The formation constant of [M(CN)6]4– is 2.50 × 1017, where M is a generic metal. A 0.170-mole quantity of M(NO3)2 is added to a liter of 1.51 M NaCN solution. What is the concentration of M2 ions at equilibrium?
To solve this problem, we need to consider the reaction between M(NO₃)₂ and NaCN. The formation constant of [M(CN)₆]⁴⁻ can be expressed as:
[M²⁺] + 6[CN⁻] ⇌ [M(CN)₆]⁴⁻
Let's denote the initial concentration of M²⁺ ions as x (in M), which will react with 6x moles of CN⁻ ions from the NaCN solution.
After the reaction reaches equilibrium, the concentration of M²⁺ ions will be x - [M(CN)₆]⁴⁻, as it is consumed to form the complex.
To determine the concentration of M²⁺ ions at equilibrium, we need to calculate the concentration of [M(CN)₆]⁴⁻. We can use the formation constant (Kf) and the initial concentrations of M²⁺ and CN⁻ to calculate it.
The equilibrium expression for our reaction is:
Kf = [M(CN)₆]⁴⁻ / ([M²⁺] * [CN⁻]⁶)
Given that Kf = 2.50 × 10¹⁷, [CN⁻] = 1.51 M, and initially, there are 0.170 moles of M(NO₃)₂ added to a 1 L solution of NaCN, we can calculate the initial concentration of M²⁺ ions ([M²⁺]initial).
[M(M²⁺)]initial = (0.170 moles) / (1 L) = 0.170 M
Now, let's calculate the concentration of [M(CN)₆]⁴⁻ at equilibrium.
2.50 × 10¹⁷ = [M(CN)₆]⁴⁻ / (0.170 M * (1.51 M)^6)
Simplifying the equation:
2.50 × 10¹⁷ = [M(CN)₆]⁴⁻ / (0.170 M * (1.51 M)^6)
[M(CN)₆]⁴⁻ = 2.50 × 10¹⁷ * 0.170 M * (1.51 M)^6
Now, we can substitute the value of [M(CN)₆]⁴⁻ into the expression for the concentration of M²⁺ ions at equilibrium:
[M²⁺]equilibrium = [M²⁺]initial - [M(CN)₆]⁴⁻
[M²⁺]equilibrium = 0.170 M - 2.50 × 10¹⁷ * 0.170 M * (1.51 M)^6
Now calculate the concentration of M²⁺ ions at equilibrium using the equation above.
To find the concentration of M²⁺ ions at equilibrium, we need to consider the reaction that occurs between M(NO₃)₂ and NaCN:
M(NO₃)₂ + 6 NaCN ⟶ [M(CN)₆]⁴⁻ + 2 NaNO₃
We know the formation constant, Kf, is equal to 2.50 × 10¹⁷. The formation constant expression for this reaction is:
Kf = [M(CN)₆]⁴⁻ / ([M²⁺] × [CN⁻]⁶)
We know the initial concentration of NaCN is 1.51 M, and since it completely dissociates, the initial CN⁻ concentration is also 1.51 M. The concentration of M²⁺ ions at equilibrium is represented as [M²⁺]eq.
Now, let's set up an expression for the equilibrium concentration of [M(CN)₆]⁴⁻ in terms of [M²⁺]eq and [CN⁻]⁶:
[M(CN)₆]⁴⁻ = Kf × [M²⁺]eq × [CN⁻]⁶
Substituting the values we know:
2.50 × 10¹⁷ = [M²⁺]eq × (1.51 M)⁶
Let's solve for [M²⁺]eq:
[M²⁺]eq = (2.50 × 10¹⁷) / (1.51 M)⁶
Calculating [M²⁺]eq:
[M²⁺]eq ≈ 7.45 × 10⁻⁷ M
Therefore, the concentration of M²⁺ ions at equilibrium is approximately 7.45 × 10⁻⁷ M.
The best way to work these is in a two step process. The first step is to assume the complex forms completely (and with a huge formation constant that is a good assumption) with the aid of an ICE chart. The second step is to start with those materials and see how much they will dissociate using a second ICE chart. As follows:
.......M^+2 + 6CN^- ==> M(CN)6^4-
I....0.170...1.51........0
C...-0.170..-6*0.170.....0.170
E....0.......0.49........0.170
=================================
......0......0.49........0.170....I
......x.....+6x...........-x......C
......x....0.49+6x.......0.170-x..E
Kf = 2.50E17 = [M(CN)6^4-]/(M^2+)(CN^-)^6
Substitute the last E line into Kf expression and solve for x.