which we can solve to get x = 6.22 × 10⁻⁶. Thus [OH⁻] = 6.22 × 10⁻⁶ M, and the pH of the final solution is 8.794 (part (a) in Figure 16.19 "The Titration of (a) a Weak Acid with a Strong Base and (b) a Weak Base with a Strong Acid"). As expected for the titration of a weak acid, the pH at the equivalence point is greater than 7.00 because the product of the titration is a base, the acetate ion, which then reacts with water to produce OH⁻.

Exercise

Calculate the pH of a solution prepared by adding 45.0 mL of a 0.213 M HCl solution to 125.0 mL of a 0.150 M solution of ammonia. The pK_b of ammonia is 4.75 at 25°C.

Answer: 9.23

As shown in part (b) in Figure 16.19 "The Titration of (a) a Weak Acid with a Strong Base and (b) a Weak Base with a Strong Acid", the titration curve for NH₃, a weak base, is the reverse of the titration curve for acetic acid. In particular, the pH at the equivalence point in the titration of a weak base is less than 7.00 because the titration produces an acid.

The identity of the weak acid or weak base being titrated strongly affects the shape of the titration curve. Figure 16.20 "Effect of Acid or Base Strength on the Shape of Titration Curves" illustrates the shape of titration curves as a function of the pK_a or the pK_b. As the acid or the base being titrated becomes weaker (its pK_a or pK_b becomes larger), the pH change around the equivalence point decreases significantly. With very dilute solutions, the curve becomes so shallow that it can no longer be used to determine the equivalence point.

Figure 16.20 Effect of Acid or Base Strength on the Shape of Titration Curves

Unlike strong acids or bases, the shape of the titration curve for a weak acid or base depends on the pK_a or pK_b of the weak acid or base being titrated. (a) Solution pH as a function of the volume of 1.00 M NaOH added to 10.00 mL of 1.00 M solutions of weak acids with the indicated pK_a values. (b) Solution pH as a function of the volume of 1.00 M HCl added to 10.00 mL of 1.00 M solutions of weak bases with the indicated pK_b values. The shapes of the two sets of curves are essentially identical, but one is flipped vertically in relation to the other. Midpoints are indicated for the titration curves corresponding to pK_a = 10 and pK_b = 10.

One point in the titration of a weak acid or a weak base is particularly important: the midpointThe point in an acid–base titration at which exactly enough acid (or base) has been added to neutralize one-half of the base (or the acid) originally present: $\left[\text{HA}\right]=\left[{\text{A}}^{-}\right]\text{.}$ of a titration is defined as the point at which exactly enough acid (or base) has been added to neutralize one-half of the acid (or the base) originally present and occurs halfway to the equivalence point. The midpoint is indicated in part (a) in Figure 16.20 "Effect of Acid or Base Strength on the Shape of Titration Curves" and part (b) in Figure 16.20 "Effect of Acid or Base Strength on the Shape of Titration Curves" for the two shallowest curves. By definition, at the midpoint of the titration of an acid, [HA] = [A⁻]. Recall from Equation 16.15 that the ionization constant for a weak acid is as follows:

$$K_{\text{a}}=\frac{[{\text{H}}_3{\text{O}}^{+}][{\text{A}}^{-}]}{\text{[HA]}}$$

If [HA] = [A⁻], this reduces to K_a = [H₃O⁺]. Taking the negative logarithm of both sides,

−log K_a = −log[H₃O⁺]

From the definitions of pK_a and pH, we see that this is identical to

Equation 16.52

pK_a = pH

Thus the pH at the midpoint of the titration of a weak acid is equal to the pK_a of the weak acid, as indicated in part (a) in Figure 16.20 "Effect of Acid or Base Strength on the Shape of Titration Curves" for the weakest acid where we see that the midpoint for pK_a = 10 occurs at pH = 10. Titration methods can therefore be used to determine both the concentration and the pK_a (or the pK_b) of a weak acid (or a weak base).

Note the Pattern

The pH at the midpoint of the titration of a weak acid is equal to the pK_a of the weak acid.

Titrations of Polyprotic Acids or Bases

When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. The most acidic group is titrated first, followed by the next most acidic, and so forth. If the pK_a values are separated by at least three pK_a units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton. A titration of the triprotic acid H₃PO₄ with NaOH is illustrated in Figure 16.21 "Titration Curve for Phosphoric Acid (H" and shows two well-defined steps: the first midpoint corresponds to pK_a1, and the second midpoint corresponds to pK_a2. Because HPO₄²⁻ is such a weak acid, pK_a3 has such a high value that the third step cannot be resolved using 0.100 M NaOH as the titrant.

Figure 16.21 Titration Curve for Phosphoric Acid (H₃PO₄), a Typical Polyprotic Acid

The curve for the titration of 25.0 mL of a 0.100 M H₃PO₄ solution with 0.100 M NaOH along with the species in solution at each K_a is shown. Note the two distinct equivalence points corresponding to deprotonation of H₃PO₄ at pH ≈ 4.6 and H₂PO₄²⁻ at pH ≈ 9.8. Because HPO₄²⁻ is a very weak acid, the third equivalence point, at pH ≈ 13, is not well defined.

