The metal ion M2+ in ammonia, NH3 forms the complex ion M(NH3)62+ (Kf = 5.1x108). Calculate the M2+ concentration when the equilibrium concentrations of NH3 and M(NH3)62+ are 0.057 and 0.45 M respectively.

Question

The metal ion M2+ in ammonia, NH3 forms the complex ion M(NH3)62+ (Kf = 5.1x108). Calculate the M2+ concentration when the equilibrium concentrations of NH3 and M(NH3)62+ are 0.057 and 0.45 M respectively.

Answer 1

To solve this problem, we need to use the equilibrium constant expression and the provided equilibrium concentrations.

The equilibrium constant expression for the formation of the complex ion M(NH3)62+ is:

Kf = [M(NH3)62+] / ([M2+] * ([NH3]^6))

Given that Kf = 5.1x10^8, [NH3] = 0.057 M, and [M(NH3)62+] = 0.45 M, we can substitute these values into the equation.

5.1x10^8 = 0.45 / ([M2+] * (0.057^6))

Now, we can rearrange the equation to solve for [M2+]:

[M2+] = 0.45 / (5.1x10^8 * (0.057^6))

Let's calculate the value:

[M2+] = 0.45 / (5.1x10^8 * (0.000058256))

[M2+] ≈ 15.973 M

Therefore, the concentration of M2+ is approximately 15.973 M when the equilibrium concentrations of NH3 and M(NH3)62+ are 0.057 and 0.45 M, respectively.