Acid-Base Titrations

Learning Objectives

By the end of this module, you will be able to:

Interpret titration curves for strong and weak acid-base systems
Compute sample pH at important stages of a titration
Explain the function of acid-base indicators

As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the changes in the concentrations of the acidic and basic species present in a solution during the process of a titration.

Titration Curve

Previously, when we studied acid-base reactions in solution, we focused only on the point at which the acid and base were stoichiometrically equivalent. No consideration was given to the pH of the solution before, during, or after the neutralization.

Example 1: Calculating pH for Titration Solutions: Strong Acid/Strong Base

A titration is carried out for 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M of a strong base NaOH the titration curve is shown in Figure 1 (below). Calculate the pH at these volumes of added base solution:

0.00 mL
12.50 mL
25.00 mL
37.50 mL

Show Answer

Since HCl is a strong acid, we can assume that all of it dissociates. The initial concentration of ${\text{H}}_{3}{\text{O}}^{\text{+}}$ is ${\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}_{0}=0.100M$ . When the base solution is added, it also dissociates completely, providing OH⁻ ions. The ${\text{H}}_{3}{\text{O}}^{\text{+}}$ and OH⁻ ions neutralize each other, so only those of the two that were in excess remain, and their concentration determines the pH. Thus, the solution is initially acidic (pH < 7), but eventually all the hydronium ions present from the original acid are neutralized, and the solution becomes neutral. As more base is added, the solution turns basic.

The total initial amount of the hydronium ions is:

$\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}={\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}_{0}\times \text{0.02500 L}=\text{0.002500 mol}$

Once X mL of the 0.100-M base solution is added, the number of moles of the OH⁻ ions introduced is:

$\text{n}{\left({\text{OH}}^{-}\right)}_{0}=0.100M\times \text{X mL}\times \left(\frac{\text{1 L}}{\text{1000 mL}}\right)$

The total volume becomes: $V=\left(\text{25.00 mL}+\text{X mL}\right)\left(\frac{\text{1 L}}{\text{1000 mL}}\right)$

The number of moles of ${\text{H}}_{3}{\text{O}}^{\text{+}}$ becomes:

$\text{n}\left({\text{H}}^{\text{+}}\right)=\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}-\text{n}{\left({\text{OH}}^{-}\right)}_{0}=\text{0.002500 mol}-0.100M\times \text{X mL}\times \left(\frac{\text{1 L}}{\text{1000 mL}}\right)$

The concentration of ${\text{H}}_{3}{\text{O}}^{\text{+}}$ is:

$\begin{array}{l}\\ \\ \left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\frac{\text{n}\left({\text{H}}^{\text{+}}\right)}{V}=\frac{\text{0.002500 mol}-0.100M\times \text{X mL}\times \left(\frac{\text{1 L}}{\text{1000 mL}}\right)}{\left(\text{25.00 mL}+\text{X mL}\right)\left(\frac{\text{1 L}}{\text{1000 mL}}\right)}\\ =\frac{\text{0.002500 mol}\times \left(\frac{\text{1000 mL}}{\text{1 L}}\right)-0.100M\times \text{X mL}}{\text{25.00 mL}+\text{X mL}}\end{array}$
$\text{pH}=-\text{log}\left(\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\right)$

The preceding calculations work if $\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}-\text{n}{\left({\text{OH}}^{-}\right)}_{0}>0$ and so n(H⁺) > 0. When $\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}=\text{n}{\left({\text{OH}}^{-}\right)}_{0}$ , the ${\text{H}}_{3}{\text{O}}^{\text{+}}$ ions from the acid and the OH⁻ ions from the base mutually neutralize. At this point, the only hydronium ions left are those from the autoionization of water, and there are no OH⁻ particles to neutralize them. Therefore, in this case:

$\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\left[{\text{OH}}^{-}\right],\left[{\text{H}}_{3}{\text{O}}^{+}\right]={K}_{\text{w}}=1.0\times {10}^{\text{-14}};\left[{\text{H}}_{3}{\text{O}}^{+}\right]=1.0\times {10}^{\text{-7}}$
$\text{pH}=-\text{log}\left(1.0\times {10}^{\text{-7}}\right)=7.00$

Finally, when $\text{n}{\left({\text{OH}}^{-}\right)}_{0}>\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}$ , there are not enough ${\text{H}}_{3}{\text{O}}^{\text{+}}$ ions to neutralize all the OH⁻ ions, and instead of $\text{n}\left({\text{H}}^{\text{+}}\right)=\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}-\text{n}{\left({\text{OH}}^{-}\right)}_{0}$ , we calculate: $\text{n}\left({\text{OH}}^{-}\right)=\text{n}{\left({\text{OH}}^{-}\right)}_{0}-\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}$

In this case:

$\begin{array}{l}\\ \\ \left[{\text{OH}}^{-}\right]=\frac{\text{n}\left({\text{OH}}^{-}\right)}{V}=\frac{0.100M\times \text{X mL}\times \left(\frac{\text{1 L}}{\text{1000 mL}}\right)-\text{0.002500 mol}}{\left(\text{25.00 mL}+\text{X mL}\right)\left(\frac{\text{1 L}}{\text{1000 mL}}\right)}\\ =\frac{0.100M\times \text{X mL}-\text{0.002500 mol}\times \left(\frac{\text{1000 mL}}{\text{1 L}}\right)}{\text{25.00 mL}+\text{X mL}}\end{array}$
$\text{pH}=14-\text{pOH}=14+\text{log}\left(\left[{\text{OH}}^{-}\right]\right)$

Let us now consider the four specific cases presented in this problem:

