The formation constant of [M(CN)4]^-2 is 7.70 × 10^16, where M is a generic metal. A 0.150-mole quantity of M(NO3)2 is added to a liter of 0.840 M NaCN solution. What is the concentration of M2 ions at equilibrium?

Well, let me juggle the numbers for you. Based on the formation constant, it seems like M really loves cyanide (CN). But who am I to judge their chemistry preferences?

Let's call the concentration of M^2+ ions x. After the M(NO3)2 dissociates, you'll have x moles of M^2+ ions. Since you added 0.150 moles of M(NO3)2, the total moles of M^2+ ions will be 0.150 + x.

Now, the concentration of M^2+ ions at equilibrium can be found using the expression for the formation constant:

[Kf] = [M^2+][CN^-]^4

Substituting the given values, we have:

7.70 × 10^16 = x * (0.840)^4

Now, I don't want to steal your thunder, but solving that equation gives me x ≈ 1.75 x 10^-4 M. So, the concentration of M^2+ ions at equilibrium is approximately 1.75 x 10^-4 M. Voila!

I hope that brought a smile to your face and made chemistry a little less intimidating. If you have any more questions, feel free to ask!

To find the concentration of M^2+ ions at equilibrium, we need to consider the reaction between M(NO3)2 and NaCN:

M(NO3)2 + 4 NaCN ⇌ M(CN)4^2- + 2 NaNO3

From the problem, we know that the formation constant (Kf) for M(CN)4^2- is 7.70 × 10^16.

Let's assume that x moles of M(NO3)2 react to form M(CN)4^2-. Therefore, x moles of M^2+ ions will be consumed, and x moles of NaNO3 will be formed at equilibrium.

The initial concentration of M^2+ ions is 0.150 mol/L, and the initial concentration of NaCN is 0.840 mol/L.

Using an ICE (Initial-Change-Equilibrium) table, we can summarize the changes in concentrations:

M(NO3)2 + 4 NaCN ⇌ M(CN)4^2- + 2 NaNO3
Initial: 0.150 mol 0.840 mol 0 0
Change: -x mol -4x mol +x mol +2x mol
Equilibrium: 0.150 - x 0.840 - 4x x 2x

From the information given, the formation constant (Kf) for M(CN)4^2- is 7.70 × 10^16. This means that at equilibrium, the concentration of M(CN)4^2- (x) multiplied by the concentration of NaNO3 (2x) should be equal to Kf:

[x] * [2x] = 7.70 × 10^16

(2x^2) = 7.70 × 10^16

Now, we can solve for x:

2x^2 = 7.70 × 10^16
x^2 = (7.70 × 10^16) / 2
x^2 = 3.85 × 10^16
x = √(3.85 × 10^16)
x ≈ 6.21 × 10^7 mol/L

Since the concentration of M^2+ ions (x) at equilibrium is obtained, we have:

[ M^2+ ] = 6.21 × 10^7 mol/L

To find the concentration of M^2+ ions at equilibrium, we can use the concept of the formation constant (K_f) and the principle of solubility equilibrium.

The formation constant (K_f) is a measure of how well a complex ion is formed in a solution. It can be defined as the equilibrium constant for the formation of a complex ion from its constituent ions.

In this case, we are given the formation constant of the complex ion [M(CN)4]^-2 as 7.70 × 10^16. This means that the complex ion is highly stable and favored to form in the given conditions.

The balanced chemical equation for the formation of [M(CN)4]^-2 from M^2+ and CN^- ions can be written as follows:

M^2+ + 4CN^- → [M(CN)4]^-2

Let's assume that at equilibrium, x moles of M^2+ ions react with 4x moles of CN^- ions to form the complex ion [M(CN)4]^-2. Since the concentration of M^2+ ions at the start is given as 0.150 moles in 1 liter of solution, the initial concentration of M^2+ ions would be 0.150 M.

Using the concept of stoichiometry, we can determine that the concentration of CN^- ions at equilibrium would be 4x. However, we are also given that the initial concentration of NaCN is 0.840 M. Since NaCN dissociates completely in the solution to give CN^- ions, the initial concentration of CN^- ions is also 0.840 M.

Now, let's use the formation constant (K_f) to set up an equation for the equilibrium concentration of the complex ion [M(CN)4]^-2. K_f is defined as the ratio of the concentration of [M(CN)4]^-2 to the concentration of M^2+ ions multiplied by the concentration of CN^- ions raised to the power of 4, as shown in the balanced chemical equation:

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

Plugging in the given values:

7.70 × 10^16 = [M(CN)4]^-2 / ([M^2+] * [CN^-]^4)

Since we assumed that x moles of M^2+ ions react with 4x moles of CN^- ions to form the complex ion, we can substitute these values in the equation:

7.70 × 10^16 = [M(CN)4]^-2 / (0.150 * (0.840 + 4x)^4)

Now we can solve this equation to find the value of x, which represents the number of moles of M^2+ ions that reacted.

Simplifying the equation, we get:

7.70 × 10^16 = [M(CN)4]^-2 / (0.150 * (0.840 + 4x)^4)

Rearranging the equation:

[M(CN)4]^-2 = 7.70 × 10^16 * 0.150 * (0.840 + 4x)^4

Since the concentration of the complex ion [M(CN)4]^-2 is not given, we assume it to be zero initially. Therefore, the equilibrium concentration of [M(CN)4]^-2 would also be x moles.

Substituting in the above equation, we have:

x = 7.70 × 10^16 * 0.150 * (0.840 + 4x)^4

Solving this equation would give us the value of x, which represents the number of moles of M^2+ ions that reacted.

After finding the value of x, we can calculate the concentration of M^2+ ions at equilibrium by dividing the number of moles of M^2+ ions by the volume of the solution (1 liter):

Concentration of M^2+ ions = x (moles) / 1 (liter)

I assume you mean M^2+ ions. Also I assume the 0.150 mol M(NO3)2 solid does not change the volume of the solution.

You do this in two steps.
.....M^2+ + 4CN^- ==> M(CN)4^2-
I..0.150...0.840.........0
C..-0.150..-4*0.150......+0.150
E....0......0.240........0.150

What we have done here is note that the Kf is so huge that essentially all of the metal ion will be used in forming the complex.

Now we work the problem backwards, starting with 0.150M M(CN)4^2- and see how much will dissociate. Althouhg I'm writing the equation backwards it DOES NOT change Kf.
.....M(CN)4^2- ==> M^2+ + 4CN
I....0.150.........0......0.240
C.....-x...........+x.....4x
E...0.150-x.........x.....0.240+4x

Kf = [M(CN)4]^2-/(M^2+)(CN^-)^4
Substitute the E line and solve for x. You can solve this much faster is you assume x is small and can be neglected in relation to 0.150 and 0.240; i.e.,
0.150-x = 0.150 and 0.240+4x = 0.240