Question

A 0.444 g sample contained only NaCl and KBr. It was dissolved in water and required 81.13 mL of 0.05664 M AgNO₃ for complete titration of both halides (giving AgCl (s) and AgBr (s)). Calculate the mass of KBr in the solid sample.

Answer 1

Answer:-
Moles AgNO3 = 0.08113 L x 0.05664 M=0.0046

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- = 0.0046
let x = moles NaCl ( molar mass = 58.44 g/mol)
let y = moles KBr ( molar mass = 119 g/mol)

x + y = 0.0046
58.44 x + 119 y = 0.0444

x = 0.0046-y

58.44 ( 0.0046-y) + 119 y= 0.444

0.268- 58.44 y + 119 y= 0.444
60.56 y = 0.1754
y=0.00290= moles KBr
x = 0.0046- 0.00290=0.00170 = moles NaCl
mass Br -= 0.00290x 79.9 g/mol=
0.23171 g

% Br- = 0.23171x 100/ 0.444=52.18