Answer to Question #176262 in General Chemistry for dumbo

Question

1. Upon analyzing a sample of silver nitrate according to Volhard’s procedure, 3.000 g of this impure sample was weighed and diluted to 100.0 mL in a volumetric flask. 25.00 mL of this solution was then titrated with a 0.05111 M KSCN requiring 35.50ml to reach the light brown color of the end point. Calculate the weight % of AgNO₃. 

Answer 1

AgNO3 + KSCN ---> AgSCN + KNO3

C₁= ?, V₁=25.00ml

C₂= 0.05111M, V₂= 35.50ml

C₁V₁=C₂V₂

C₁=C₂V₂/V₁= 0.05111x35.50/25.00=0.0725M

No of moles of AgNO3= CxV= 0.0725x100/1000= 0.00725moles

Molar mass of AgNO3= 170g/mol

Mass of AgNO3= mole x molar mass

= 0.00725x170=1.2325g

Mass of sample=3.000g

%weight of AgNO3= 1.2325/3.000 x 100%=41.08%