You will usually be introduced to titration and back titration through the context of inorganic chemistry e.g., acid-base, redox and complexometric titrations. However, titrations can also be useful in organic chemistry! In the example below, a titration is used to determine the amount of ester produced in an esterification reaction: Question 1: 0.20 mol of a…
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Titration in Organic Chemistry II
You will usually be introduced to titration and back titration through the context of inorganic chemistry e.g., acid-base, redox and complexometric titrations. However, titrations are also useful in organic chemistry! In the example below, a back-titration is used to investigate the structure of a diester: Example 1: 7.74 g of a dimethyl diester is hydrolysed…
Titration in Organic Chemistry I
You will usually be introduced to titration and back titration through the context of inorganic chemistry e.g., acid-base, redox and complexometric titrations. However, titrations are also useful in organic chemistry! In the example below, titration is used to investigate the structure of an unsaturated carboxylic acid: Example 1: 4.05 g of an unsaturated carboxylic acid (RCOOH)…
Using percentage yield to determine the amount of reactant needed
Percentage yield is defined as the actual mass of product divided by the theoretical maximum mass of product, multiplied by 100%. We can similarly express percentage yield in terms of moles. There are essentially three types of questions: 1. Calculate the % yield for a reaction Here, you would divide the actual moles of product by the…
Using percentage yield with limiting reagents
Percentage yield is defined as the actual mass of product divided by the theoretical maximum mass of product, multiplied by 100%. Equivalently, we can express this as the actual moles of product divided by the theoretical maximum moles of product, multiplied by 100%. The theoretical maximum moles of product depends on the moles of the…
Calculations with parts per million
Parts per million (ppm) is a rather awkward unit of concentration and for this reason often appears in questions! To work with this unit, you need to be confident with converting between various SI units of mass e.g., Mg, kg, g, mg and μg Unit Multiplier Mg = 1 metric tonne x 106 kg x…
Redox titration with iodine-thiosulphate to determine the amount of other oxidising agents
In the most simple case, an iodine-thiosulphate titration can be used to determine the amount of iodine. However, the techique is actually a lot more powerful than that and can be used to determine the amount of a much broader range of oxidising agent. The basic principle is that we use an oxidising agent to…
Redox titration with iodine-thiosulfate to determine the amount of iodine
Thiosulfate (S2O32-) is a strong reducing agent and so can be used in titrations to determine the amount of oxidising agent present: Reduction: 2 S2O32- -> S4O62- + 2 e– Unfortunately both S2O32- and S4O62- are colourless and there isn’t any suitable indicator for either. This means that, for a useful titration method, the other…
Redox Titration with manganate (VII)
MnO4– is a strong oxidising agent and so it be used in titrations to determine the amount of a reducing agent present. MnO4– has a very intense purple colour and is reduced to Mn2+ which is a very pale pink and almost colourless at low concentrations. In a typical titration, MnO4– is added from the…
Back titration to determine the amount of an inert solid
You have read the exam question carefully and decided that the following are the key bits of information: 100 cm3 of 0.1 mol/dm3 HCl is added to 0.240 g of a mixture containing Mg(OH)2 and an inert solid. The resulting solution is made up to 250 cm3 with distilled water. 25 cm3 of the solution is…