Question

A 0.4571 g sample of a chloride containing unknown was titrated with 0.1013 M AgNO3 using the Fajans' method, requiring 42.33 mL to reach the faint pink endpoint. Calculate the percent chloride in the sample.

Answer 1

Since in in Fajan's method there is no chance of over consumption of AgNO₃ , SO amount of AgNO₃ require to tritrate chloride ion directly proposional to chloride ion present in solution.

Reaction

AgNO₃ + Ci^- = AgCl + NO₃^-

one equivalent AgNO₃ react with one equivqlent chloride ion. Now 0.1013 M AgNO₃ will nutralise 0.1013 M chloride ion . let total volume of sample 42.33 mL since unknown sample volume is given.

that means 1000 ml sample contain 0.1013 mole chloride ion

so, 42.33 mL sample contain (0.1013 / 1000) * 42.33 = 4.2 * 10^-3 mole

now molecular weight of chloride ion 17 gm / mole

so , amount of chloride ion = molecular weight * mole

= 4.2 * 10^-3 mole * 17 = 0.0729 gm

now 0.4571 gm unknown sample contain 0.0729 gm chloride ion

so, percentage of chloride ion = (0.0729 / 0.4571 ) * 1000 = 15.94 %