Analytical Chem istry - DePauw University

542 Analytical Chemistry 2.0or a pCl of 1.93. To find the concentration of Ag + we use the K sp forAgCl; thusK−+ sp 18 . × 10[ Ag ] = =−[ Cl ] 118 . × 1010−2= 15 . × 10or a pAg of 7.82. The following table summarizes additional results forthis titration.−8Volume ofNaCl (mL) pAg pCl0 1.30 –5.00 1.44 8.3110.0 1.60 8.1415.0 1.81 7.9320.0 2.15 7.6025.0 4.89 4.8930.0 7.54 2.2035.0 7.82 1.9340.0 7.97 1.7845.0 8.07 1.6850.0 8.14 1.60Click here to return to the chapter.Practice Exercise 9.23The titration uses0.1078MKSCNLM× 0. 02719 L= 2.931× 10 −3mol KSCNThe stoichiometry between SCN – and Ag + is 1:1; thus, there are− 3+ 107.87 gAg2.931× 10 molAg × = 0.3162 gAgmolAgin the 25.00 mL sample. Because this represents ¼ of the total solution,there are 0.3162 × 4 or 1.265 g Ag in the alloy. The %w/w Ag in the alloyis1.265 gAg× 100 = 64. 44%w/wAg1.963 gsampleClick here to return to the chapter.