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) is dissolved in a suitable solvent to form 250 cm3 of solution. A 50 cm3 sample is titrated with 16.2 cm3 of 0.2 mol dm-3 aqueous NaOH and phenolphthalein until the mixture becomes pink. A second 50 cm3 sample is titrated with 38.9 cm3 of 0.25 mol dm-3 aqueous Br2 until the mixture becomes brown. Deduce what you can about the structure of the molecule. Answer Starting with the first tiration: n NaOH = 0.2 x 16.2 / 1000 = 3.24 x 10-3 mol We know that there will be a 1:1 ratio between the NaOH and RCOOH so we can work out the RCOOH moles in the sample: NaOH + RCOOH –> RCOONa + H2O n RCOOH = 3.24 x 10-3 mol (in 50 cm3) The total is 5x bigger than the sample, so we can work out the RCOOH moles in total: n RCOOH = 5 x 3.24 x 10-3 = 0.0162 mol (overall) Next, we can work out the Mr of RCOOH: Mr RCOOH = m / n = 4.05 / 0.0162 = 250 If we subtract the Mr of COOH (45), this means that the Mr of the R group is 205. Now, turning to the second titration: n Br2 = 0.25 x 38.9 / 1000 = 9.73 x 10-3 mol The total is 5x bigger than the sample, so we can work out the Br2 moles in total: n Br2 = 5 x 9.73 x 10-3 = 0.0486 mol (overall) We can now work out the molar ratio between Br2 and RCOOH from the two calculated amounts: n Br2 / n RCOOH = 0.0486 / 0.0162 = 3 This tells us that there must be 3 C=C in RCOOH, since each C=C reacts with Br2 in a 1:1 ratio by electrophillic addition. To summarise, we know that RCOOH has an Mr of 250 and R has a Mr of 205 and contains 3 C=C bonds. The general formula of an alkyl group is CnH2n+1 and each double bond reduces the number of hydrogens by 2. This means that the general formula of an R group with 3 C=C would be CnH2n-5. This allows us to determine the value of n 12 x n + 1 x (2n – 5) = 205 14 n – 5 = 205 14 n = 210 n = 15, so therefore R = -C15H25 |