Analytical Chem istry - DePauw University

Chapter 9 Titrimetric Methods529Kb− +[ OH ][ NH ]4( x)( x)= =NH 0.125− x= × −175 . 10 5[ ]3x = −−[ OH ] = 147 . × 10 3 MThe pH at the beginning of the titration, therefore, is 11.17.Before the equivalence point the pH is determined by an NH 3 /NH 4+buffer. For example, after adding 10.0 mL of HCl(0.125 M)(25.0 mL) (0.0625 M)(10.0 mL[ NH ] = − )= 0.0714 M325.0 mL + 10.0 mL[ NH ]+ =4( 0.0625 M)(10.0 mL)=25.0 mL + 10.0 mL0. 0179 M0 0714 MpH = 9. 244 + log . = 984 .0.0179 MAt the equivalence point the predominate ion in solution is NH 4 + . Tocalculate the pH we first determine the concentration of NH 4+[ NH ]+ =4and then calculate the pHKaobtaining a value of 5.31.( 0.125 M)(25.0 mL)= 0.0417 M25.0 mL + 50.0 mL+[ HO ][ NH ]3 3( x)( x)= =NH 0.0417 − x= 570 . × 10 −10+[ ]4x = [ HO+ ] = 488 . × 10 −6M3After the equivalence point, the pH is determined by the excess HCl. Forexample, after adding 70.0 mL of HCl( 0. 0625 M)(70.0 mL) −(0.125 M)(25.0 mL)[ HCl]=25.0 mL + 70.0 mLand the pH is 1.88. Some additional results are shown here.= 0.0132Volume of HCl (mL) pH Volume of HCl (mL) pH0 11.17 60 2.1310 9.84 70 1.8820 9.42 80 1.7530 9.07 90 1.6640 8.64 100 1.6050 5.31Click here to return to the chapter.M