Answer to Question #159117 in General Chemistry for Dexther Villanueva

Question

A 500.0mg of butter was warmed and shaken vigorously with water. The undissolved material was removed by filtering and the aqueous portion was made 1.0M in HNO3 and 0.025M in Fe(NO3)3. This acidified solution was treated with 10.00ml of 0.1755M AgNO3 to precipitate the chloride ion and, after the addition of a small amount of nitrobenzene, 14.22ml of 0.1006M KSCN was required to back titrated the excess Ag+ . Calculate the % NaCl in the butter.

Answer 1

Back titration:

AgNO₃ + KSCN → AgSCN + KNO₃

(14.22 / 1000) L x 0.1006 M = 0.00143 mol of KSCN

Mole of AgNO₃ reacted: 0.00143 mol

Total mole of AgNO₃:

(10.00 / 1000) L x 0.1755 M = 0.001755 mol

Δ = 0.001755 - 0.00143 = 0.000325 mol – reacted with Cl^-

Ag⁺ + Cl^- → AgCl

Moles of Cl^-: 0.000325 mol

Molar mass of Cl^-: 35.45 g/mol

Mass = 0.000325 x 35.45 = 0.0115 g

0.0115 / 0.500 x 100% = 2.3% Cl^- in the sample.