Analytical Chem istry - DePauw University

448 Analytical Chemistry 2.00. 1183 MNaOH× 0.02284 LNaOH×1 molHClmolNaOH= 2.702× 10 −3molHClreact with NaOH. The difference between the total moles of HCl and themoles of HCl reacting with NaOH5. 235× 10 −3−molHCl − 2. 702× 10 3 molHCl = 2.533× 10 −3molHClis the moles of HCl reacting with NH 3 . Because all the nitrogen in NH 3comes from the sample of cheese, we use a conservation of mass to determinethe grams of nitrogen in the sample.31 molNH 14.01 g2.533× 10− molHCl× ×molHCl molNHThe mass of protein, therefore, is3N3= 0.03549638 . gprotein0.03549 gN× = 0.2264 gproteingNand the % w/w protein is0.2264 gprotein× 100 = 23.%1 w/w protein0.9814 gsamplegNPractice Exercise 9.8Limestone consists mainly of CaCO 3 , with traces of iron oxides andother metal oxides. To determine the purity of a limestone, a 0.5413-gsample is dissolved using 10.00 mL of 1.396 M HCl. After heating toexpel CO 2 , the excess HCl was titrated to the phenolphthalein end point,requiring 39.96 mL of 0.1004 M NaOH. Report the sample’s purity as%w/w CaCO 3 .Click here to review your answer to this exercise.Earlier we noted that we can use an acid–base titration to analyze amixture of acids or bases by titrating to more than one equivalence point.The concentration of each analyte is determined by accounting for its contributionto each equivalence point.Example 9.5The alkalinity of natural waters is usually controlled by OH – , HCO 3 – , andCO 3 2– , which may be present singularly or in combination. Titrating a100.0-mL sample to a pH of 8.3 requires 18.67 mL of a 0.02812 M HCl.A second 100.0-mL aliquot requires 48.12 mL of the same titrant to reach