The titration curve for the reaction of a polyprotic base with a strong acid is the mirror image of the curve shown in Figure 16.21 "Titration Curve for Phosphoric Acid (H". The initial pH is high, but as acid is added, the pH decreases in steps if the successive pK_b values are well separated. Table 16.4 "Values of p" lists the ionization constants and pK_a values for some common polyprotic acids and bases.

Example 13

Calculate the pH of a solution prepared by adding 55.0 mL of a 0.120 M NaOH solution to 100.0 mL of a 0.0510 M solution of oxalic acid (HO₂CCO₂H), a diprotic acid (abbreviated as H₂ox). Oxalic acid, the simplest dicarboxylic acid, is found in rhubarb and many other plants. Rhubarb leaves are toxic because they contain the calcium salt of the fully deprotonated form of oxalic acid, the oxalate ion (⁻O₂CCO₂⁻, abbreviated ox²⁻).Oxalate salts are toxic for two reasons. First, oxalate salts of divalent cations such as Ca²⁺ are insoluble at neutral pH but soluble at low pH, as we shall see in Chapter 17 "Solubility and Complexation Equilibriums". As a result, calcium oxalate dissolves in the dilute acid of the stomach, allowing oxalate to be absorbed and transported into cells, where it can react with calcium to form tiny calcium oxalate crystals that damage tissues. Second, oxalate forms stable complexes with metal ions, which can alter the distribution of metal ions in biological fluids.

Given: volume and concentration of acid and base

Asked for: pH

Strategy:

A Calculate the initial millimoles of the acid and the base. Use a tabular format to determine the amounts of all the species in solution.

B Calculate the concentrations of all the species in the final solution. Use Equation 16.16 to determine [H⁺] and convert this value to pH.

Solution:

A Table 16.4 "Values of p" gives the pK_a values of oxalic acid as 1.25 and 3.81. Again we proceed by determining the millimoles of acid and base initially present:

$$100.0\cancel{\text{mL}}\left(\frac{0.0510{\text{ mmol H}}_2\text{ox}}{\cancel{\text{mL}}}\right)=5.10{\text{ mmol H}}_2\text{ox}$$ $$55.0\cancel{\text{mL}}\left(\frac{0.120\text{ mmol NaOH}}{\cancel{\text{mL}}}\right)=6.60\text{ mmol NaOH}$$

The strongest acid (H₂ox) reacts with the base first. This leaves (6.60 − 5.10) = 1.50 mmol of OH⁻ to react with Hox⁻, forming ox²⁻ and H₂O. The reactions can be written as follows:

$$\underset{5.10\text{ mmol}}{{\text{H}}_2\text{ox}}+\underset{6.60\text{ mmol}}{{\text{OH}}^{-}}\to \underset{5.10\text{ mmol}}{{\text{Hox}}^{-}}+\underset{5.10\text{ mmol}}{{\text{H}}_2\text{O}}$$ $$\underset{5.10\text{ mmol}}{{\text{Hox}}^{-}}+\underset{1.50\text{ mmol}}{{\text{OH}}^{-}}\to \underset{1.50\text{ mmol}}{{\text{ox}}^{2-}}+\underset{1.50\text{ mmol}}{{\text{H}}_2\text{O}}$$

In tabular form,

	H2ox	OH−	Hox−	ox2−
initial	5.10 mmol	6.60 mmol	0 mmol	0 mmol
change (step 1)	−5.10 mmol	−5.10 mmol	+5.10 mmol	0 mmol
final (step 1)	0 mmol	1.50 mmol	5.10 mmol	0 mmol
change (step 2)	—	−1.50 mmol	−1.50 mmol	+1.50 mmol
final	0 mmol	0 mmol	3.60 mmol	1.50 mmol

B The equilibrium between the weak acid (Hox⁻) and its conjugate base (ox²⁻) in the final solution is determined by the magnitude of the second ionization constant, K_a2 = 10^−3.81 = 1.6 × 10⁻⁴. To calculate the pH of the solution, we need to know [H⁺], which is determined using exactly the same method as in the acetic acid titration in Example 12:

final volume of solution = 100.0 mL + 55.0 mL = 155.0 mL

Thus the concentrations of Hox⁻ and ox²⁻ are as follows:

$$\begin{array}{cc}\hfill [{\text{Hox}}^{-}]& =\frac{3.60{\text{ mmol Hox}}^{-}}{155.0\text{ mL}}=2.32\times {10}^{-2}\text{ M}\hfill \\ \hfill [{\text{ox}}^{2-}]& =\frac{1.50{\text{ mmol ox}}^{2-}}{155.0\text{ mL}}=9.68\times {10}^{-3}\text{ M}\hfill \end{array}$$

