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# Lecture 34 - Measuring Ksp Step 1 Set-up the cell...

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4/10/10 1 36 Step 1: Set-up the cell Measuring K sp Pb(s)| Pb 2+ (???? M) || Sn + (1.00 M) | Sn(s) From PbSO 4 (s) Use common-ion: 1.0 M SO 4 2- from Na 2 SO 4 (aq) Half-cell for which we know E o Doesn’t have to be used Cathode (RED CAT) Anode (AN OX) 0.219 V = -0.137 ! 0.0592 2 log 1 1.0 " # \$ % & ! -0.125 ! 0.0592 2 log 1 [Pb 2 + ] " # \$ % & E cell = E C o ! 0.0592 n logQ C " # \$ % & ! E A o ! 0.0592 n logQ A " # \$ % & Half-rxns as reductions : Solve: [Pb 2+ ] = 1.6 ! 10 -8 M 37 Step 2 : Measure cell potential and solve for concentration of metal ion Take potential measurement : E cell = 0.219 V C: Sn 2+ (aq) + 2 e - ! Sn (s) E° = -0.137 V A: Pb 2+ (aq) + 2 e - ! Pb(s) E° = -0.125 V 38 Step 3 : Calculate Ksp using above [Pb 2+ ] PbSO 4 (s) ! Pb 2+ (aq) + SO 4 2- (aq) i. -- c. -- e. -- Know: s = 1.6 ! 10 -8 M K sp = [Pb 2+ ][SO 4 2- ] Check assumption... 0 1.0 +s +s s 1.0 + s Common ion Assume: 1.0 + s " 1.0 = (1.6 ! 10 -8 )(1.0) = 1.6 ! 10 -8 39 Using that K sp value, can you determine what the cell potential would be if Na 2 SO 4 (aq) were replaced by KCl(aq)? E cell = 0.103 V No common ion Step 1: Calculate [Pb 2+ ] using K sp Step 2 : Determine cell potential using Nernst equation Non-standard conditions 5.6. Electrolysis (Section 21-7)

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