X = 0 mL
$\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\frac{\text{n}\left({\text{H}}^{\text{+}}\right)}{V}=\frac{\text{0.002500 mol}\times \left(\frac{\text{1000 mL}}{\text{1 L}}\right)}{\text{25.00 mL}}=0.1M$ pH = −log(0.100) = 1.000
X = 12.50 mL
$\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\frac{\text{n}\left({\text{H}}^{\text{+}}\right)}{V}=\frac{\text{0.002500 mol}\times \left(\frac{\text{1000 mL}}{\text{1 L}}\right)-0.100M\times \text{12.50 mL}}{\text{25.00 mL}+\text{12.50 mL}}=0.0333M$ pH = −log(0.0333) = 1.477
X = 25.00 mL
Since the volumes and concentrations of the acid and base solutions are the same: $\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}=\text{n}{\left({\text{OH}}^{-}\right)}_{0}$ , and pH = 7.000, as described earlier.
X = 37.50 mL
In this case: $\text{n}{\left({\text{OH}}^{-}\right)}_{0}>\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}$ $\left[{\text{OH}}^{-}\right]=\frac{\text{n}\left({\text{OH}}^{-}\right)}{V}=\frac{0.100M\times \text{35.70 mL}-\text{0.002500 mol}\times \left(\frac{\text{1000 mL}}{\text{1 L}}\right)}{\text{25.00 mL}+\text{37.50 mL}}=0.0200M$ pH = 14 − pOH = 14 + log([OH⁻]) = 14 + log(0.0200) = 12.30

Check Your Learning

Calculate the pH for the strong acid/strong base titration between 50.0 mL of 0.100 M HNO₃(aq) and 0.200 M NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 40.0 mL.

Show Answer

In Example 1, we calculated pH at four points during a titration. Table 1 shows a detailed sequence of changes in the pH of a strong acid and a weak acid in a titration with NaOH.

Table 1. pH Values in the Titrations of a Strong Acid with a Strong Base and of a Weak Acid with a Strong Base
Volume of 0.100 M NaOH Added (mL)	Moles of NaOH Added	pH Values 0.100 M HCl^[1]	pH Values 0.100 M CH₃CO₂H^[2]
0.0	0.0	1.00	2.87
5.0	0.00050	1.18	4.14
10.0	0.00100	1.37	4.57
15.0	0.00150	1.60	4.92
20.0	0.00200	1.95	5.35
22.0	0.00220	2.20	5.61
24.0	0.00240	2.69	6.13
24.5	0.00245	3.00	6.44
24.9	0.00249	3.70	7.14
25.0	0.00250	7.00	8.72
25.1	0.00251	10.30	10.30
25.5	0.00255	11.00	11.00
26.0	0.00260	11.29	11.29
28.0	0.00280	11.75	11.75
30.0	0.00300	11.96	11.96
35.0	0.00350	12.22	12.22
40.0	0.00400	12.36	12.36
45.0	0.00450	12.46	12.46
50.0	0.00500	12.52	12.52

The simplest acid-base reactions are those of a strong acid with a strong base. Table 1 shows data for the titration of a 25.0-mL sample of 0.100 M hydrochloric acid with 0.100 M sodium hydroxide. The values of the pH measured after successive additions of small amounts of NaOH are listed in the first column of this table, and are graphed in Figure 1, in a form that is called a titration curve. The pH increases slowly at first, increases rapidly in the middle portion of the curve, and then increases slowly again. The point of inflection (located at the midpoint of the vertical part of the curve) is the equivalence point for the titration. It indicates when equivalent quantities of acid and base are present. For the titration of a strong acid with a strong base, the equivalence point occurs at a pH of 7.00 and the points on the titration curve can be calculated using solution stoichiometry (Table 1 and Figure 1).

Two graphs are shown. The first graph on the left is titled “Titration of Weak Acid.” The horizontal axis is labeled “Volume of 0.100 M N a O H added (m L).” Markings and vertical gridlines are provided every 5 units from 0 to 50. The vertical axis is labeled “p H” and is marked every 1 unis beginning at 0 extending to 14. A red curve is drawn on the graph which increases steadily from the point (0, 3) up to about (20, 5.5) after which the graph has a vertical section from (25, 7) up to (25, 11). The graph then levels off to a value of about 12.5 from about 40 m L up to 50 m L. The midpoint of the vertical segment of the curve is labeled “Equivalence point p H, 8.72.” The second graph on the right is titled “Titration of Strong Acid.” The horizontal axis is labeled “Volume of 0.100 M N a O H added (m L).” Markings and vertical gridlines are provided every 5 units from 0 to 50. The vertical axis is labeled “p H” and is marked every 1 units beginning at 0 extending to 14. A red curve is drawn on the graph which increases gradually from the point (0, 1) up to about (22.5, 2.2) after which the graph has a vertical section from (25, 4) up to nearly (25, 11). The graph then levels off to a value of about 12.4 from about 40 m L up to 50 m L. The midpoint of the vertical segment of the curve is labeled “Equivalence point p H, 7.00.”

Figure 1. (a) The titration curve for the titration of 25.00 mL of 0.100 M CH3CO2H (weak acid) with 0.100 M NaOH (strong base) has an equivalence point of 7.00 pH. (b) The titration curve for the titration of 25.00 mL of 0.100 M HCl (strong acid) with 0.100 M NaOH (strong base) has an equivalence point of 8.72 pH.

The titration of a weak acid with a strong base (or of a weak base with a strong acid) is somewhat more complicated than that just discussed, but it follows the same general principles. Let us consider the titration of 25.0 mL of 0.100 M acetic acid (a weak acid) with 0.100 M sodium hydroxide and compare the titration curve with that of the strong acid. Table 1 gives the pH values during the titration, Figure 1 shows the titration curve.

Although the initial volume and molarity of the acids are the same, there are important differences between the two titration curves. The titration curve for the weak acid begins at a higher value (less acidic) and maintains higher pH values up to the equivalence point. This is because acetic acid is a weak acid, which is only partially ionized. The pH at the equivalence point is also higher (8.72 rather than 7.00) due to the hydrolysis of acetate, a weak base that raises the pH:

${\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(l\right)+{\text{OH}}^{-}\left(aq\right)$

After the equivalence point, the two curves are identical because the pH is dependent on the excess of hydroxide ion in both cases.

