Question

A 0.2432-g sample contained only NaCl and KBr. It was dissolved in water and required 47.85 mL of 0.04858 M AgNO3 for complete titration of both halides [giving AgCl(s) and AgBr(s)]. Calculate the weight percent of Br in the solid sample.

Answer 1

AgNO3 reacts with Cl- and Br- to give AgCl and AgBr precipitate

total moles of AgNO3 used = 0.04858 M x 47.85 ml = 2.325 mmol = 0.002325 mol

total moles = moles NaCl + moles KBr

let x be the moles of NaCl and y be moles of KBr

0.002325 mol = x + y

x = 0.002325 - y ------------------(1)

and,

moles x molar mass = grams of solute

molar mass NaCl = 58.44 g/mol and molar mass KBr = 119 g/mol

So,

total mass of solid sample = mass of NaCl + mass of KBr

0.2432 g = 58.44(x mol) + 119(y mol)

0.2432 g = 58.44x + 119y

feed (1) in above equation,

0.2432 g = 58.44(0.002325 - y) + 119y

0.2432 = 0.1360 - 58.44y + 119y

60.56y = 0.1072

moles of KBr = y = 0.002 mol

mass of Br = 0.002 mol x 80 g/mol = 0.160 g

weight percent Br in the solid sample = 0.160 g x 100/0.2432 g = 65.8%