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 0.074 and 0.22 M respectively
M^2+ + 6NH3 ==> [M(NH3)6]^2+
Write the Kf expression for the above reaction, plug in the equilibrium concentrations of NH3 and the complex and solve for (M^2+). This is a simple one to do.
To calculate the concentration of M2+ in the given equilibrium, we can use the equilibrium constant expression (Kf) to set up an equation and solve for the unknown concentration of M2+.
The formation of the complex ion M(NH3)62+ can be represented by the following equation:
M2+ + 6NH3 ⇌ M(NH3)62+
We are given the equilibrium concentrations of NH3 and M(NH3)62+, which are 0.074 M and 0.22 M, respectively. Let's assume the initial concentration of M2+ is x M.
According to the equation, for every one mole of M2+, six moles of NH3 are required to form one mole of M(NH3)62+. Therefore, at equilibrium, the concentration of NH3 will be (0.074 - 6x) M, and the concentration of M(NH3)62+ will be (0.22 - x) M.
The equilibrium constant expression for this reaction is given by:
Kf = ([M(NH3)62+] / [M2+][NH3]^6)
Substituting the given equilibrium concentrations:
5.1x108 = (0.22 - x) / (x * (0.074 - 6x)^6)
Now, we can solve this equation for x, which will give the concentration of M2+ at equilibrium. However, the equation involves a polynomial to the sixth power, which can make the calculations complex.
To simplify the calculation, we can assume that the value of x is small compared to 0.074. This assumption is valid because the equilibrium constant is very large (Kf = 5.1x108), indicating that the formation of the complex is favored and that the concentration of M2+ will be significantly lower than the concentration of NH3.
Under this assumption, we can approximate (0.074 - 6x)^6 as approximately 0.074^6, as the contribution of the term involving x will be negligible.
Now we have:
5.1x108 ≈ (0.22 - x) / (x * 0.074^6)
Simplifying further:
(x * 0.074^6) * 5.1x108 ≈ 0.22 - x
Rearranging the equation:
5.1x108 * x * 0.074^6 + x ≈ 0.22
Solving this equation will give us the value of x and consequently the concentration of M2+ at equilibrium.
Please note that while this approximation simplifies the calculation, it introduces some error. A more accurate calculation would involve solving the equation without assuming the concentration of M2+ is small compared to NH3.