Example 2: Titration of a Weak Acid with a Strong Base

The titration curve shown in Figure 3 is for the titration of 25.00 mL of 0.100 M CH₃CO₂H with 0.100 M NaOH. The reaction can be represented as ${\text{CH}}_{3}{\text{CO}}_{2}\text{H}+{\text{OH}}^{-}\longrightarrow {\text{CH}}_{3}{\text{CO}}_{2}{}^{-}+{\text{H}}_{2}\text{O}$ .

What is the initial pH before any amount of the NaOH solution has been added? K_a = 1.8 $\times$ 10⁻⁵ for CH₃CO₂H.
Find the pH after 25.00 mL of the NaOH solution have been added.
Find the pH after 12.50 mL of the NaOH solution has been added.
Find the pH after 37.50 mL of the NaOH solution has been added.

Show Answer

Assuming that the dissociated amount is small compared to 0.100 M, we find that:
${K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}\approx \frac{{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}^{\text{2}}}{{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}_{0}}$ , and $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\sqrt{{K}_{a}\times \left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}=\sqrt{1.8\times {10}^{-5}\times 0.100}=1.3\times {10}^{-3}$ $\text{pH}=-\text{log}\left(1.3\times {10}^{-3}\right)=2.87$
After 25.00 mL of NaOH are added, the number of moles of NaOH and CH₃CO₂H are equal because the amounts of the solutions and their concentrations are the same. All of the CH₃CO₂H has been converted to ${\text{CH}}_{3}{\text{CO}}_{2}{}^{-}$ . The concentration of the ${\text{CH}}_{3}{\text{CO}}_{2}{}^{-}$ ion is: $\frac{\text{0.00250 mol}}{\text{0.0500 L}}=\text{0.0500 M}{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}$ The equilibrium that must be focused on now is the basicity equilibrium for ${\text{CH}}_{3}{\text{CO}}_{2}{}^{-}:$ ${\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$ so we must determine K_b for the base by using the ion product constant for water: ${K}_{\text{b}}=\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}$ ${K}_{\text{a}}=\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]},\text{so}\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}=\frac{\left[{\text{H}}^{\text{+}}\right]}{{K}_{\text{a}}}$ .Since K_w = [H⁺][OH⁻]: ${K}_{\text{b}}=\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]}{{K}_{\text{a}}}=\frac{{K}_{\text{w}}}{{K}_{\text{a}}}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}=5.6\times {10}^{-10}$ Let us denote the concentration of each of the products of this reaction, CH₃CO₂H and OH⁻, as x. Using the assumption that x is small compared to 0.0500 M, ${K}_{\text{b}}=\frac{{x}^{\text{2}}}{0.0500M}$ , and then: $x=\left[{\text{OH}}^{-}\right]=5.3\times {10}^{-6}$
$\text{pOH}=-\text{log}\left(5.3\times {10}^{-6}\right)=5.28$
$\text{pH}=14.00 - 5.28=8.72$ Note that the pH at the equivalence point of this titration is significantly greater than 7.
In (1), 25.00 mL of the NaOH solution was added, and so practically all the CH₃CO₂H was converted into ${\text{CH}}_{3}{\text{CO}}_{2}{}^{-}$ . In this case, only 12.50 mL of the base solution has been introduced, and so only half of all the CH₃CO₂H is converted into ${\text{CH}}_{3}{\text{CO}}_{2}{}^{-}$ . The total initial number of moles of CH₃CO₂H is 0.02500L $\times$ 0.100 M = 0.00250 mol, and so after adding the NaOH, the numbers of moles of CH₃CO₂H and ${\text{CH}}_{3}{\text{CO}}_{2}{}^{-}$ are both approximately equal to $\frac{\text{0.00250 mol}}{2}=\text{0.00125 mol}$ , and their concentrations are the same.Since the amount of the added base is smaller than the original amount of the acid, the equivalence point has not been reached, the solution remains a buffer, and we can use the Henderson-Hasselbalch equation: $\text{pH}=p{K}_{\text{a}}+\text{log}\frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]}=-\text{log}\left({K}_{\text{a}}\right)+\text{log}\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}=-\text{log}\left(1.8\times {10}^{-5}\right)+\text{log}\left(1\right)$ (as the concentrations of ${\text{CH}}_{3}{\text{CO}}_{2}{}^{-}$ and CH₃CO₂H are the same)Thus: $\text{pH}=-\text{log}\left(1.8\times {10}^{-5}\right)=4.74$ (the pH = the pK_a at the halfway point in a titration of a weak acid)
After 37.50 mL of NaOH is added, the amount of NaOH is 0.03750 L $\times$ 0.100 M = 0.003750 mol NaOH. Since this is past the equivalence point, the excess hydroxide ions will make the solution basic, and we can again use stoichiometric calculations to determine the pH: $\left[{\text{OH}}^{-}\right]=\frac{\left(\text{0.003750 mol}-\text{0.00250 mol}\right)}{\text{0.06250 L}}=2.00\times {10}^{-2}M$ So: $\text{pOH}=-\text{log}\left(2.00\times {10}^{-2}\right)=\text{1.70, and pH}=14.00 - 1.70=12.30$ Note that this result is the same as for the strong acid-strong base titration example provided, since the amount of the strong base added moves the solution past the equivalence point.

Check Your Learning

Calculate the pH for the weak acid/strong base titration between 50.0 mL of 0.100 M HCOOH(aq) (formic acid) and 0.200 M NaOH (titrant) at the listed volumes of added base: 0.00 mL, 15.0 mL, 25.0 mL, and 30.0 mL.

