Question

2. A student dissolved 2.636 g of a powder containing unknown masses of NaCl and KBr in 30.00 mL of de-ionized water and titrated the solution with 0.0500 M AgNO3.

If it takes 33.62 mL of Ag

the Fajan’s end point, what is the mass of KBr in the powder?   

Answer 1

Moles of 33.6ml of 0.05M AgNO3 = 0.00168 moles

The reactions are NaCl+AgNO3---> AgCl +NaNO3   and KBr+AgNO3---> KNO3+AgBr

let x= mass of NaCl, Mass of KBr= 2.636-x

1 moles of NaCl requires 1 mole of AgNO3 and 1 mole of KBr requires 1 mole of AgNO3'

Moles = Mass/ Molecular weight Molecular weights : NaCl =58.5 and KBr= 119

Moles of NaCl in 30ml =x/ 58.5   Moles of KBr= (2.636-x)/119

x/58.5 moles of NaCl requires x/58.5 moles of NaCl and (2.636-x)/119 mole of KBr

total moles of AgNO3= x/58.5+(2.636-x)/119= 0.00168

x*0.0171 +(2.636-x)*0.0084= 0.00168

x*0.0171+0.022151-x*0.0084= 0.00168

0.022151-0.00168= x*(0.0171-0.0084)

x= 2.353 gms Mass of KBr= 2.636-2.353=0.283 gms