Calculate the silver ion concentration, [Ag+] for the reaction [Ag(CN2)]- --> Ag+ + 2CN-. The formation constant is 1x10^21. The silver ion concentration was in a solution which was originally 5.89x10^-3 M in AgNO3 and 1.54x10^-1 M in KCN.
..................Ag^+ + 2CN^- ==> [Ag(CN)2]^-
Initial....0.00589......0.308..............0
change -0.00589...-0.0118...........0.00589
equilib....0...............0.296............0.00589
With such a large Kf of 1E21 for the complex then the reaction goes far to the right. So the easy way to see this is let's take that same equation and see it from the right to the left. Here is the ICE chart. Remember we are going from the right to the left.
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..................Ag^+ + 2CN^- <== [Ag(CN)2]^-
..................0..........0.296............0.00598 M...........Initial
..................+x.........+2x................ -x........................Change
..................+x......0.296+2x.........0.00598-x.............equilibrium
Now plug those numbers into the expression for the formation constant.
Kf = 1E21 = [Ag(CN)2]^-/[Ag^+][CN^-]^2
Post your work if you get stuck. [Ag^+] will be a very small number of course.
To calculate the silver ion concentration, [Ag+], we need to apply the principles of equilibrium and use the provided information about the initial concentrations and formation constant.
First, let's define the given information:
- Initial concentration of AgNO3 = 5.89x10^-3 M
- Initial concentration of KCN = 1.54x10^-1 M
- Formation constant, Kf = 1x10^21 (This represents the equilibrium constant for the reaction)
Now, let's assume that the silver ion concentration at equilibrium is x M.
According to the balanced equation, the molar ratio between Ag+ and [Ag(CN)2]− is 1:1. Therefore, the concentration of [Ag(CN)2]− at equilibrium would be also x M.
Using the formation constant, Kf, we can set up an equation using the following expression:
Kf = [Ag+][CN-]^2 / [Ag(CN)2]−
Substituting the values:
1x10^21 = x * [CN-]^2 / x
Simplifying the equation:
1x10^21 = [CN-]^2
Next, let's consider the initial concentrations of KCN and AgNO3. Since KCN dissociates completely to give CN- ions, the initial concentration of CN- is equal to the initial concentration of KCN. Therefore, [CN-] = 1.54x10^-1 M.
Now, we can solve for x, which represents the silver ion concentration at equilibrium.
1x10^21 = (1.54x10^-1)^2
Taking the square root of both sides, we get:
1x10^21 = 2.3716x10^-2
Thus, the silver ion concentration, [Ag+], is approximately 2.3716x10^-2 M.