Show Answer

Acid-Base Indicators

Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than 5.0 $\times$ 10⁻⁹M (pH < 8.3). In more basic solutions where the hydronium ion concentration is less than 5.0 $\times$ 10⁻⁹M (pH > 8.3), it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called acid-base indicators. Acid-base indicators are either weak organic acids or weak organic bases.

The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use HIn as a simple representation for the complex methyl orange molecule:

$\begin{array}{ccc}\text{HIn}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)& \rightleftharpoons & {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{In}}^{-}\left(aq\right)\\ \text{red}& &\text{yellow}\end{array}$

${K}_{a}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{In}}^{-}\right]}{\left[\text{HIn}\right]}=4.0\times {10}^{-4}$

The anion of methyl orange, In⁻, is yellow, and the nonionized form, HIn, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Châtelier’s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers.

An indicator’s color is the visible result of the ratio of the concentrations of the two species In⁻ and HIn. If most of the indicator (typically about 60−90% or more) is present as In⁻, then we see the color of the In⁻ ion, which would be yellow for methyl orange. If most is present as HIn, then we see the color of the HIn molecule: red for methyl orange. For methyl orange, we can rearrange the equation for K_a and write:

$\frac{\left[{\text{In}}^{-}\right]}{\left[\text{HIn}\right]}=\frac{\left[\text{substance with yellow color}\right]}{\left[\text{substance with red color}\right]}=\frac{{K}_{\text{a}}}{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}$

This shows us how the ratio of $\frac{\left[{\text{In}}^{-}\right]}{\left[\text{HIn}\right]}$ varies with the concentration of hydronium ion.

The above expression describing the indicator equilibrium can be rearranged:

$\begin{array}{rll}\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{{K}_{\text{a}}}&=&\frac{\left[\text{HIn}\right]}{\left[{\text{In}}^{-}\right]}\\\text{log}\left(\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}{{K}_{\text{a}}}\right)&=&\text{log}\left(\frac{\left[\text{HIn}\right]}{\left[{\text{In}}^{-}\right]}\right)\\\text{log}\left(\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\right)-\text{log}\left({K}_{\text{a}}\right)&=&-\text{log}\left(\frac{\left[{\text{In}}^{-}\right]}{\left[\text{HIn}\right]}\right)\\-\text{pH}+\text{p}{K}_{\text{a}}&=&-\text{log}\left(\frac{\left[{\text{In}}^{-}\right]}{\left[\text{HIn}\right]}\right)\\\text{pH}=\text{p}K\text{a}+\text{log}\left(\frac{\left[{\text{In}}^{-}\right]}{\left[\text{HIn}\right]}\right)&\text{or}&\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\left(\frac{\left[\text{base}\right]}{\left[\text{acid}\right]}\right)\end{array}$

The last formula is the same as the Henderson-Hasselbalch equation, which can be used to describe the equilibrium of indicators.

When $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ has the same numerical value as K_a, the ratio of [In⁻] to [HIn] is equal to 1, meaning that 50% of the indicator is present in the red form (HIn) and 50% is in the yellow ionic form (In⁻), and the solution appears orange in color. When the hydronium ion concentration increases to 8 $\times$ 10⁻⁴M (a pH of 3.1), the solution turns red. No change in color is visible for any further increase in the hydronium ion concentration (decrease in pH). At a hydronium ion concentration of 4 $\times$ 10⁻⁵M (a pH of 4.4), most of the indicator is in the yellow ionic form, and a further decrease in the hydronium ion concentration (increase in pH) does not produce a visible color change. The pH range between 3.1 (red) and 4.4 (yellow) is the color-change interval of methyl orange; the pronounced color change takes place between these pH values.

There are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. Figure 2 presents several indicators, their colors, and their color-change intervals.

This figure provides a graphical representation of indicators and color ranges. A horizontal axis is labeled “p H.” This axis begins at zero and increases by ones up to 13. The left side of the graphic provides a column with the names of indicators. To the right of each indicator name is either one or two colored bars that are shaded according to the indicator color at various p H ranges. From the top, the first row is labeled “Crystal violet.” The associated colored bar is yellow at its left end at a p H of 0 and changes to green and blue moving right to its endpoint at a p H of 1.8. The second row is labeled “Cresol red.” The associated colored bar is red at its left end at a p H of 1 and changes to orange and yellow moving right to its endpoint at a p H of just over 2. A second bar to its right is yellow at a p H of around 7 and proceeds through orange to red at a p H of about 9. The third row is labeled “Thymol blue.” The associated colored bar is red at its left end at a p H of nearly 1.2 and changes to orange and red moving right to its endpoint at a p H of 2.8. A second bar begins in yellow at a p H of 8 and proceeds through green and blue to its end at a p H of around 9.1. The fourth row is labeled “Erythrosin B.” The associated colored bar is red from a p H of 2.2 to its endpoint at a p H of 3.6. The fifth row is labeled “2 comma 4 dash Dinitrophenol.” The associated colored bar is white at its left end at a p H of 2.6 and changes to yellow at its endpoint at a p H of 4. The sixth row is labeled “Bromophenol blue.” The associated colored bar is yellow at its left end at a p H of 3 and changes to green and blue moving right to its endpoint at a p H of 4.5. The seventh row is labeled “Methyl orange.” The associated colored bar is red-orange at its left end at a p H of 4.2 and changes to yellow moving right to its endpoint at a p H of 6.3. The eighth row is labeled “Bromocresol green.” The associated colored bar is yellow at its left end at a p H of 3.8 and changes to green and blue moving right to its endpoint at a p H of 5.4. The ninth row is labeled “Methyl red.” The associated colored bar is orange at its left end at a p H of 4.2 and changes to yellow moving right to its endpoint at a p H of 6.3. The tenth row is labeled “Eriochrome * Black T.” The associated colored bar is red at its left end at a p H of 5 and changes to purple and blue moving right to its endpoint at a p H of 6.5. The eleventh row is labeled “Bromocresol purple.” The associated colored bar is yellow at its left end at a p H of 5.2 and changes to purple moving right to its endpoint at a p H of 6.8. The twelfth row is labeled “Alizarin.” The first associated colored bar is yellow-orange at its left end at a p H of 5.7 and changes to red moving right to its endpoint at a p H of 7.2. A second bar begins in red at a p H of 11 and changes to purple, then dark blue at its right end at a p H of 12.4. The thirteenth row is labeled “Bromothymol blue.” The associated colored bar is yellow at its left end at a p H of 6 and changes to green and blue moving right to its endpoint at a p H of 7.6. The fourteenth row is labeled “Phenol red.” The associated colored bar is yellow-orange at its left end at a p H of 6.8 and changes to orange and red moving right to its endpoint at a p H of 8.2. The fifteenth row is labeled “m dash Nitrophenol.” The associated colored bar is white at its left end at a p H of 6.8 and changes to yellow moving right to its endpoint at a p H of 8.6. The sixteenth row is labeled “o dash Cresolphthalein.” The associated colored bar is white at its left end at a p H of 8.3 and changes to red moving right to its endpoint at a p H of 9.8. The seventeenth row is labeled “Phenolphthalein.” The associated colored bar is white at its left end at a p H of 8 and changes to pink moving right to its endpoint at a p H of 10. The eighteenth row is labeled “Thymolphthalein.” The associated colored bar is light blue at its left end at a p H of 9.3 and changes to a deep, dark blue moving right to its endpoint at a p H of 10.5. The nineteenth row is labeled “Alizarin yellow R.” The associated colored bar is yellow-orange at its left end at a p H of 10 and changes to red moving right to its endpoint at a p H of 12.

