After mixing but before reacting the concns are (Ag^+) = 0.0023/2 = 0.00115M
(CN^-) = 0.21/2 = 0.105M
I've divided by 2 since equal volumes of each have been added which means both dilute each other by a factor of 2. Then we do an ICE chart.
..........Ag^+ + 2CN^- --> Ag(CN)2^-
I....0.00115....0.105........0
C...-0.00115...-0.00115....+0.0115
E......0........0.10385....+0.0115
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We can make this assumption (that with such a huge Kf that the reaction goes to completion and there is no free Ag^+ floating around. This isn't EXACTLY right but with K of 10^20 it isn't far from right. THEN we start with another ICE chart but this time work backwards,(from right to left) like this.
......Ag^+ + 2CN^- <-- Ag(CN)2^-
......0.....0.10385....0.00115.....I
......x......+x.........-x.........C
......x.....0.10385+x...0.00115-x..E
Now substitute the E line of this last ICE chart into Kf expression and solve for x = (Ag^+). x should be a very small number.
Silver ion reacts with excess CN− to form a colorless complex ion, Ag(CN)−2, which has a formation constant Kf=3.0×1020.
Calculate the concentration of Ag+ in a solution prepared by mixing equal volumes of 2.3×10−3M AgNO3 and 0.21M NaCN.
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