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 using an excess of NaOH (5 g). Each NaOH reacts with one of the -COOCH3 ester linkages. The product mixture is then titrated with 25.9 cm3 of 0.3 mol dm3 HCl. Deduce the Mr and structure of the diester. Answer Let’s start by writing out some reactions to establish the molar ratios. The diester has two ester linkages so requires 2 NaOH to be hydrolysed. Since we don’t yet know its molecular formula, we can’t write out a balanced symbol equation. Step 1: diester + 2 NaOH –> dicarboxylate salt + 2 alcohol Step 2: HCl + NaOH –> NaCl + H2O We can then work out the molar quantities using the data given in the question: n NaOH = 5 / 40 = 0.125 mol (at start of Step 1) n HCl = 25.9 / 1000 x 0.3 = 7.77 x 10-3 mol (at start of Step 2) Using the balanced symbol equation in Step 2, we know that HCl and the NaOH (remaining at end of Step 1) react in a 1:1 ratio n NaOH = 7.77 x 10-3 mol (remaining at end of Step 1) We can then determine the amount of NaOH used in Step 1 and therefore the amount of diester used in Step 1: n NaOH = 0.125 – 7.77 x 10-3 = 0.117 mol (used in Step 1) n diester = 1/2 x 0.117 = 0.0586 mol (used in Step 1) We can then work out the Mr of the diester Mr = 7.74 / 0.0586 = 132 We know that a diester must contain two -COOCH3 so we can subtract the Mr of 2 -COOCH3 (i.e., 2 x 59 = 118) to find the Mr for the remaining parts of the molecule (i.e., 132 – 118 = 14). The only solution is therefore CH3OOCCH2COOCH3 i.e., the dimethyl ester of propan-1,3-dioic acid. |