Figure 2. This chart illustrates the ranges of color change for several acid-base indicators.

Titration curves help us pick an indicator that will provide a sharp color change at the equivalence point. The best selection would be an indicator that has a color change interval that brackets the pH at the equivalence point of the titration.

The color change intervals of three indicators are shown in Figure 3. The equivalence points of both the titration of the strong acid and of the weak acid are located in the color-change interval of phenolphthalein. We can use it for titrations of either strong acid with strong base or weak acid with strong base.

A graph is shown which is titled “Titration of Weak Acid.” The horizontal axis is labeled “Volume of 0.100 M N a O H added (m L)” and begins at 0 with markings every 5 units up to 50. The vertical axis is labeled “p H” and begins at 0 and increases by single units up to 14. A red curve is drawn on the graph. The curve begins at (0, 3) and passes through the points (5, 4.1), (10, 4.7), (15, 5), (20, 5.5), and (22.5, 6), after which it rapidly increases, forming a vertical section centered at the point (25, 8.7). The rapid increase of the curve then levels off and the curve passes through the points (30, 12), (35, 12.4), (40, 12.5), (45, 12.6), and (50, 12.6). A brown rectangle extends horizontally across the graph covering the p H of 3 to 4.2 range. To the right, this rectangle is labeled “Methyl orange p H range.” A blue rectangle extends horizontally across the graph covering the p H of 4.6 to 8 range. To the right, this rectangle is labeled “Litmus p H range.” A purple rectangle extends horizontally across the graph covering the p H of 8.4 to 10 range. To the right, this rectangle is labeled “Phenolphthalein p H range.” The midpoint of the vertical segment of the curve is labeled “Equivalence point p H, 8.72.”

Figure 3. The graph shows a titration curve for the titration of 25.00 mL of 0.100 M CH3CO2H (weak acid) with 0.100 M NaOH (strong base) and the titration curve for the titration of HCl (strong acid) with NaOH (strong base). The pH ranges for the color change of phenolphthalein, litmus, and methyl orange are indicated by the shaded areas.

Litmus is a suitable indicator for the HCl titration because its color change brackets the equivalence point. However, we should not use litmus for the CH₃CO₂H titration because the pH is within the color-change interval of litmus when only about 12 mL of NaOH has been added, and it does not leave the range until 25 mL has been added. The color change would be very gradual, taking place during the addition of 13 mL of NaOH, making litmus useless as an indicator of the equivalence point.

We could use methyl orange for the HCl titration, but it would not give very accurate results: (1) It completes its color change slightly before the equivalence point is reached (but very close to it, so this is not too serious); (2) it changes color, as Figure 3 shows, during the addition of nearly 0.5 mL of NaOH, which is not so sharp a color change as that of litmus or phenolphthalein; and (3) it goes from yellow to orange to red, making detection of a precise endpoint much more challenging than the colorless to pink change of phenolphthalein. Figure 3 shows us that methyl orange would be completely useless as an indicator for the CH₃CO₂H titration. Its color change begins after about 1 mL of NaOH has been added and ends when about 8 mL has been added. The color change is completed long before the equivalence point (which occurs when 25.0 mL of NaOH has been added) is reached and hence provides no indication of the equivalence point.

We base our choice of indicator on a calculated pH, the pH at the equivalence point. At the equivalence point, equimolar amounts of acid and base have been mixed, and the calculation becomes that of the pH of a solution of the salt resulting from the titration.

Key Concepts and Summary

A titration curve is a graph that relates the change in pH of an acidic or basic solution to the volume of added titrant. The characteristics of the titration curve are dependent on the specific solutions being titrated. The pH of the solution at the equivalence point may be greater than, equal to, or less than 7.00. The choice of an indicator for a given titration depends on the expected pH at the equivalence point of the titration, and the range of the color change of the indicator.

