# 15.06: Acid-Base Titrations

Learning Objectives

• Describe a titration experiment.
• Explain what an indicator does.
• Perform a titration calculation correctly.

The reaction of an acid with a base to make a salt and water is a common reaction in the laboratory, partly because so many compounds can act as acids or bases. Another reason that acid-base reactions are so prevalent is because they are often used to determine quantitative amounts of one or the other. Performing chemical reactions quantitatively to determine the exact amount of a reagent is called a titration. A titration can be performed with almost any chemical reaction for which the balanced chemical equation is known. Here, we will consider titrations that involve acid-base reactions.

In a titration, one reagent has a known concentration or amount, while the other reagent has an unknown concentration or amount. Typically, the known reagent (the titrant) is added to the unknown quantity and is dissolved in solution. The unknown amount of substance (the analyte) may or may not be dissolved in solution (but usually is). The titrant is added to the analyte using a precisely calibrated volumetric delivery tube called a burette (also spelled buret; see Fig. 12.4.1). The burette has markings to determine how much volume of solution has been added to the analyte. When the reaction is complete, it is said to be at the equivalence point; the number of moles of titrant can be calculated from the concentration and the volume, and the balanced chemical equation can be used to determine the number of moles (and then concentration or mass) of the unknown reactant. Fig. 12.4.1 Equipment for Titrations. A burette is a type of liquid dispensing system that can accurately indicate the volume of liquid dispensed

For example, suppose 25.66 mL (or 0.02566 L) of 0.1078 M HCl was used to titrate an unknown sample of NaOH. What mass of NaOH was in the sample? We can calculate the number of moles of HCl reacted:

# mol HCl = (0.02566 L)(0.1078 M) = 0.002766 mol HCl

We also have the balanced chemical reaction between HCl and NaOH:

HCl + NaOH → NaCl + H2O

So we can construct a conversion factor to convert to number of moles of NaOH reacted:

$0.002766\,\cancel{mol\, HCl}\times \frac{1\, mol\, NaOH}{1\,\cancel{mol\, HCl}}=0.002766\, mol\, NaOH$

Then we convert this amount to mass, using the molar mass of NaOH (40.00 g/mol):

$0.002766\,\cancel{mol\, HCl}\times \frac{40.00\, g\, NaOH}{1\,\cancel{mol\, HCl}}=0.1106\, g\, NaOH$

This is type of calculation is performed as part of a titration.

Example $$\PageIndex{1}$$:

What mass of Ca(OH)2 is present in a sample if it is titrated to its equivalence point with 44.02 mL of 0.0885 M HNO3? The balanced chemical equation is as follows:

2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O

Solution

In liters, the volume is 0.04402 L. We calculate the number of moles of titrant:

# moles HNO3 = (0.04402 L)(0.0885 M) = 0.00390 mol HNO3

Using the balanced chemical equation, we can determine the number of moles of Ca(OH)2 present in the analyte:

$0.00390\,\cancel{mol\, HNO_{3}}\times \frac{1\, mol\, Ca(OH)_{2}}{2\,\cancel{mol\, HNO_{3}}}=0.00195\, mol\, Ca(OH)_{2}$

Then we convert this to a mass using the molar mass of Ca(OH)2:

$0.00195\,\cancel{mol\, Ca(OH)_{2}}\times \frac{74.1\, g\, Ca(OH)_{2}}{\cancel{mol\, Ca(OH)_{2}}}=0.144\, g\, Ca(OH)_{2}$

Exercise $$\PageIndex{1}$$

What mass of H2C2O4 is present in a sample if it is titrated to its equivalence point with 18.09 mL of 0.2235 M NaOH? The balanced chemical reaction is as follows:

H2C2O4 + 2NaOH → Na2C2O4 + 2H2O