The formation constant* of [M(CN)4]2– is 7.70 × 1016, where M is a generic metal. A 0.150-mole quantity of M(NO3)2 is added to a liter of 0.870 M NaCN solution. What is the concentration of M2 ions at equilibrium?
This is a little hard to show in writing but here goes.
......M^2+ + 4CN^- ==> [M(CN)4]^2-
I..0.150....0.870.......0
C.-0.150...-0.600....+0.150
E.....0.....0.270.....0.150
The first equilibrium shows the formation of the complex on the right. With such a large Kf of over 10^16, it is obvious that essentially ALL of the M^2+ (from the M(NO3)2) reacts to form the complex.
Now that the complex is formed, we reverse the equilibrium (so it appears we are starting with [M(CN)4]^2- to form M^+ and CN^-. In this respect it's just like a Ka or Kb problem in which you're looking for the products. The E line for the first equilibrium is the I line for the reverse equilibrium. So it looks this way.
M^+ + 4CN^- ==> [M(CN)4]^2-
0....0.270......0.150.......I
x......4x........-x.........C
x..0.270+4x....0.150-x......E
Plug the E line into the Kf expression and solve for x = (M^2+).
Post your work if you get stuck and I can help you through it.
Where can I find the Kf expression?
I don't understand, DrBob222.
To find the concentration of M2 ions at equilibrium, we need to first determine the concentration of [M(CN)4]2– ions. We will use the given information about the formation constant and the initial concentrations of M(NO3)2 and NaCN solution.
Let's follow the step-by-step process to solve this problem:
1. Write the balanced chemical equation for the reaction:
M(NO3)2 + 4NaCN ⇌ [M(CN)4]2– + 2NaNO3
2. Define the equilibrium expression for the formation of [M(CN)4]2–:
Kf = [M(CN)4]2– / ([M(NO3)2] * [NaCN]^4)
3. Identify the given quantities:
Initial moles of M(NO3)2 = 0.150 moles
Initial concentration of NaCN = 0.870 M
Formation constant (Kf) = 7.70 × 1016
4. Determine the equilibrium concentration of [M(CN)4]2–:
[M(CN)4]2– = (Kf * [M(NO3)2] * [NaCN]^4) / (1 + Kf * [NaCN]^4)
= (7.70 × 1016 * 0.150 * (0.870)^4) / (1 + 7.70 × 1016 * (0.870)^4)
5. Calculate the concentration of M2 ions at equilibrium:
Concentration of M2 ions at equilibrium = 2 * [M(CN)4]2–
= 2 * [(7.70 × 1016 * 0.150 * (0.870)^4) / (1 + 7.70 × 1016 * (0.870)^4)]
By substituting the given values into the equations and performing the calculations, you will find the concentration of M2 ions at equilibrium.