Exercises

Explain how to choose the appropriate acid-base indicator for the titration of a weak base with a strong acid.
Explain why an acid-base indicator changes color over a range of pH values rather than at a specific pH.
Why can we ignore the contribution of water to the concentrations of ${\text{H}}_{3}{\text{O}}^{\text{+}}$ in the solutions of following acids: (1) 0.0092 M HClO, a weak acid, (2) 0.0810 M HCN, a weak acid, and (3) 0.120 M $\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}{}^{2+}$ a weak acid, K_a = 1.6 $\times$ 10⁻⁷, but not the contribution of water to the concentration of OH⁻?
We can ignore the contribution of water to the concentration of OH⁻ in a solution of the following bases (1) 0.0784 M C₆H₅NH₂, a weak base and (2) 0.11 M (CH₃)₃N, a weak base, but not the contribution of water to the concentration of ${\text{H}}_{3}{\text{O}}^{\text{+}}?$
Draw a curve similar to that shown in Figure 3 for a series of solutions of NH₃. Plot [OH⁻] on the vertical axis and the total concentration of NH₃ (both ionized and nonionized NH₃ molecules) on the horizontal axis. Let the total concentration of NH₃ vary from 1 $\times$ 10⁻¹⁰M to 1 $\times$ 10⁻²M.
The indicator dinitrophenol is an acid with a K_a of 1.1 $\times$ 10⁻⁴. In a 1.0 $\times$ 10⁻⁴–M solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from 10% ionized (colorless) to 90% ionized (yellow).

Show Selected Answers

Exercises

Draw a curve for a series of solutions of HF. Plot ${\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}_{\text{total}}$ on the vertical axis and the total concentration of HF (the sum of the concentrations of both the ionized and nonionized HF molecules) on the horizontal axis. Let the total concentration of HF vary from 1 $\times$ 10⁻¹⁰M to 1 $\times$ 10⁻²M.

Show Answer

$\text{HF}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{F}}^{-}\left(aq\right){K}_{\text{a}}=7.2\times {10}^{-4}$

${K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{F}}^{-}\right]}{\left[\text{HF}\right]}$

The methodology for conducting the calculations is as follows: Because HF is a weak acid, the ionization is not complete; thus the ${\text{H}}_{3}{\text{O}}^{\text{+}}$ concentration will always be less than the initial molarity of the HF concentration. Thus for initial concentrations from 10⁻¹⁰M to 1 $\times$ 10⁻⁷M, the contribution of ${\text{H}}_{3}{\text{O}}^{\text{+}}$ ions to the solution will be smaller than the contribution of ${\text{H}}_{3}{\text{O}}^{\text{+}}$ ions from the self-ionization of water. Thus the ${\text{H}}_{3}{\text{O}}^{\text{+}}$ concentration in HF solutions of 10⁻¹⁰ to 10⁻⁸M is approximately 1 $\times$ 10⁻⁷M. Calculate the concentration of ${\text{H}}_{3}{\text{O}}^{\text{+}}$ in a 1 $\times$ 10⁻⁷M solution of HF. Setting up a table for the changes in concentration, we find:

	[HF]	[H₃O⁺]	[F⁻]
Initial concentration (M)	1 × 10⁻⁷	1 × 10⁻⁷	0
Change (M)	−x	x	x
Equilibrium (M)	1 × 10⁻⁷ − x	1 × 10⁻⁷+ x	x

Putting the concentrations into the equilibrium expression gives:

${K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{F}}^{-}\right]}{\left[\text{HF}\right]}=\frac{\left(1\times {10}^{-7}+x\right)x}{1\times {10}^{-7}-x}=7.2\times {10}^{-4}$

Recognizing that the initial concentration of HF, 1 $\times$ 10⁻⁷M, is very small and that K_a is not extremely small, we would expect that x cannot be neglected. Therefore, we will solve for x using the quadratic formula:

x² + 7.207 $\times$ 10⁻⁴x − 7.2 $\times$ 10⁻¹¹ = 0

$\begin{array}{ll}x\hfill & =\frac{-7.201\times {10}^{-4}\pm \sqrt{{\left(7.201\times {10}^{-4}\right)}^{\text{2}}-4\left(1\right)\left(-7.2\times {10}^{-11}\right)}}{2}\hfill \\ \hfill & =\frac{-7.201\times {10}^{-4}\pm 7.202999\times {10}^{-4}}{2}=9.995\times {10}^{-8}\hfill \end{array}$

As suspected, x is of the same order of magnitude as 1.0 $\times$ 10⁻⁷; therefore, it was necessary for us to use the quadratic formula. The ${\text{H}}_{3}{\text{O}}^{\text{+}}$ concentration in a 1.0 $\times$ 10⁻⁷M HF solution is:

${\text{H}}_{3}{\text{O}}^{\text{+}}$ = 1.0 $\times$ 10⁻⁷ + x = 1.0 $\times$ 10⁻⁷ + 0.9995 $\times$ 10⁻⁷ = 1.999 $\times$ 10⁻⁷M

Let’s calculate the ${\text{H}}_{3}{\text{O}}^{\text{+}}$ concentration in 1 $\times$ 10⁻⁶M HF solution. Again, because the concentration of HF is so small, we will consider the initial $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ to be 1 $\times$ 10⁻⁷M from the ionization of water. The change in concentrations is:

	[HF]	[H₃O⁺]	[F⁻]
Initial concentration (M)	1.0 × 10⁻⁶	1.0 × 10⁻⁶	0
Change (M)	−x	x	x
Equilibrium (M)	1.0 × 10⁻⁶ − x	1.0 × 10⁻⁶+ x	x

As we found in the last calculation, we would expect that x is not small relative to 1 $\times$ 10⁻⁶ and 1 $\times$ 10⁻⁷. Therefore, we will use the quadratic formula to solve for x:

${K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{F}}^{-}\right]}{\left[\text{HF}\right]}=\frac{\left(1.0\times {10}^{-7}+x\right)x}{1.0\times {10}^{-6}-x}=7.2\times {10}^{-4}$

x² + 7.201 $\times$ 10⁻⁴x − 7.2 $\times$ 10⁻¹⁰ = 0

$\begin{array}{ll}x\hfill & =\frac{-7.201\times {10}^{-4}\pm \sqrt{{\left(7.201\times {10}^{-4}\right)}^{\text{2}}-4\left(1\right)\left(-7.2\times {10}^{-10}\right)}}{2}\hfill \\ \hfill & =\frac{-7.201\times {10}^{-4}\pm 7.22097\times {10}^{-4}}{2}=9.98\times {10}^{-7}\hfill \end{array}$

