Chemistry » Acid-Base Equilibria » Acid-Base Titrations

Titration Curve

Acid-Base Titrations

As seen in the tutorial 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

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 the figure below. Calculate the pH at these volumes of added base solution:

(a) 0.00 mL

(b) 12.50 mL

(c) 25.00 mL

(d) 37.50 mL

Solution

Since HCl is a strong acid, we can assume that all of it dissociates. The initial concentration of H3O+ is \({\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}_{0}=0.100\phantom{\rule{0.4em}{0ex}}M.\) When the base solution is added, it also dissociates completely, providing OH ions. The H3O+ 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:

\(\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}={\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}_{0}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{0.02500 L}=\text{0.002500 mol}\)

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

\(\text{n}{\left({\text{OH}}^{\text{−}}\right)}_{0}=0.100\phantom{\rule{0.4em}{0ex}}M\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{X mL}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 L}}{\text{1000 mL}}\right)\phantom{\rule{0.2em}{0ex}}\)

The total volume becomes: \(V=\left(\text{25.00 mL}+\text{X mL}\right)\left(\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 L}}{\text{1000 mL}}\right)\phantom{\rule{0.2em}{0ex}}\)

The number of moles of H3O+ becomes:

\(\text{n}\left({\text{H}}^{\text{+}}\right)=\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}-\text{n}{\left({\text{OH}}^{\text{−}}\right)}_{0}=\text{0.002500 mol}-0.100\phantom{\rule{0.4em}{0ex}}M\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{X mL}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 L}}{\text{1000 mL}}\right)\phantom{\rule{0.2em}{0ex}}\)

The concentration of H3O+ is:

\(\begin{array}{}\\ \\ \left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{n}\left({\text{H}}^{\text{+}}\right)}{V}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{0.002500 mol}-0.100\phantom{\rule{0.4em}{0ex}}M\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{X mL}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 L}}{\text{1000 mL}}\right)\phantom{\rule{0.2em}{0ex}}}{\left(\text{25.00 mL}+\text{X mL}\right)\left(\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 L}}{\text{1000 mL}}\right)\phantom{\rule{0.2em}{0ex}}}\phantom{\rule{0.2em}{0ex}}\\ =\phantom{\rule{0.2em}{0ex}}\cfrac{\text{0.002500 mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1000 mL}}{\text{1 L}}\right)\phantom{\rule{0.2em}{0ex}}-0.100\phantom{\rule{0.4em}{0ex}}M\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{X mL}}{\text{25.00 mL}+\text{X mL}}\phantom{\rule{0.2em}{0ex}}\end{array}\)

\(\text{pH}=\text{−log}\left(\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\right)\)

The preceding calculations work if \(\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}-\text{n}{\left({\text{OH}}^{\text{−}}\right)}_{0}>0\) and so n(H+) > 0. When \(\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}=\text{n}{\left({\text{OH}}^{\text{−}}\right)}_{0},\) the H3O+ 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:

\(\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\left[{\text{OH}}^{\text{−}}\right],\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]={K}_{\text{w}}=1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−14}};\phantom{\rule{0.2em}{0ex}}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−7}}\)

\(\text{pH}=\text{−log}\left(1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{\text{−7}}\right)=7.00\)

Finally, when \(\text{n}{\left({\text{OH}}^{\text{−}}\right)}_{0}>\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0},\) there are not enough H3O+ ions to neutralize all the OH ions, and instead of \(\text{n}\left({\text{H}}^{\text{+}}\right)=\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}-\text{n}{\left({\text{OH}}^{\text{−}}\right)}_{0},\) we calculate: \(\text{n}\left({\text{OH}}^{\text{−}}\right)=\text{n}{\left({\text{OH}}^{\text{−}}\right)}_{0}-\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}\)

In this case:

\(\begin{array}{}\\ \\ \left[{\text{OH}}^{\text{−}}\right]=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{n}\left({\text{OH}}^{\text{−}}\right)}{V}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{0.100\phantom{\rule{0.4em}{0ex}}M\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{X mL}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 L}}{\text{1000 mL}}\right)\phantom{\rule{0.2em}{0ex}}-\text{0.002500 mol}}{\left(\text{25.00 mL}+\text{X mL}\right)\left(\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 L}}{\text{1000 mL}}\right)\phantom{\rule{0.2em}{0ex}}}\phantom{\rule{0.2em}{0ex}}\\ =\phantom{\rule{0.2em}{0ex}}\cfrac{0.100\phantom{\rule{0.4em}{0ex}}M\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{X mL}-\text{0.002500 mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1000 mL}}{\text{1 L}}\right)\phantom{\rule{0.2em}{0ex}}}{\text{25.00 mL}+\text{X mL}}\phantom{\rule{0.2em}{0ex}}\end{array}\)

