Question

An analytical chemist weighs out 0.260g of an unknown diprotic acid into a 250mL volumetric flask and dilutes to the mark with distilled water. She then titrates this solution with 0.1400M NaOH solution. When the titration reaches the equivalence point, the chemist finds she has added 24.7mL of NaOH solution. Calculate the molar mass of the unknown acid. Round your answer to 3 significant digits.

Answer 1

Let the diprotic acid be denoted as H₂A.

Hence, the reaction of the acid with NaOH can be written as

$H_2A_{(aq)} + 2NaOH_{(aq)} \rightleftharpoons Na_2A_{(aq)} + 2H_2O _{(l)}$

Hence, 2 moles of NaOH reacts with 1 mol of diprotic acid to reach equivalence point.

Now, concentration of NaOH being used = 0.1400 M = 0.1400 mol/L

Volume of NaOH used to reach equivalence point = 24.7 mL = 0.0247 L

Hence, the number of moles of NaOH used is

$moles = \frac{0.1400 \ mol}{L} \times 0.0247 \ L = 0.003458 \ mol$

Hence, the number of moles of acid that must be present in the solution can be calculated as

$\frac{1 \ mol \ H_2A}{2 \ mol \ NaOH} \times 0.003458 \ mol \ NaOH = 0.001729 \ mol \ H_2A$

Note that it is given that 0.260 g of diprotic acid is used.

Hence, 0.260 g of the acid must be equivalent to 0.001729 mol of the acid.

Hence, the molar mass of the acid can be calculated as follows:

$molar \ mass = \frac{0.260 \ g}{0.001729 \ mol} \approx 150.4 \frac{g}{mol} \approx {\color{Red} 150. \ g/mol}$

Hence, the molar mass of the acid is 150. g/mol (Rounded to three significant figure).