The ${\text{H}}_{3}{\text{O}}^{\text{+}}$ concentration in a 1 $\times$ 10⁻⁶M HF solution is:

$\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = 1.0 $\times$ 10⁻⁷ + 9.98 $\times$ 10⁻⁷ = 1.10 $\times$ 10⁻⁶M

From this calculation we can see that the contribution of ${\text{H}}_{3}{\text{O}}^{\text{+}}$ from the self-ionization of water is becoming insignificant relative to the concentration of ${\text{H}}_{3}{\text{O}}^{\text{+}}$ generated from the ionization of HF. Thus, for all subsequent concentrations of HF greater than 10⁻⁶M, we will not have to consider the ionization of water. To make the plot indicated in this exercise, it is necessary to choose at least two more concentrations between 10⁻⁶M and 10⁻²M. We will do one more calculation of $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ at an HF concentration of 10⁻²M. However, this calculation will be done the same way for any concentration greater than 10⁻⁶M. The change in concentrations is:

	[HF]	[H₃O⁺]	[F⁻]
Initial concentration (M)	10⁻²	~0	0
Change (M)	−x	x	x
Equilibrium (M)	10⁻² − x	x	x

Putting these values in the equilibrium expression gives:

${K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{F}}^{-}\right]}{\left[\text{HF}\right]}=\frac{\left(x\right)\left(x\right)}{{10}^{-2}-x}=7.2\times {10}^{-4}$

x² + 7.2 $\times$ 10⁻⁴x − 7.2 $\times$ 10⁻⁶ = 0

$\begin{array}{ll}x\hfill & =\frac{-7.2\times {10}^{-4}\pm \sqrt{{\left(7.2\times {10}^{-4}\right)}^{\text{2}}-4\left(1\right)\left(-7.2\times {10}^{-6}\right)}}{2}\hfill \\ \hfill & =\frac{-7.2\times {10}^{-4}\pm 5.415\times {10}^{-3}}{2}=2.4\times {10}^{-3}\hfill \end{array}$

This value of x justifies our use of the quadratic formula rather than using the approximation method, for x, 2.35 $\times$ 10⁻³, is 23% of 10⁻²M.

Calculating the ${\text{H}}_{3}{\text{O}}^{\text{+}}$ concentration for a 10⁻²–M solution of HF, we find:

$\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = x = 2.34 $\times$ 10⁻³M

Plotting the values of $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ that we have calculated gives the following:

Question 2

Calculate the pH at the following points in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (K_a = 9.8 $\times$ 10⁻⁵) with 0.100 M KOH.

no KOH added
20 mL of KOH solution added
39 mL of KOH solution added
40 mL of KOH solution added
41 mL of KOH solution added

Show Answer

1. Let HA represent barbituric acid and A⁻ represent the conjugate base. The reaction and equilibrium constant are:

$\text{HA}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right){K}_{\text{a}}=9.8\times {10}^{-5}$

${K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=9.8\times {10}^{-5}$

The initial and equilibrium concentrations for this system can be written as follows:

	[HA]	[H₃O⁺]	[A⁻]
Initial concentration (M)	0.100	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.100 − x	1 × 10⁻⁷+ x	x

Substituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (0.100 − x) ≈ 0.100, gives:

$\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.100-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.100}=9.8\times {10}^{-5}$

Solving for x gives 3.13 $\times$ 10⁻³M. Because this value is less than 5% of 0.100, our assumption is correct. Therefore, $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = 3.13 $\times$ 10⁻³M:

pH = −log(3.13 $\times$ 10⁻³) = 2.504 = 2.50;

2. The moles of OH⁻ added are given by mol OH⁻ = M $\times$ V = (0.100 M) $\times$ (0.020 L) = 0.00200 mol

The initial moles of barbituric acid are given by mol HA = M $\times$ V = (0.100 M) $\times$ (0.040 L) = 0.00400 mol

Assume that the added hydroxide ion reacts completely with an equal number of moles of HA, forming an equal number of moles of A⁻ in the process. Thus, the moles of the ions are given by:

mol HA = 0.00400 − 0.00200 = 0.00200 mol
mol A⁻ = 0.00200 mol

The total volume is: 40.0 mL + 20.0 mL = 60.0 mL = 0.0600 L

The initial concentrations of the ions are given by:

$\begin{array}{l}{ }\left[\text{HA}\right]=\frac{0.00200\text{mol}}{0.0600\text{L}}=0.0333M\\ \left[{\text{A}}^{-}\right]=\frac{0.00200\text{mol}}{0.0600\text{L}}=0.0333M\end{array}$

The initial and equilibrium concentrations for this system can be written as follows:

	[HA]	[H₃O⁺]	[A⁻]
Initial concentration (M)	0.0333	0	0.0333
Change (M)	−x	x	x
Equilibrium (M)	0.0333 − x	x	0.0333 + x

Substituting the equilibrium concentrations into the equilibrium expression, and making the assumptions that (0.0333 − x) ≈ 0.0333 and (0.0333 + x) ≈ 0.0333, gives:

$\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{\left(x\right)\left(0.0333+x\right)}{\left(0.0333-x\right)}\approx \frac{\left(x\right)\left(0.0333\right)}{0.0333}=9.8\times {10}^{-5}$

Solving for x gives 9.8 $\times$ 10⁻⁵M. Because this value is less than 5% of 0.0333, our assumptions are correct. Therefore, $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = 9.8 $\times$ 10⁻⁵M:

pH = −log(9.8 $\times$ 10⁻⁵) = 4.009 = 4.01

3. The moles of OH⁻ added are given by mol OH⁻ = M $\times$ V = (0.100 M) $\times$ (0.039 L) = 0.00390 mol

The initial moles of barbituric acid are given by mol HA = M $\times$ V = (0.100 M) $\times$ (0.040 L) = 0.00400 mol

