Analytical Chem istry - DePauw University

532 Analytical Chemistry 2.0Practice Exercise 9.8The total moles of HCl used in this analysis isOf this,1.396molNaOHL× 0. 01000 L= 1.396× 10 −2mol HCl0.1004 molNaOH1 molHCl× 0.03996 L×= 4.012× 10 −3molHClLmolNaOHare consumed in the back titration with NaOH, which means that−2 −31. 396× 10 molHCl − 4. 012× 10 molHCl = 995 . × 10 −3molHClreact with the CaCO 3 . Because CO 3 2– is dibasic, each mole of CaCO 3consumes two moles of HCl; thus31 molCaCO995 . × 10−3molHCl× ×2mol HCl100.09 gCaCO3= 0.498 gCaCOmolCaCO0.498 gCaCO 3× 100 = 96.%8 w/w CaCO0.5143 gsample3Click here to return to the chapter.Practice Exercise 9.9Of the two analytes, 2-methylanilinium is the stronger acid and is the firstto react with the titrant. Titrating to the bromocresol purple end point,therefore, provides information about the amount of 2-methylaniliniumin the sample.0.200molNaOHL31 molC H NCl7 10× 0.01965 L×mol NaOH143.61 gC H NCl7 10× = 0. 564 gC H NCl7 10molC H NCl7 100.564 gC H NCl7 10× 100 = 28.%1 w/ wC H NCl7 102.006 gsampleTitrating from the bromocresol purple end point to the phenolphthaleinend point, a total of 48.41 mL – 19.65 mL, or 28.76 mL, gives theamount of NaOH reacting with 3-nitrophenol. The amount of 3-nitrophenolin the sample, therefore, is3