ma Kf problem - 0.370M So instead of FORMING the Ag(NH 3 2...

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Preset Chapter 17 Beta Problems #2 This problem is testing your knowledge of K f but smartwork does this the ICE table in a very strange manner. I think that it is easier to think of this problem with some backwards equilibrium than follow the way smartwork does this problem. In a solution of ammonia, Ag + doesn’t actually exist. All of the silver ion will complex with ammonia to form the Ag(NH 3 ) 2 + complex. This means that the initial value for the silver ion in the ICE table is 0, and the initial value of the complex is 0.170M because we are in 1L of solution This leaves the question of what to do with the initial NH 3 concentration. With the balanced equation, you can find that 2 moles of ammonia will react with one mole of silver ion; you can also note that with this stoichiometry, ammonia is in excess, while silver is your limiting reagent. Before the complex forms, there is 0.710M ammonia; the 0.170M of silver ion becoming the metal complex will react with 0.340M ammonia, so the initial ammonia concentration is actually
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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 C-x +x +2x E 0.170-x 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*10-8 , which is the equilibrium concentration of Ag + in the solution....
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