Question

If all of the silver in a 7.563 g sample of a mineral, which was dissolved by using nitric acid, is precipitated as silver thiocyanate (AgCNS) by using 82.46 mL of a 0.2500 M potassium thiocyanate solution, what is the percent silver (by mass) in the sample?

Answer 1

Ans. Potassium thiocyanate (KSCN) dissociates in aqueous medium as follow-

            KSCN --------------> K⁺(aq) + SCN^-(aq)

Thiocyanate ion further associates with silver cation in nitric acid to form the precipitate (silver thiocyanate) as follow-

            Ag⁺(aq) + SCN^-(aq) ---------> AgSCN(s)

Stoichiometry: 1 mol KSCN forms 1 mol AgSCN.

# Moles of KSCN used = Molarity x Volume of solution in liters

                                    = 0.2500 M x 0.08246 L

                                    = 0.020615 mol

# Following stoichiometry,

            No. of moles of Ag⁺ in sample = Moles of KSCN consumed

So,

            Moles of Ag in the sample = 0.020615 mol

            Mass of Ag in sample = Moles x Molar mass

                                                = 0.020615 g x (107.8682 g/ mol)

                                                = 2.224 g

Therefore, mass of Ag in sample = 2.224 g

# % Ag = (Mass of Ag / Mass of sample) x 100

                        = (2.224 g / 7.563 g) x 100

                        = 29.41 %