Question

We need five working standards covering the range between 50.0 and 500.0 ppb Pb2+ for the analysis of Pb in drinking water. A stock solution of Pb2+ was made by dissolving 0.1588g of Pb(NO3)2 in distilled water and dilute to 100 mL mark of volumetric flask. How should we make these working standards? Hint: Since the target concentrrations are so low here, you may use a dilution ratio of 1 to 1000, i.e., 1mL to 1000mL

Answer 1

According given data weight of Pb(NO3)20.1588g, volume of soluti on(V1)-100mL concentrati on of solution(M) weight mol.wt 1000 び(mL) 0.1588 1000 331 100 Normality of solution = 2 ×0. 0048-0. 0096N so concentration of Pb2+ions in solution = 0.00961 we want make 50 and 500ppb solutions (a)50 ppb means 0.05ppm that means 0.05mg dissolvedin one litre Concentration of Pb"+ ions in soltu on (N2) 5x101-0.048x10*N 1041L According to diluti on lavw 0.0096xⅤ1-0.048x 10-5 ×1L V1-5x10-5L-0.05mL (b) 500 ppb means 0.5ppm that means 0.5mg dissolved in one litre Concentrati on of Pb** ions in soltuion (N2)- According to dilution law 0.0096 ×V1-0. 048x 10t ×1L V1=5x10-4L-0.5mL 1041L