Analytical Chem istry - DePauw University

Chapter 6 Equilibrium Chemistry247+ −[ NH ][ OH ]4−5K b= = 175 . × 10 6.42[ NH ]3K w 3= [ HO+ ][ OH− ] = 100 . × 10 −146.43[ Ag(NH ) ]= = 17 . × 10+3 2β 2 +2[ Ag ][ NH ]376.44We still need three additional equations. The first of these equation is a massbalance for NH 3 .+ +C = [ NH ] + [ NH ] + 2×[ Ag(NH) ]NH 3 3 4 3 2 6.45In writing this mass balance equation we multiply the concentration ofAg(NH 3 ) 2 + by two since there are two moles of NH 3 per mole of Ag(NH 3 ) 2 + .The second additional equation is a mass balance between iodide and silver.Because AgI is the only source of I - and Ag + , each iodide in solution musthave an associated silver ion, which may be Ag + or Ag(NH 3 ) 2 + ; thusFinally, we include a charge balance equation.[ I − ] = [ Ag + ] + [ Ag(NH ) + ]3 26.46+ + + + − −[ Ag ] + [ Ag(NH ) ] + [ NH ] + [ H O ] = [ OH ] + [ I ]3 2 4 36.47Although the problem looks challenging, three assumptions greatlysimplify the algebra.Assumption One. Because the formation of the Ag(NH 3 ) 2 + complex is sofavorable (b 2 is 1.7 × 10 7 ), there is very little free Ag + and it is reasonableto assume that[Ag + ]