Question

A 0.2386-g sample contained only NaCl and KBr. It was dissolved in water and required 48.40 mL of 0.04837 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

Moles AgNO3 = 0.04840 L x 0.04837 M=0.002341

AgNO3 is a strong salt : AgNO3 => Ag+ + NO3-

Ag+ reacts with Cl- and Br- to give a precipitate :
Ag+ (aq) + Cl- (aq) = AgCl
Ag+ (aq) + Br- (aq) = AgBr (s)

moles Cl-+ moles Br- = 0.002341
let x = moles NaCl ( molar mass = 58.44 g/mol)
let y = moles KBr ( molar mass = 119 g/mol)

x + y = 0.002341
58.44 x + 119 y = 0.2386

x = 0.002341-y

58.44 ( 0.002341-y) + 119 y= 0.2386

0.137 - 58.44 y + 119 y= 0.2386
60.56 y = 0.102
y=0.00168 = moles KBr
x = 0.002341 - 0.00168=0.000661 = moles NaCl
mass Br -= 0.00168 x 79.9 g/mol= 0.134 g

% Br- = 0.134 x 100/ 0.2386=56.2