Question

A solution contains Ag+ and Hg2+ ions. The addition of 0.100 L of 1.49 M NaI solution is just enough to precipitate all the ions as AgI and HgI2. The total mass of the precipitate is 34.4 g.

Find the mass of AgI in the precipitate.

Express your answer to two significant figures

Answer 1

The reaction is Ag⁺ +NaI---> AgI + Na+ (1) and HgI2⁺² + 2NaI----> HgI2 +2Na+ (2)

moles of NaI added from 0.149M and 0.1 L= 0.149*0.1= 0.0149 moles =0.0149

let x= moles of NaI used for precipitatino of AgI ( Reaction 1) then 0.0149-x =moles of NaI through reaction 2

Molecular weight of NaI= 150

x moles of AgI produces x moles of AgI ( from reaction)= x*235 gm of AgI ( Molecular weight of AgI)=235x

2 moles of AgI produces 1 mole of HgI2

(0.149-x) moles produces (0.149-x)/2 moles of HgI2 =454{(0.149-x)/2} ( Molecular weight of HgI2= 454) =227*(0.149-x) gms

total mass = 235x+ 227(0.149-x)= 34.4 ( amount of AgI and HgI2 precipiated)

(235-227)x +227*0.149= 34.4

8x+33.823= 34.4 therefore 8x= 34.4-33.823=0.577

x= 0.577/8 =0.072125 moles mass of AgI precipitated = 235x = 235*0.072125=16.94 gms