\(\text{pH}=14-\text{pOH}=14+\text{log}\left(\left[{\text{OH}}^{\text{−}}\right]\right)\)

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

(a) X = 0 mL

\(\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{n}\left({\text{H}}^{\text{+}}\right)}{V}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{0.002500 mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1000 mL}}{\text{1 L}}\right)\phantom{\rule{0.2em}{0ex}}}{\text{25.00 mL}}\phantom{\rule{0.2em}{0ex}}=0.1\phantom{\rule{0.4em}{0ex}}M\)

pH = −log(0.100) = 1.000

(b) X = 12.50 mL

\(\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{n}\left({\text{H}}^{\text{+}}\right)}{V}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{0.002500 mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1000 mL}}{\text{1 L}}\right)\phantom{\rule{0.2em}{0ex}}-0.100\phantom{\rule{0.4em}{0ex}}M\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{12.50 mL}}{\text{25.00 mL}+\text{12.50 mL}}\phantom{\rule{0.2em}{0ex}}=0.0333\phantom{\rule{0.4em}{0ex}}M\)

pH = −log(0.0333) = 1.477

(c) X = 25.00 mL

Since the volumes and concentrations of the acid and base solutions are the same: \(\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}=\text{n}{\left({\text{OH}}^{\text{−}}\right)}_{0},\) and pH = 7.000, as described earlier.

(d) X = 37.50 mL

In this case:

\(\text{n}{\left({\text{OH}}^{\text{−}}\right)}_{0}>\text{n}{\left({\text{H}}^{\text{+}}\right)}_{0}\)

\(\left[{\text{OH}}^{\text{−}}\right]=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{n}\left({\text{OH}}^{\text{−}}\right)}{V}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{0.100\phantom{\rule{0.4em}{0ex}}M\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{35.70 mL}-\text{0.002500 mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1000 mL}}{\text{1 L}}\right)\phantom{\rule{0.2em}{0ex}}}{\text{25.00 mL}+\text{37.50 mL}}\phantom{\rule{0.2em}{0ex}}=0.0200\phantom{\rule{0.4em}{0ex}}M\)

pH = 14 − pOH = 14 + log([OH]) = 14 + log(0.0200) = 12.30

In the example, we calculated pH at four points during a titration. The table below shows a detailed sequence of changes in the pH of a strong acid and a weak acid in a titration with NaOH.

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 AddedpH Values 0.100 M HClpH Values 0.100 M CH3CO2H
0.00.01.002.87
5.00.000501.184.14
10.00.001001.374.57
15.00.001501.604.92
20.00.002001.955.35
22.00.002202.205.61
24.00.002402.696.13
24.50.002453.006.44
24.90.002493.707.14
25.00.002507.008.72
25.10.0025110.3010.30
25.50.0025511.0011.00
26.00.0026011.2911.29
28.00.0028011.7511.75
30.00.0030011.9611.96
35.00.0035012.2212.22
40.00.0040012.3612.36
45.00.0045012.4612.46
50.00.0050012.5212.52

The simplest acid-base reactions are those of a strong acid with a strong base. the table above 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 the figure below, 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 observed when the amount of base added is equivalent to the amount of acid in the sample according to the stoichiometry of the titration reaction. The volume of titrant containing this stoichiometric amount of base is called the equivalence point for the titration (see the earlier tutorial on stoichiometry). 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 (the table above and the figure below).

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.”

(a) 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 7.00 pH. (b) The titration curve for the titration of 25.00 mL of 0.100 M acetic acid (weak 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. The table above gives the pH values during the titration, the figure above 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:

\({\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(l\right)+{\text{OH}}^{\text{−}}\left(aq\right)\)

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

Example

Titration of a Weak Acid with a 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.”

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.

The titration curve shown in the figure above is for the titration of 25.00 mL of 0.100 M CH3CO2H with 0.100 M NaOH. The reaction can be represented as:

\({\text{CH}}_{3}{\text{CO}}_{2}\text{H}+{\text{OH}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}+{\text{H}}_{2}\text{O}\)

(a) What is the initial pH before any amount of the NaOH solution has been added? Ka = 1.8 \(×\) 10−5 for CH3CO2H.

