Answer to Question #175790 in General Chemistry for Kiana

Question

A Sodium Chloride (0.1920g) NaCl
was assayed using volhard Method,
using 52mL of 0.0975N AgNO3 and
12.3 mL of 0.1192Ammonium
thioscyanate. Calculate the sample in
%.

Answer 1

moles of AgCl = 0.1920/58.5 = 0.00328 = moles of Cl^-

moles of Ag⁺ = (52/1000)×0.0975 = 0.00507 moles of Ag⁺

Ag⁺ + Cl^- $\to$ AgCl

0.00507__0.00328

Remaining Ag⁺ = 0.00507-0.00328 = 0.00179 moles

Moles of SCN^- = (12.3/1000)×0.1192 = 0.00146 moles

Ag⁺ + SCN^- $\to$ AgSCN

0.00179__0.00146

Remaining Ag⁺ = 0.00179-0.00146 = 0.00033 moles

%Ag remaining = (0.00033/0.00507)×100

= 6.50%