Question

A 5.5984 g sample of soil was analyzed for lead. The sample was digested in hot perchloric acid for 5 hours, cooled, filtered to remove insoluble silicates, transferred to a 250-ml volumetric flask and diluted to the mark with distilled water. A 50.00-ml portion of the resulting solution was treated with phosphoric acid (H₃PO₄) forming 0.0862 g of precipitate. What was the weight % of lead?

                  3 Pb²⁺ + 2 PO₄^3- ® Pb₃(PO₄)₂(s)

Answer 1

precipitate Pb3(PO4)2 formed = 0.0862 g

Precipitate moles = mass / Molar mass of it                  where Pb3(PO4)2 has molar mass 811.54 g/mol

                   = 0.0862 / 811.54 = 0.000106

Pb2+ moles = 3 x Pb3(PO4)2 moles             ( as per balanced reaction coefficents)

Pb2+ moles = 3 x 0.000106 = 0.0003186534

This is Pb2+ moles present in 50 ml part

Original 250 ml contains 5 times more Pb2+

Pb2+ moles = 5 x 0.0003186534 = 0.001593267

Pb2+ si fomed by perchoric acid

Pb moles = 0.001593267

Pb mass = moles x atomic mass of Pb

         = 0.001593267 x 207.2 g/mol

   = 0.33 g

weight % of Pb = (100 x Pb mass / sample mass)

            = ( 100 x 0.33 /5.5984)

        = 5.9 %