Now that we know how to titrate, let's look at some ways it is used in the lab.  All you need is some sort of unknown acid and a standardized base solution.  You might want to review the procedure for standardizing a base from the last lab, because that's going to be the first thing you need to do in this lab.

Last week's lab was pretty straightforward because the unknown was the Molarity of an acid solution and there was a simple equation to solve for that.  This week, we'll try some other things.  First, we'll take a solid crystalline acid and use titration to find its molar mass.  Then we'll take an acid that is part of a mixture and use titration to find the percent acidity of the solution.

Part A: Identification of a Carboxylic Acid

Carboxylic acids are organic compounds of the form

R – C = O
|
    OH

where R represents some carbon-based chain.  The H is easily ionized from the OH part of the molecule, producing H+ ions and making the compound an acid.  Many carboxylic acids are crystalline solids, although some like acetic acid (found in vinegar) are liquids.

Carboxylic acids can be titrated with a base just like any other acid.  We can use the equation

n = MV

to find the number of moles of carboxylic acid present in the sample.  If we already know the mass in grams of the sample, we can calculate the sample’s molecular mass using the equation

M = m/n

where M is the molecular mass, m is the mass of the sample in grams, and n is the number of moles in the sample.  We can then compare the molecular mass with known carboxylic acids to identify our sample.

The calculations are relatively straightforward as long as the carboxylic acid is monoprotic.  We are then assured of a one-to-one stoichiometric ratio in the titration and the mole calculation works fine.  But what if the acid is diprotic?  Then, two moles of base are needed to neutralize each mole of acid.  The mole calculation (n = MV) actually gives us the number of moles of base, so we actually have half as many moles of acid as we expected.  Plugging that into the molecular mass equation (M = m/n) we find that the actual molecular mass is two times the calculated value.  If the acid is triprotic, the molecular mass is three times larger.

Methodology

1.	Prepare 250 mL of a 0.100 M NaOH solution in a volumetric flask.  Fill a buret with the solution.

2.	Standardize the solution with approximately 0.40 g of KHP.

3.	Obtain approximately 0.50 g of an unknown carboxylic acid in an Erlenmeyer flask.  Dissolve the acid in approximately 100 mL of distilled water.  Some acids will require additional water or even gentle heating if they do not dissolve.  When the acid is completely dissolved, add 2 drops of phenolphthalein.

4.	Titrate the NaOH solution into the carboxylic acid until a pink color persists.  Record the volume of NaOH used and calculate the molecular mass of the acid.

5.	If time permits, repeat steps 3 and 4 with a different unknown carboxylic acid.

Analysis

Find the moles of carboxylic acid using

n = MV

where M is the Molarity of your NaOH solution and V is the volume used from the buret.  Make sure you express your volume in liters.  Now you can find the molecular mass by dividing the grams of sample by the number of moles.

Your unknown will be one of the following carboxylic acids:

Monoprotic: ascorbic acid, benzoic acid, salicylic acid
Diprotic: adipic acid, oxalic acid dehydrate, tartaric acid
Triprotic: citric acid monohydrate

You will need to look up the molecular masses of these acids and compare your experimental molecular mass to these known values.  Remember that you will need to multiply your molecular mass by 2 if you suspect that it is diprotic or by 3 if you suspect that it is triprotic.

Part B: Percent Acetic Acid in Vinegar

Vinegar is a dilute solution of acetic acid (CH3COOH) in water.  Acetic acid is a weak monoprotic acid that can be titrated against a base like any other acid.  Most commercial vinegar brands are somewhere around 5% acetic acid by mass.  Since they go by mass, we'll need to measure out our vinegar sample by grams rather than milliliters.  Then we can titrate the sample against your standardized NaOH.

The math is easy in this one.  We can use the equation n = MV to find the number of moles of acetic acid present in the vinegar sample.  Once we know the moles of acetic acid, we can convert to grams of acetic acid and compare it to the total mass of the vinegar to find the percent of acetic acid in the sample.

Methodology

1.	Obtain 2.50 g of vinegar in an Erlenmeyer flask.  Add approximately 100 mL of distilled water and 2 drops of phenolphthalein.

2.	Titrate your standardized NaOH solution into the vinegar until a pink color persists.  Record the volume of NaOH used and calculate the percent acetic acid in the vinegar.

3.	If time permits, repeat steps 1 and 2 and average your results.

Analysis

Find the moles of acetic acid using

n = MV

where M is the Molarity of your NaOH solution and V is the volume used from the buret in liters.  Now convert the moles to grams using

m = nM

where m is the mass in grams, n is the number of moles, and M is the molar mass of acetic acid (60.0 g/mol).  The percent acetic acid is just the mass of acetic acid divided by the total mass of the sample, then multiplied by 100.  Find the actual percent acetic acid from the container and calculate your relative error.

Questions

1.	Could you have done part A if your carboxylic acid had been completely insoluble in water?  Why or why not?

2.	Why don't we need to know the Molarity of the carboxylic acid solution in part A?

3.	If acetic acid was diprotic, how would your calculations in part B change?

4.	In this lab we used distilled vinegar, which is colorless.  How would it have affected your experiment if we had used apple cider vinegar, which is darkly colored?

5.	We've titrated acids that are solids and acids that are liquids.  Can you come up with a way to titrate an acid that is a gas?