This document provides an outline of the lab procedures and examples for calculations. It also provides a template of a data table suitable for the work students will do.
Very precise measurements require great care. It is possible to use a 50 mL buret to make measurements of differences in volume to the nearest 0.01 mL in a procedure called a titration. For a total volume larger than 10 mL this gives four significant figures. This degree of precision depends on the skill of the user, which takes time and practice to acquire. The main challenge is knowing when to stop adding the solution that is in the buret to the solution underneath it. This is normally determined by the change in color of an indicator at what is called the endpoint of a titration. An indicator is selected by the designer of a procedure to change color over a range of pH values that includes the value at the moment of stoichiometric equivalence. The person performing a titration aims to add the final drop that causes a permanent color change in such a way as to be sure not to have gone beyond equivalence. This is accomplished by adding the final amounts slowly, in small increments.
Solid sodium hydroxide (NaOH) attracts water from the water vapor in air, which adds to the mass measured when it is placed on a lab balance. In addition, the presence of this water allows minute amounts of carbon dioxide (CO2) to dissolve, which then reacts with the sodium hydroxide to make sodium carbonate (Na2CO3):
2NaOH + CO2 → Na2CO3 + H2OBecause the mass of solid sodium hydroxide as measured in the lab cannot be guaranteed to be 100% NaOH it is impossible to make a solution with a precise concentration. This is important because if a sodium hydroxide solution is going to be used to measure the concentration of an acid then it must have a precise concentration. To overcome this obstacle the solution must be standardized. This means that its concentration must be determined by reacting the solution with a precise amount of an acid so that both the volume of the sodium hydroxide solution and the number of moles in that volume may be known.
The usual primary standard used to standardize a sodium hydroxide solution is solid KHP (potassium hydrogen phthalate, KHC8H4O4). When dried in an oven KHP can be weighed on a balance and its mass can be used to precisely calculate the number of moles. Then it is possible to react the KHP with a sodium hydroxide solution to determine the number of moles of sodium hydroxide in a precisely measured volume of solution. A buret is used to measure the volume of solution to the nearest 0.01 mL. Phenolphthalein is an acid-base indicator which changes color at just the right pH to show show when the reaction of KHP with NaOH is complete. The chemical reaction below shows the neutralization of KHP by NaOH (the KHP is shown as the anion of the salt, leaving out the potassium ion and sodium ions).
OH– + HC8H4O4– → C8H4O42– + H2OHere is an example calculation to find the concentration of an approximately 0.5 M solution of sodium hydroxide using a mass of KHP of 1.057 g. The measured volume of NaOH solution is 10.52 mL.
1 mol KHP 1 mol NaOH 1.057 g × ------------ × ---------- = 5.176 × 10–3 204.22 g 1 mol KHP mol NaOH |
so the concentration is 5.176 × 10–3 mol NaOH ----------------------- = 0.4920 M 0.01052 L |
Note that since your sodium hydroxide solution should have a concentration near 0.5 M the volume of that solution needed to react with approx. 1 gram of KHP is near 10 mL. This suggests that you add up to 8.5 or even 9 mL before slowly dripping in the last of the solution to precisely determine the end point volume of the titration.
Vinegar is a solution of acetic acid (CH3COOH) in water. As a pure substance, acetic acid is extremely corrosive and has an overwhelming odor of vinegar. It is sold in stores in diluted form with a concentration of 5% by mass. This means that 5 g out of every 100 g of the mixture is acetic acid. You will use the sodium hydroxide solution you standardized in Part 1, along with a stoichiometric calculation, to measure by titration the number of moles of acetic acid per liter of vinegar solution. You will also determine the mass % of acetic acid in the vinegar by weighing the sample of vinegar, calculating the mass of acetic acid dissolved in the sample, and calculating a percent. The neutralization reaction is written below, followed by an example showing how to calculate the concentration of acetic acid and an example showing how to calculate the mass % of acetic acid.
CH3COOH + NaOH → NaCH3COO + H2OGiven a standardized sodium hydroxide solution with a concentration of 0.4920 M and a volume of 10.00 mL of vinegar, what is the concentration of the acetic acid in the vinegar if 16.56 mL of sodium hydroxide solution are required to reach the endpoint of the titration?
1 L 0.4920 mol NaOH 1 mol CH3COOH 16.56 mL × ---------- × ---------------- × ---------------- = 8.148 × 10–3 mol CH3COOH 1000 mL 1 L 1 mol NaOH
So the concentration of acetic acid in the vinegar is: (8.148 × 10–3 mol CH3COOH)/(0.0100 L) = 0.8148 M.
To calculate the mass % and compare your result with the label on the bottle it is necessary also to know the mass of the vinegar solution. Usually, this is very nearly precisely 10.00 g since the acetic acid adds little mass at only 5% and the density of the water it is dissolved in is 1.00 g/mL. For the purposes of this example, the sample was found to have a mass of 10.023 g. The first step is to find the number of grams of acetic acid in the sample. Then the mass % may be calculated:
60.035 g 8.148 × 10–3 mol × ------------ = 0.4891 g CH3COOH 1 mol |
so the mass % is 0.4891 g ---------- × 100% = 4.880% 10.023 g |
Since there is a lot to keep track of, here is a template for a data table you may use for this lab. It may be simplest to cut it out and paste it into your lab notebook.
Standardization of NaOH |
Trial 1 | Trial 2 |
Determination of Acetic Acid |
Trial 1 | Trial 2 | |
Mass of KHP (g) | Mass of Flask (g) | |||||
mol KHP | Mass of Flask + Vinegar (g) | |||||
Initial Vol. NaOH (Vi) (mL) | Initial Vol. NaOH (Vi) (mL) | |||||
Final Vol. NaOH (Vf) (mL) | Final Vol. NaOH (Vf) (mL) | |||||
Vol. of NaOH (ΔV = Vf – Vi) (mL) | Vol. of NaOH (ΔV = Vf – Vi) (mL) | |||||
mol NaOH | mol NaOH | |||||
Conc. NaOH (M) | mol CH3COOH | |||||
Average Conc. NaOH (M) | Conc. CH3COOH (M) | |||||
% Diff. in Conc. NaOH | Average Conc. CH3COOH (M) | |||||
% Diff. in Conc. CH3COOH | ||||||
Mass Percent CH3COOH | ||||||
% Error compared to 5% |
Calculating Percent Difference This compares the absolute value of the difference between two experimental values to their average: |(value 1) – (value 2)| ------------------------ × 100% average value |
Calculating Percent Error This compares the experimental value to the accepted or standard value: |(std. value) – (exp. value)| ------------------------ × 100% std. value |
In addition to the questions in the lab handout, please answer the following questions.