Exercise on Volhard Method

Question

In a Volhard analysis of Cl^- an analyst weighed 1.6g sample (ca. 45%m/m Cl^-) and made it up to 250.00mL of solution. He pipetted 25.00mL of this solution into a conical flask. Calculate the volume of ca. 0.1M AgNO₃ that must be added to precipitate the chloride if the method requires a 60% excess of AgNO₃.

[Molar mass Cl^- = 35.453 g/mol)

Answer: Total Volume = 36mL

Answer 1

Let's look at the reaction taking place:

Cl^-(aq) + AgNO₃(aq) ==> AgCl(s) + NO₃^-(aq)

approximate mols Cl^- present = 1.6 g x 0.45 x 1 mol Cl^- / 35.45 g = 0.02031 mols Cl^- in original sample

mols Cl- present in sample used for analysis: 0.02031 mols / 250 ml x 25 mls = 0.002031 mols Cl^-

mols AgNO₃ needed = 0.002031 mols AgNO₃ since they react in a 1:1 mol ratio (see balance equation)

volume of 0.1 M AgNO₃ needed: (x L)(0.1 mol/L) = 0.002031 mols and x = 0.02031 L = 20.31 mls AgNO₃

To account for a 60% excess requirement, we would thus need 20.31 x 1.60 = 32.50 mls AgNO₃ needed