Analytical Chem istry - DePauw University

664 Analytical Chemistry 2.0C Cu0.16220. 00 mg Cu 100 . mL+ ×L 10.00 mL0.118=CCuleaves us with an equation in which C Cu is the only variable. Solving forC Cu gives its value asCCu0.162 0.118=+ 200 . mg Cu/L CCu0. 162C= 0. 118C+ 0.236 mg Cu/LCu Cu0. 044 = 0.236 mg Cu/LC CuC Cu= 54 .mg Cu/LClick here to return to the chapter.Practice Exercise 10.6Substituting into equation 10.11 and equation 10.12 givesA = 0. 336 = 15. 2C + 5.60C400 CrA = 0. 187 = 0. 533C + 507 . C505 CrTo determine C Cr and C Co we solve the first equation for C CoCCo0. 336−15.2C=560 .and substitute the result into the second equation.0. 336−15.2CCr0. 187 = 0. 533C+ 507 . ×= 0.3042− 1323 . CCr560 .CrSolving for C Cr gives the concentration of Cr 3+ as 8.86 10 –3 M. Substitutingthis concentration back into the equation for the mixture’s absorbanceat 400 nm gives the concentration of Co 2+ as 3.60 10 –2 M.Click here to return to the chapter.Practice Exercise 10.7Letting X represent MnO 4 – and Y represent Cr 2 O 7 2– , we plot the equationAAmixSXCrC C AX Y= + ×C C ASXSYSYSXCoCo