Analytical Chem istry - DePauw University

Chapter 9 Titrimetric Methods509So l u t i o nTo find the moles of titrant reacting with the sample, we first need to correctfor the reagent blank; thusV Ag= 36. 85 mL − 071 . mL = 3614 . mL(0.1120 MAgNO ) × ( 0. 03614 LAgNO ) = 4.048×10 −3 molAgNO3 33Titrating with AgNO 3 produces a precipitate of AgCl and AgBr. In formingthe precipitates, each mole of KCl consumes one mole of AgNO 3 andeach mole of NaBr consumes one mole of AgNO 3 ; thusmolesKCl + molesNaBr = 4.048× 10 −3We are interested in finding the mass of KCl, so let’s rewrite this equationin terms of mass. We know thatgKClmolesKCl =74.551 gKCl/mol KClgNaBrmolesNaBr=102.89 gNaBr/molNaBrwhich we substitute back into the previous equationgKCl74.551 gKCl/mol KClgNaBr+102.89 gNaBr/mol NaBr= 4.048× 10 −3Because this equation has two unknowns—g KCl and g NaBr—we needanother equation that includes both unknowns. A simple equation takesadvantage of the fact that the sample contains only KCl and NaBr; thus,gNaBr= 0.3172 g−gKClgKCl74.551 gKCl/mol KCl0.3172 g−gKCl+102.89 gNaBr/molNaBr= 4.048× 10 −31. 341× 10 −2 (g KCl) + 3. 083× 10 −3 − 9. 719×10−3( gKCl) = 4.048× 10 −3−369 . × 10 (g KCl) = 9.65×103 − 4The sample contains 0.262 g of KCl and the %w/w KCl in the sample is0.262 gKCl× 100 = 82.%6 w/wKCl0.3172 gsampleThe analysis for I – using the Volhard method requires a back titration.A typical calculation is shown in the following example.