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Unformatted text preview: 0.370M. So instead of FORMING the Ag(NH 3 ) 2 + complex and finding Ag + concentration, you are BREAKING DOWN the Ag(NH 3 ) 2 + complex to find the equilibrium Ag + concentration. This is the ICE chart that I used. I know these aren’t the right arrows, but understand that this is at equilibrium. Ag(NH 3 ) 2 + Ag + 2 NH 3 I 0.170 0.370 Cx +x +2x E 0.170x x 2x+0.370 Approach this like a normal equilibrium problem, but remember that the K f is for the reaction for FORMING complex ions. If you are breaking a complex ion down like we are in this problem, you need to take the inverse of the K f given. Your K f expression for FORMATION is normally K f = [Ag(NH 3 ) 2 + ] [Ag + ][NH 3 ] 2 BUT, since we are going backwards, our expression has to change to this: 1 = [Ag + ][NH 3 ] 2 K f [Ag(NH 3 ) 2 + ] When you plug and chug the numbers given, you get that x= 7.30*108 , which is the equilibrium concentration of Ag + in the solution....
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 Winter '08
 Herron,S
 Chemistry, Equilibrium, Solvent, Lone pair, 0.170M, 0.340M, 0.370M

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