29

# Titrations II

### Learning Objectives

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

• Interpret titration curves for strong and weak acid-base systems
• Compute sample pH at important stages of a titration

### EXAMPLE 29.0.1

#### Titration of a Weak Acid with a Strong Base

Consider the titration of 25.00 mL of 0.100 M CH3CO2H with 0.100 M NaOH. The reaction can be represented as:
${\text{CH}}_{3}{\text{CO}}_{2}\text{H}+{\text{OH}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}+{\text{H}}_{2}\text{O}$

Calculate the pH of the titration solution after the addition of the following volumes of NaOH titrant:

(a) 0.00 mL

(b) 25.00 mL

(c) 12.50 mL

(d) 37.50 mL

#### Solution

(a) The initial pH is computed for the acetic acid solution in the usual ICE approach:

${K}_{\text{a}}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}\phantom{\rule{0.2em}{0ex}}\approx \phantom{\rule{0.2em}{0ex}}\frac{{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}^{\text{2}}}{{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}_{0}}\phantom{\rule{0.2em}{0ex}},$ 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) The acid and titrant are both monoprotic and the sample and titrant solutions are equally concentrated; thus, this volume of titrant represents the equivalence point. Unlike the strong-acid example above, however, the reaction mixture in this case contains a weak conjugate base (acetate ion). The solution pH is computed considering the base ionization of acetate, which is present at a concentration of

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

Base ionization of acetate is represented by the equation

${\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{OH}}^{\text{−}}\left(aq\right)$
${K}_{\text{b}}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{\text{−}}\right]}{{K}_{\text{a}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{{K}_{\text{w}}}{{K}_{\text{a}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}5.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-10}$

Assuming x << 0.0500, the pH may be calculated via the usual ICE approach: ${K}_{\text{b}}=\phantom{\rule{0.2em}{0ex}}\frac{{x}^{\text{2}}}{0.0500\phantom{\rule{0.2em}{0ex}}M}$

$x=\left[{\text{OH}}^{\text{−}}\right]=5.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}$
$\text{pOH}=\text{−log}\left(5.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\right)=5.28$
$\text{pH}=14.00-5.28=8.72$

Note that the pH at the equivalence point of this titration is significantly greater than 7, as expected when titrating a weak acid with a strong base.

(c) Titrant volume = 12.50 mL. This volume represents one-half of the stoichiometric amount of titrant, and so one-half of the acetic acid has been neutralized to yield an equivalent amount of acetate ion. The concentrations of these conjugate acid-base partners, therefore, are equal. A convenient approach to computing the pH is use of the Henderson-Hasselbalch equation:

$\text{pH}=p{K}_{\text{a}}+\text{log}\phantom{\rule{0.2em}{0ex}}\frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]}\phantom{\rule{0.2em}{0ex}}=\text{−log}\left({K}_{\text{a}}\right)+\text{log}\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}\phantom{\rule{0.2em}{0ex}}=\text{−log}\left(1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\right)+\text{log}\left(1\right)$
$\text{pH}=\text{−log}\left(1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\right)=4.74$

(pH = pKa at the half-equivalence point in a titration of a weak acid)

(d) Titrant volume = 37.50 mL. This volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant. In such solutions, the solution pH is determined primarily by the amount of excess strong base:

$\left[{\text{OH}}^{\text{−}}\right]=\phantom{\rule{0.2em}{0ex}}\frac{\left(\text{0.003750 mol}-\text{0.00250 mol}\right)}{\text{0.06250 L}}\phantom{\rule{0.2em}{0ex}}=2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}M$
$\text{pOH}=\text{−log}\left(2.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\right)=\text{1.70, and pH}=14.00-1.70=12.30$

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.

0.00 mL: 2.37; 15.0 mL: 3.92; 25.00 mL: 8.29; 30.0 mL: 12.097

Performing calculations similar to those in the preceding example permits a more full assessment of titration curves. A summary of pH/volume data pairs for the strong and weak acid titrations is provided in Table 29.1 and plotted as titration curves in Figure 29.1. A comparison of these two curves illustrates several important concepts that are best addressed by identifying the four stages of a titration:

initial state (added titrant volume = 0 mL): pH is determined by the acid being titrated; because the two acid samples are equally concentrated, the weak acid will exhibit a greater initial pH

pre-equivalence point (0 mL < V < 25 mL): solution pH increases gradually and the acid is consumed by reaction with added titrant; composition includes unreacted acid and the reaction product, its conjugate base

equivalence point (V = 25 mL): a drastic rise in pH is observed as the solution composition transitions from acidic to either neutral (for the strong acid sample) or basic (for the weak acid sample), with pH determined by ionization of the conjugate base of the acid

postequivalence point (V > 25 mL): pH is determined by the amount of excess strong base titrant added; since both samples are titrated with the same titrant, both titration curves appear similar at this stage.

Table 29.1

pH Values in the Titrations of a Strong Acid and of a Weak Acid

Volume of 0.100 M NaOH Added (mL) Moles of NaOH Added pH Values 0.100 M HCl1 pH Values 0.100 M CH3CO2H2
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
Figure 29.2

(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.

### Suggested Citation

General College Chemistry. https://edtechbooks.org/general_college_chemistry_2

Previous Version(s)

Flowers, P., et al. (2019). Chemistry: Atoms First 2e. https://openstax.org/details/books/chemistry-atoms-first-2e (14.7)

CC BY: This work is released under a CC BY license, which means that you are free to do with it as you please as long as you properly attribute it.

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