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.
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.
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 %
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