8.4: 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 many compounds can act as acids or bases. The acid-base reaction can be used to determine quantitative amounts of an acid or a base. 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 dissolved in solution and added to the unknown quantity. 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 Figure 8.4.1). 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.
For example, suppose 25.66 mL (or 0.02566 L) of 0.1078 M HCl(aq) was used to titrate an NaOH(aq) sample of unknown concentration. 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(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
So we can construct a conversion factor to convert to number of moles of NaOH reacted:
[latex]0.002766\cancel{mol\, HCl}\times \frac{1\, mol\, NaOH}{1\cancel{mol\, HCl}}=0.002766\, mol\, NaOH\nonumber[/latex]
Then we convert this amount to mass, using the molar mass of NaOH (40.00 g/mol):
[latex]0.002766\cancel{mol\, HCl}\times \frac{40.00\, g\, NaOH}{1\cancel{mol\, HCl}}=0.1106\, g\, NaOH\nonumber[/latex]
This type of calculation is performed as part of a titration.
Example 8.4.1
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:
2 HNO3 (aq) + Ca(OH)2 (aq) → Ca(NO3)2 (aq) + 2 H2O (l)
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:
[latex]0.00390\cancel{mol\, HNO_{3}}\times \frac{1\, mol\, Ca(OH)_{2}}{2\cancel{mol\, HNO_{3}}}=0.00195\, mol\, Ca(OH)_{2}\nonumber[/latex]
Then we convert this to a mass using the molar mass of Ca(OH)2:
[latex]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}\nonumber[/latex]
Exercise 8.4.1
What mass of H2C2O4 (oxalic acid) 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 (aq)+ 2 NaOH (aq) → Na2C2O4 (aq) + 2 H2O (l)
Answer
0.182 g
How does one know if a reaction is at its equivalence point? Usually, the person performing the titration adds a small amount of an indicator, a substance that changes color depending on the acidity or basicity of the solution. Because different indicators change colors at different levels of acidity, choosing the correct one is important in performing an accurate titration.
The videos that follow the Key Takeaway show some tips on how to properly perform a titration, the glassware involved, and an example calculation.
