Analytical Chem istry - DePauw University

528 Analytical Chemistry 2.09ISolutions to Practice ExercisesPractice Exercise 9.1The volume of HCl needed to reach the equivalence point isVeq= MVV = b bM)(25.0 mL)aM= ( 0.1250.0625 M= 50.0 mLaBefore the equivalence point, NaOH is present in excess and the pH isdetermined by the concentration of unreacted OH – . For example, afteradding 10.0 mL of HCl[ OH ]− =( 0. 125 M)(25.0 mL) −(0.0625 M)(10.0 mL)25.0 mL + 10.0 mL= 0.0714Mthe pH is 12.85.−+Kw100 . × 10[ HO ] = =3−[ OH ] 0.0714 M14= 140 . × 10− 13 MFor the titration of a strong base with a strong acid the pH at the equivalencepoint is 7.00.For volumes of HCl greater than the equivalence point, the pH is determinedby the concentration of excess HCl. For example, after adding 70.0mL of titrant the concentration of HCl is( 0. 0625 M)(70.0 mL) −(0.125 M)(25.0 mL)[ HCl]=70.0 mL + 25.0 mLgiving a pH of 1.88. Some additional results are shown here.= 0.0132Volume of HCl (mL) pH Volume of HCl (mL) pH0 13.10 60 2.1310 12.85 70 1.8820 12.62 80 1.7530 12.36 90 1.6640 11.98 100 1.6050 7.00Click here to return to the chapter.Practice Exercise 9.2The volume of HCl needed to reach the equivalence point isVeq= MVV = b bM)(25.0 mL)aM= ( 0.1250.0625 M= 50.0 mLaBefore adding HCl the pH is that for a solution of 0.100 M NH 3 .M