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 1.31 M NaCN solution. What is the concentration of M2 ions at equilibrium?

Oh boy, we're getting into the world of chemistry! Don't worry, I'm here to make it fun. So we have this fancy formation constant for [M(CN)6]4–, what a mouthful! It's 2.50 × 10^17, which is just a really big number.

Now, we're adding 0.170 moles of M(NO3)2 to a liter of 1.31 M NaCN solution. Let's see what we can do with that.

Since we have 1.31 M NaCN, that means we have 1.31 moles of NaCN in 1 liter of solution.

So when we add 0.170 moles of M(NO3)2, we can assume that all of the M2 ions will react with the CN- ions from the NaCN to form [M(CN)6]4–. That's a party you don't want to miss!

Now, if all the M2 ions react, we can say that the concentration of M2 ions is 0.170 M. Because all the M2 ions got together with those CN- ions to form [M(CN)6]4–. And they're probably having a great time, doing the Macarena or something.

So, the concentration of M2 ions at equilibrium is 0.170 M. And there you have it, the chemistry dance party is complete!

To find the concentration of M²⁺ ions at equilibrium, we need to calculate the equilibrium concentration of [M(CN)₆]⁴⁻ ions first.

Let's assume that x moles of M²⁺ ions react with the 0.170 moles of NaCN. Since the stoichiometry is 1:6 between M²⁺ and [M(CN)₆]⁴⁻, the concentration of [M(CN)₆]⁴⁻ ions at equilibrium is 6x.

The concentration of NaCN is 1.31 M, so the initial concentration of CN⁻ ions is also 1.31 M.

Using the expression for the formation constant, we have:

Kf = [M(CN)₆]⁴⁻ /[M²⁺] x [CN⁻]⁶

Substituting the given values:

2.50 × 10¹⁷ = (6x) / (x)⁶

Simplifying further:

2.50 × 10¹⁷ = 6 / (x)⁵

Rearranging the equation:

(x)⁵ = 6 / (2.50 × 10¹⁷)

Taking the fifth root of both sides:

x = (6 / (2.50 × 10¹⁷))^(1/5)

Evaluating the above expression, we find that x ≈ 1.92 × 10⁻⁴ M.

Since the concentration of M²⁺ is x moles/liter, the concentration of M²⁺ ions at equilibrium is 1.92 × 10⁻⁴ M.

To find the concentration of M^2+ ions at equilibrium, we need to consider the reaction between M^2+ ions and CN^− ions to form [M(CN)6]^4− ions. The reaction equation is as follows:

M^2+ + 6CN^- ⇌ [M(CN)6]^4-

The formation constant (K_f) for this reaction is given as 2.50 × 10^17.

Let's assume that x is the concentration of M^2+ ions that react with CN^- ions to form [M(CN)6]^4- ions. Since 1 mole of M(NO3)2 produces 1 mole of M^2+ ions, the initial concentration of M^2+ ions is 0.170 M.

Therefore, the concentration of M^2+ ions at equilibrium is (0.170 - x) M.

At equilibrium, the concentration of [M(CN)6]^4- ions is equal to x M, as it is formed by the reaction between M^2+ and CN^- ions in a 1:6 molar ratio.

The concentration of CN^- ions at equilibrium is (6x) M, as 6 moles of CN^- ions are consumed to form 1 mole of [M(CN)6]^4- ions.

Now, let's apply the equilibrium constant expression for the reaction:

K_f = [M(CN)6]^4- / (M^2+ * [CN^-]^6)

Substituting the given values, we have:

2.50 × 10^17 = x / ((0.170 - x) * (6x)^6)

Simplifying the equation, we have:

2.50 × 10^17 = x / ((0.170 - x) * 46656x^6)

Multiply both sides by (0.170 - x) and rearrange the equation:

2.50 × 10^17 * (0.170 - x) * 46656x^6 = x

Solving this nonlinear equation can be complex and require numerical methods such as iteration or using a software program specific for solving equations.

Once you solve the equation, you will find the value of x, which is the concentration of M^2+ ions at equilibrium.

You do these by noting that the formation constant is huge so that the complex essentially forms at 100%, then turn the reaction around as if it were a weak acid and calculate the (M^2+).

..........M^2+ + 6CN^- ==> M(CN)6^4-
I.....0.170M....1.31M........0
C....-0.170...-6*0.170.....+0.170
E........0.....0.290........0.170
Of course the M^2+ is not zero but it is a very small number. Also, of course the complex is not quite 0.170 but it is very close to that. So now we say we have a complex of 0.170 with 0.290M CN^- and ask what the (M^2+) is.
-------------------
I.........0......0.290.........0.170
C.........x......+6x............-x
E.........x......6x+0.290......0.170-x

Kf = [M(CN)6]^4-/(M^2+)(CN^-)^6
2.5E17 = (0.170-x)/(x)(0.290+6x)^6
Solve for x = (M^2+)