Analytical Chem istry - DePauw University

Chapter 7 Collecting and Preparing Samples3497Lorganic solvent needed to extract 99.5% of the Pb 2+ in a single extraction?(d) Under the conditions of your extraction for Zn 2+ , how manyextractions are needed to remove 99.5% of the Zn 2+ if each extractionuses 25.0 mL of organic solvent?Solutions to Practice ExercisesPractice Exercise 7.1To reduce the overall variance by improving the method’s standard deviationrequires thats 2 2500 s 2 s 22= . ppm = + = ( 21 . ppm) + ssampmeth2methSolving for s meth gives its value as 0.768 ppm. Relative to its original valueof 1.1 ppm, this is reduction of 3.0 × 10 1 %. To reduce the overall varianceby improving the standard deviation for sampling requires that2 2 2 2 2s = 500 . ppm = s + s = s + ( 1.1 ppm)samp meth sampSolving for s samp gives its value as 1.95 ppm. Relative to its original valueof 2.1 ppm, this is reduction of 7.1%.Click here to return to the chapter.Practice Exercise 7.2The analytical method’s standard deviation is 1.96 × 10 –3 g/cm 3 as this isthe standard deviation for the analysis of a single sample of the polymer.The sampling variance is2 2 2 2 2 3 2s = s − s = 365× 10 −−( . ) − ( 1. 96× 10 ) = 133 . × 10 −3sampmethConverting the variance to a standard deviation gives s meth as 3.64 × 10 –2g/cm 3 .Click here to return to the chapter.Practice Exercise 7.3To determine the sampling constant, K s , we need to know the averagemass of the samples and the percent relative standard deviation for theconcentration of olaquindox in the feed. The average mass for the fivesamples is 0.95792 g. The average concentration of olaquindox in thesamples is 23.14 mg/kg with a standard deviation of 2.200 mg/kg. Thepercent relative standard deviation, R, isssamp2.200 mg/kgR = × 100 = × 100 = 95 .07 ≈ 951 .X 23.14 mg/kgSolving for K s gives its value asKs22= mR = ( 0. 95792 g)(9.507) = 86.58 g ≈86.6 g2