Assume that the added hydroxide ion reacts completely with an equal number of moles of HA, forming an equal number of moles of A⁻ in the process. Thus, the moles of the ions are given by:

mol HA = 0.00400 − 0.00390 = 0.00010 mol
mol A⁻ = 0.00930 mol

The total volume is 40.0 mL + 39.0 mL = 79.0 mL = 0.0790 L

The initial concentrations of the ions are given by:

$\begin{array}{l}{ }\left[\text{HA}\right]=\frac{0.00010\text{mol}}{0.0790\text{L}}=0.00127M\\ \left[{\text{A}}^{-}\right]=\frac{0.00390\text{mol}}{0.0790\text{L}}=0.0494M\end{array}$

The initial and equilibrium concentrations for this system can be written as follows:

	[HA]	[H₃O⁺]	[A⁻]
Initial concentration (M)	0.00127	0	0.0494
Change (M)	−x	x	x
Equilibrium (M)	0.00127 − x	x	0.0494 + x

Substituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (0.00127 − x) ≈ 0.00127 and (0.0494 + x) ≈ 0.0494, gives:

$\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}=\frac{\left(x\right)\left(0.0494+x\right)}{\left(0.00127-x\right)}\approx \frac{\left(x\right)\left(0.0494\right)}{0.00127}=9.8\times {10}^{-5}$

Solving for x gives 2.52 $\times$ 10⁻⁶M. Because this value is less than 5% of 0.00127 and 0.0494, our assumptions are correct. Therefore, $\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]$ = 2.52 $\times$ 10⁻⁶M:

pH = −log(2.52 $\times$ 10⁻⁶) = 5.599 = 5.60

4. The moles of OH⁻ added are given by mol OH⁻ = M $\times$ V = (0.100 M) $\times$ (0.040 L) = 0.00400 mol

The initial moles of barbituric acid are given by mol HA = M $\times$ V = (0.100 M) $\times$ (0.040 L) = 0.00400 mol

This is the equivalence point, where the moles of base added equal the moles of acid present initially. At the equivalence point, mol A⁻ = 0.00400 mol

The total volume is 40.0 mL + 40.0 mL = 80.0 mL = 0.0800 L

The initial concentration of the conjugate base is $\left[{\text{A}}^{-}\right]=\frac{0.00400\text{mol}}{0.0800\text{L}}=0.0500M$

The reaction and equilibrium constant are:

$\begin{array}{l}{ }{\text{A}}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons \text{HA}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\\ \\ {K}_{\text{b}}=\frac{\left[\text{HA}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{A}}^{-}\right]}=\frac{{K}_{\text{w}}}{{K}_{\text{a}}}=\frac{1.0\times {10}^{-14}}{9.8\times {10}^{-4}}=1.08\times {10}^{-10}\end{array}$

The initial and equilibrium concentrations for this system can be written as follows:

	[A⁻]	[HA]	[OH⁻]
Initial concentration (M)	0.0500	0	0
Change (M)	−x	x	x
Equilibrium (M)	0.0500 − x	x	x

Substituting the equilibrium concentrations into the equilibrium expression, and making the assumption that (0.0500 − x) ≈ 0.0500, gives:

$\frac{\left[\text{HA}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{A}}^{-}\right]}=\frac{\left(x\right)\left(x\right)}{\left(0.0500-x\right)}\approx \frac{\left(x\right)\left(x\right)}{0.0500}=1.02\times {10}^{-10}$

Solving for x gives 2.26 $\times$ 10⁻⁶M. Because this value is less than 5% of 0.0500, our assumption is correct. Therefore, [OH⁻] = 2.26 $\times$ 10⁻⁶M:

pOH = −log(2.26 $\times$ 10⁻⁶) = 5.646
pH = 14.000 − pOH = 14.000 − 5.646 = 8.354 = 8.35;

5. The moles of OH⁻ added are given by mol OH⁻ = M $\times$ V = (0.100 M) $\times$ (0.041 L) = 0.00410 mol

The initial moles of barbituric acid are given by mol HA = M $\times$ V = (0.100 M) $\times$ (0.040 L) = 0.00400 mol

This is past the equivalence point, where the moles of base added exceed the moles of acid present initially. The excess moles of hydroxide ion are given by mol OH⁻ = 0.00410 − 0.00400 = 0.00010 mol

The total volume is 40.0 mL + 41.0 mL = 81.0 mL = 0.0810 L

The concentration of OH⁻ is $\left[{\text{OH}}^{-}\right]=\frac{0.00010\text{mol}}{0.0810\text{L}}=0.0012M$

pOH = −log(0.0012) = 2.921
pH = 14.000 − pOH = 14.000 − 2.921 = 11.079 = 11.08

Glossary

acid-base indicator: organic acid or base whose color changes depending on the pH of the solution it is in

color-change interval: range in pH over which the color change of an indicator takes place

titration curve: plot of the pH of a solution of acid or base versus the volume of base or acid added during a titration

Titration of 25.00 mL of 0.100 M HCl (0.00250 mol of HCI) with 0.100 M NaOH. ↵
Titration of 25.00 mL of 0.100 M CH₃CO₂H (0.00250 mol of CH₃CO₂H) with 0.100 M NaOH. ↵

Acid-Based Equilibria

Learning Objectives

Titration Curve

Example 1: Calculating pH for Titration Solutions: Strong Acid/Strong Base

Check Your Learning

Example 2: Titration of a Weak Acid with a Strong Base

Check Your Learning

Acid-Base Indicators

Key Concepts and Summary

Exercises

Exercises

Question 2

Glossary

Candela Citations