# 14.7 Acid-base titrations  (Page 4/8)

 Page 4 / 8
$\text{pH}=p{K}_{\text{a}}+\text{log}\phantom{\rule{0.2em}{0ex}}\frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]}\phantom{\rule{0.2em}{0ex}}=\text{−log}\left({K}_{\text{a}}\right)+\text{log}\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\right]}{\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}\phantom{\rule{0.2em}{0ex}}=\text{−log}\left(1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\right)+\text{log}\left(1\right)$

(as the concentrations of ${\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}$ and CH 3 CO 2 H 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 p K a 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:

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

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.

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

## 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 $×$ 10 −9 M (pH<8.3). In more basic solutions where the hydronium ion concentration is less than 5.0 $×$ 10 −9 M (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:

$\begin{array}{ccc}\text{HIn}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)& \phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}& {\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{In}}^{\text{−}}\left(aq\right)\\ \phantom{\rule{0.5em}{0ex}}\text{red}\hfill & & \phantom{\rule{5.5em}{0ex}}\text{yellow}\hfill & \end{array}$
${K}_{a}=\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{In}}^{\text{−}}\right]}{\left[\text{HIn}\right]}\phantom{\rule{0.2em}{0ex}}=4.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}$

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:

$\frac{\left[{\text{In}}^{\text{−}}\right]}{\left[\text{HIn}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\left[\text{substance with yellow color}\right]}{\left[\text{substance with red color}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{{K}_{\text{a}}}{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}$

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

The above expression describing the indicator equilibrium can be rearranged:

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