Analytical Chem istry - DePauw University

Chapter 9 Titrimetric Methods537are consumed in the back titration with Mg 2+ , which means that−5. 028× 10 molEDTA− 4.708×104 −5mol EDTA= 45 . 57× 10 −4molEDTAreact with the BaSO 4 . Each mole of BaSO 4 reacts with one mole ofEDTA; thus41 molBaSO4. 557× 10−4molEDTA × ×molEDTA1 mol Na SO 142 gNaSO2 4× . 04molBaSO molNaSO422 44= 0.064730.06473 gNaSO2 4× 100 = 41. 23%w/wNaSO0.1557 gsampleClick here to return to the chapter.Practice Exercise 9.172 4The volume of Tl 3+ needed to reach the equivalence point isVeqM VSn SnM)(50.0 mL)= V = = ( 0.050 = 25.0 mLTlM (0.100 M)TlgNaSO2 4Before the equivalence point, the concentration of unreacted Sn 2+ andthe concentration of Sn 4+ are easy to calculate. For this reason we findthe potential using the Nernst equation for the Sn 4+ /Sn 2+ half-reaction.For example, the concentrations of Sn 2+ and Sn 4+ after adding 10.0 mLof titrant are2 ( 0. 050 M)(50.0 mL) − ( 0.100 M)(10.0 mL)[ Sn+ ] ==50.0 mL + 10.0 mLand the potential is4 ( 0.100 M)(10.0 mL)[ Sn+ ] == 0.0167 M50.0 mL + 10.0 mLE =+ 0.139 V −0.059162⎛0.0250 M⎞log⎝⎜0.0167 M⎠⎟ 0.0250=+0 . 134 VAfter the equivalence point, the concentration of Tl + and the concentrationof excess Tl 3+ are easy to calculate. For this reason we find the potentialusing the Nernst equation for the Tl 3+ /Tl + half-reaction. For example,after adding 40.0 mL of titrant, the concentrations of Tl + and Tl 3+ are[ Tl ]+ =( 0.0500 M)(50.0 mL)= 00278 . M50.0 mL + 40.0mLM