(b) Find the pH after 25.00 mL of the NaOH solution have been added.

(c) Find the pH after 12.50 mL of the NaOH solution has been added.

(d) Find the pH after 37.50 mL of the NaOH solution has been added.

Solution(a) Assuming that the dissociated amount is small compared to 0.100 M, we find that:

\({K}_{\text{a}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}\phantom{\rule{0.2em}{0ex}}\approx \phantom{\rule{0.2em}{0ex}}\cfrac{{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}^{\text{2}}}{{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}_{0}}\phantom{\rule{0.2em}{0ex}},\) and \(\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\sqrt{{K}_{a}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}=\sqrt{1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0.100}=1.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\)

\(\text{pH}=\text{−log}\left(1.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\right)=2.87\)

(b) 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 \({\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}.\) The concentration of the \({\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\) ion is:

\(\cfrac{\text{0.00250 mol}}{\text{0.0500 L}}\phantom{\rule{0.2em}{0ex}}=\text{0.0500 M}{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\)

The equilibrium that must be focused on now is the basicity equilibrium for \({\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}:\)

\({\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)\)

so we must determine Kb for the base by using the ion product constant for water:

\({K}_{\text{b}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]\left[{\text{OH}}^{\text{−}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]}\)

\({K}_{\text{a}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}\phantom{\rule{0.2em}{0ex}},\phantom{\rule{0.2em}{0ex}}\text{so}\phantom{\rule{0.4em}{0ex}}\cfrac{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{H}}^{\text{+}}\right]}{{K}_{\text{a}}}.\)

Since Kw = [H+][OH]:

\({K}_{\text{b}}=\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{−}}\right]}{{K}_{\text{a}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{K}_{\text{w}}}{{K}_{\text{a}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\cfrac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}\)

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, \({K}_{\text{b}}=\phantom{\rule{0.2em}{0ex}}\cfrac{{x}^{\text{2}}}{0.0500\phantom{\rule{0.2em}{0ex}}M}\phantom{\rule{0.2em}{0ex}},\) and then:

\(x=\left[{\text{OH}}^{\text{−}}\right]=5.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\)

\(\text{pOH}=\text{−log}\left(5.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\right)=5.28\)

\(\text{pH}=14.00-5.28=8.72\)

Note that the pH at the equivalence point of this titration is significantly greater than 7.

(c) In (b), 25.00 mL of the NaOH solution was added, and so practically all the CH3CO2H was converted into \({\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}.\) 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 \({\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}.\) The total initial number of moles of CH3CO2H is 0.02500L \(×\) 0.100 M = 0.00250 mol, and so after adding the NaOH, the numbers of moles of CH3CO2H and \({\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\) are both approximately equal to \(\cfrac{\text{0.00250 mol}}{2}\phantom{\rule{0.2em}{0ex}}=\text{0.00125 mol},\) 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:

\(\text{pH}=p{K}_{\text{a}}+\text{log}\phantom{\rule{0.2em}{0ex}}\cfrac{\left[\text{Base}\right]}{\left[\text{Acid}\right]}\phantom{\rule{0.2em}{0ex}}=\text{−log}\left({K}_{\text{a}}\right)+\text{log}\phantom{\rule{0.2em}{0ex}}\cfrac{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}\phantom{\rule{0.2em}{0ex}}=\text{−log}\left(1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\right)+\text{log}\left(1\right)\)

(as the concentrations of \({\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\) and CH3CO2H are the same)

Thus:

\(\text{pH}=\text{−log}\left(1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\right)=4.74\)

(the pH = the pKa at the halfway point in a titration of a weak acid)

(d) After 37.50 mL of NaOH is added, the amount of NaOH is 0.03750 L \(×\) 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:

\(\left[{\text{OH}}^{\text{−}}\right]=\phantom{\rule{0.2em}{0ex}}\cfrac{\left(\text{0.003750 mol}-\text{0.00250 mol}\right)}{\text{0.06250 L}}\phantom{\rule{0.2em}{0ex}}=2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}M\)

So:

\(\text{pOH}=\text{−log}\left(2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\right)=\text{1.70, and pH}=14.00-1.70=12.30\)

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.

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