# 14.7 Acid-base titrations

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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
• 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.

## 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 [link] . 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 H 3 O + 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 H 3 O + 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:

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

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

The number of moles of H 3 O + becomes:

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

The concentration of H 3 O + is:

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

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 H 3 O + 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 H 3 O + 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:

$\begin{array}{}\\ \\ \left[{\text{OH}}^{\text{−}}\right]=\phantom{\rule{0.2em}{0ex}}\frac{\text{n}\left({\text{OH}}^{\text{−}}\right)}{V}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{0.100\phantom{\rule{0.4em}{0ex}}M\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{X mL}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\frac{\text{1 L}}{\text{1000 mL}}\right)\phantom{\rule{0.2em}{0ex}}-\text{0.002500 mol}}{\left(\text{25.00 mL}+\text{X mL}\right)\left(\phantom{\rule{0.2em}{0ex}}\frac{\text{1 L}}{\text{1000 mL}}\right)\phantom{\rule{0.2em}{0ex}}}\phantom{\rule{0.2em}{0ex}}\\ =\phantom{\rule{0.2em}{0ex}}\frac{0.100\phantom{\rule{0.4em}{0ex}}M\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{X mL}-\text{0.002500 mol}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\phantom{\rule{0.2em}{0ex}}\frac{\text{1000 mL}}{\text{1 L}}\right)\phantom{\rule{0.2em}{0ex}}}{\text{25.00 mL}+\text{X mL}}\phantom{\rule{0.2em}{0ex}}\end{array}$
$\text{pH}=14-\text{pOH}=14+\text{log}\left(\left[{\text{OH}}^{\text{−}}\right]\right)$

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

(a) X = 0 mL

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

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