Question

Compute the volume of 0.20M silver nitrate needed to precipitate the chloride ion in a 0.60g sample of an unknown that is 45% chloride and the mass of the solid nitrate used to prepare such a solution.

Answer 1

Calculations

(i) Chloride present in 0.60 g. of sample = 0.60 x (45 / 100)

..................................................................= 0.27 g

.............................................................or, = (0.27 / 35.5) moles

...................................................................= 7.61 x 10^-3 moles  

(ii) study the stoichiometry of precipitation of Cl^- with Ag⁺  & find out the number of moles

of Ag⁺ required for precipitation as AgCl

a) AgCl Is precipitated as per ionic reaction-

...............................Ag⁺ (aq ) + Cl^- (aq ) --------------> AgCl (s)

b ) Since one mole of Cl^- is precipitated by one mole of Ag⁺ in their aqueos solutions

therefore 7.61 x 10^-3 moles ---------------7.61 x 10^-3 moles of Ag⁺ -----------------------

(iii) calculate the volume of 0.2M AgNO₃ which contains 7.61 x 10^-3 moles of Ag⁺ -

...........0.20 moles of Ag NO₃  are present in a volume of 0.20 M AgNO₃  = 1000 ml

7.61 x 10^-3 moles ---------------------------------------------------------------------- = [ ( 1000 x 7.61 x 10^-3 ) / 0.20 ] ml

.....................................................................................................................= 38.05 ml   

Thus , the volume of 0.20 M AgNO₃ required for precipitation of chloride in 0.60 g of sample = 38.05 ml

2. Calculation for weight of nitrate (ie. AgNO₃ ) required to prepare such AgNO₃ solution *

The calculation is based upon the concept that 1000 ml of a molar (ie. 1M )

solution of AgNO₃ contains a mass = its gram molar mass = 143.5 gms

Therefore ,the solution should contain a weight Ag NO₃= [ ( 38.5 x 0.20 x 143.5 ) /1000 ]

...........................................................................= 1.10495 gms / L