Counting the number of end groups: Acid-base titrations

Recall from general chemistry laboratory those titration experiments? This analysis method works by neutralizing an unknown concentration of acid or base with a known concentration of acid or base. For example, we may have a polymer that has terminal carboxylic acid groups, which would give an acidic solution when dissolved in water. We could add a strong base, like NaOH, to our sample and neutralize the carboxylic acid groups. If we know how much NaOH is required to neutralize the acid, then we can figure out how many carboxylic end groups there must have been in the first place. Usually, a colorimetric pH indicator is used to determine the end point of the titration. Easy groups to distinguish by acid-base titration are carboxylic acids and amines. Sometimes the polymer as its originally produced does not have an acidic or basic end group, but we can do a reaction to make that functional group amenable to titration. An example would be terminal hydroxyl groups, which are not distinguishable by acid-base titration but can be converted via multiple pathways to carboxylic acids:

molecular diagrams of
Figure 11.4: Terminal hydroxyl groups can be converted to carboxylic acids
Source: Lauren Zarzar

PROBLEM


For end-group analysis, 0.8632 g of a carboxyl terminated polybutadiene sample of the general structure shown below, dissolved in a mixture of ethanol and toluene, consumed 5.2 mL of 0.1242 M alcoholic potassium hydroxide solution in titration using phenolphthalein as the indicator. Calculate the molar mass of the polymer.

molecular diagram of a carboxyl terminated polybutadiene

ANSWER


First we find the number of carboxylic acid groups, which we know must equal the number of hydroxyl groups:

moles COOH groups=mol KOH

=5.2ml 1L 1000mL 0.1242 mol L 

=6.46 10 4 mol

Given the structure of the polymer, we see there are 2 carboxylic acid groups per molecule. Since we know how many carboxylic acids there are, we can figure out how many polymer molecules there are:

moles polymer= 6.46 10 4 mol 2 

=3.23 10 4 mol

To find the number average molar mass, we simply divide:

M ¯ n = 0.8632g 3.23 10 4 mol 

=2672g/ mol