We can now calculate [H⁺] at equilibrium using the following equation:

$$K_{\text{a2}}=\frac{[{\text{ox}}^{2-}][{\text{H}}^{+}]}{[{\text{Hox}}^{-}]}$$

Rearranging this equation and substituting the values for the concentrations of Hox⁻ and ox²⁻,

$$[{\text{H}}^{+}]=\frac{K_{\text{a2}}[{\text{Hox}}^{-}]}{[{\text{ox}}^{2-}]}=\frac{(1.6\times {10}^{-4})(2.32\times {10}^{-2})}{9.68\times {10}^{-3}}=3.7\times {10}^{-4}\text{ M}$$

So

pH = −log[H⁺] = −log(3.7 ×10⁻⁴) = 3.43

This answer makes chemical sense because the pH is between the first and second pK_a values of oxalic acid, as it must be. We added enough hydroxide ion to completely titrate the first, more acidic proton (which should give us a pH greater than pK_a1), but we added only enough to titrate less than half of the second, less acidic proton, with pK_a2. If we had added exactly enough hydroxide to completely titrate the first proton plus half of the second, we would be at the midpoint of the second step in the titration, and the pH would be 3.81, equal to pK_a2.

Exercise

Piperazine is a diprotic base used to control intestinal parasites (“worms”) in pets and humans. A dog is given 500 mg (5.80 mmol) of piperazine (pK_b1 = 4.27, pK_b2 = 8.67). If the dog’s stomach initially contains 100 mL of 0.10 M HCl (pH = 1.00), calculate the pH of the stomach contents after ingestion of the piperazine.

Answer: 4.9

Indicators

In practice, most acid–base titrations are not monitored by recording the pH as a function of the amount of the strong acid or base solution used as the titrant. Instead, an acid–base indicatorA compound added in small amounts to an acid–base titration to signal the equivalence point by changing color. is often used that, if carefully selected, undergoes a dramatic color change at the pH corresponding to the equivalence point of the titration. Indicators are weak acids or bases that exhibit intense colors that vary with pH. The conjugate acid and conjugate base of a good indicator have very different colors so that they can be distinguished easily. Some indicators are colorless in the conjugate acid form but intensely colored when deprotonated (phenolphthalein, for example), which makes them particularly useful.

We can describe the chemistry of indicators by the following general equation:

where the protonated form is designated by HIn and the conjugate base by In⁻. The ionization constant for the deprotonation of indicator HIn is as follows:

The pK_in (its pK_a) determines the pH at which the indicator changes color.

Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH (Figure 16.22 "Naturally Occurring pH Indicators in Red Cabbage Juice"). In all cases, though, a good indicator must have the following properties:

• The color change must be easily detected.
• The color change must be rapid.
• The indicator molecule must not react with the substance being titrated.
• To minimize errors, the indicator should have a pK_in that is within one pH unit of the expected pH at the equivalence point of the titration.

Figure 16.22 Naturally Occurring pH Indicators in Red Cabbage Juice

Red cabbage juice contains a mixture of substances whose color depends on the pH. Each test tube contains a solution of red cabbage juice in water, but the pH of the solutions varies from pH = 2.0 (far left) to pH = 11.0 (far right). At pH = 7.0, the solution is blue.

Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. Figure 16.23 "Some Common Acid–Base Indicators" shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values.

Figure 16.23 Some Common Acid–Base Indicators

Approximate colors are shown, along with pK_in values and the pH range over which the color changes.

It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In⁻ increases, the color of the solution slowly changes from the characteristic color of HIn to that of In⁻. As we will see in Section 16.6 "Buffers", the [In⁻]/[HIn] ratio changes from 0.1 at a pH one unit below pK_in to 10 at a pH one unit above pK_in. Thus most indicators change color over a pH range of about two pH units.

We have stated that a good indicator should have a pK_in value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in Figure 16.24 "Choosing the Correct Indicator for an Acid–Base Titration". This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M NaOH. The pH ranges over which two common indicators (methyl red, pK_in = 5.0, and phenolphthalein, pK_in = 9.5) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the HCl titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of NaOH has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the HCl titration, the phenolphthalein indicator will turn pink when about 50 mL of NaOH has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the midpoint of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of NaOH will therefore cause the methyl red indicator to change color, resulting in a huge error.

Figure 16.24 Choosing the Correct Indicator for an Acid–Base Titration

The graph shows the results obtained using two indicators (methyl red and phenolphthalein) for the titration of 0.100 M solutions of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M NaOH. Due to the steepness of the titration curve of a strong acid around the equivalence point, either indicator will rapidly change color at the equivalence point for the titration of the strong acid. In contrast, the pK_in for methyl red (5.0) is very close to the pK_a of acetic acid (4.76); the midpoint of the color change for methyl red occurs near the midpoint of the titration, rather than at the equivalence point.

In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a pK_in between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with pK_in > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pK_in < 7.0, should be used.

The existence of many different indicators with different colors and pK_in values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container (Figure 16.25 "pH Paper").

Figure 16.25 pH Paper

pH paper contains a set of indicators that change color at different pH values. The approximate pH of a solution can be determined by simply dipping a paper strip into the solution and comparing the color to the standards provided.