Question

Consider the complex ion \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+} .\) Its \(K_{\mathrm{f}}\) is \(2.1 \times 10^{18} .\) At what concentration of \(e n\) is \(67 \%\) of the \(\mathrm{Ni}^{2+}\) converted to \(\left.\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\right\\}\)

Short answer

Answer: The concentration of ethylenediamine (en) required to convert 67% of the Ni(2+) ions to the [Ni(en)3](2+) complex is approximately 6.69 x 10^-7 M.

Step-by-step solution

Write the chemical equilibrium equation

For the complex formation of \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\), the chemical equilibrium equation will be: \(\mathrm{Ni}^{2+}(aq) + 3 \mathrm{en}(aq) \rightleftharpoons \left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(aq)\)

Write the expression for the formation constant K\(_f\)

Now, let's find the expression for the formation constant \(K_\mathrm{f}\) using the equilibrium concentrations of the species: \(K_\mathrm{f} = \frac{[\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^ {2+}]}{[\mathrm{Ni}^{2+}][\mathrm{en}]^3}\) where \([\mathrm{Ni}^{2+}]\) and \([\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}]\) are equilibrium concentrations of \(\mathrm{Ni}^{2+}\) and \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\), respectively, and \([\mathrm{en}]\) is the concentration of the complexing agent ethylenediamine (en). Solving this expression for \([\mathrm{en}]^3\): \([\mathrm{en}]^3 = \frac{[\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}]}{[\mathrm{Ni}^{2+}] \times K_\mathrm{f}}\)

Use the given 67% conversion to find the relation between concentrations

Since 67% of the \(\mathrm{Ni}^{2+}\) is converted to the complex, we can write: \([\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}] = 0.67 \times [\mathrm{Ni}^{2+}]_0\) where \([\mathrm{Ni}^{2+}]_0\) is the initial concentration of \(\mathrm{Ni}^{2+}\). The uncomplexed \(\mathrm{Ni}^{2+}\) concentration will then be 33%, so: \([\mathrm{Ni}^{2+}] = 0.33 \times [\mathrm{Ni}^{2+}]_0\) Now substitute these concentrations into the equation from Step 2: \([\mathrm{en}]^3 = \frac{0.67 \times [\mathrm{Ni}^{2+}]_0}{0.33 \times [\mathrm{Ni}^{2+}]_0 \times K_\mathrm{f}}\)

Solve for the concentration of en ([en])

Now, solve for \([\mathrm{en}]\) using the given value of \(K_\mathrm{f} = 2.1 \times 10^{18}\): \([\mathrm{en}]^3 = \frac{0.67 \times [\mathrm{Ni}^{2+}]_0}{0.33 \times [\mathrm{Ni}^{2+}]_0 \times 2.1 \times 10^{18}}\) Notice that the initial concentration of \(\mathrm{Ni}^{2+}\) (\([\mathrm{Ni}^{2+}]_0\)) cancels out, so the final expression is: \([\mathrm{en}]^3 = \frac{0.67}{0.33 \times 2.1 \times 10^{18}}\) Now find the cube root of both sides to obtain the concentration of en: \([\mathrm{en}] = \sqrt[3]{\frac{0.67}{0.33 \times 2.1 \times 10^{18}}}\)

Calculate the concentration of en

Compute the numerical value: \([\mathrm{en}] = \sqrt[3]{\frac{0.67}{0.33 \times 2.1 \times 10^{18}}} \approx 6.69 \times 10^{-7} \ \mathrm{M}\) Thus, the concentration of ethylenediamine (en) required to convert 67% of the \(\mathrm{Ni}^{2+}\) to the complex \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\) is about \(6.69 \times 10^{-7}\) M.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial when studying complex ion formation. In simple terms, chemical equilibrium is the state in which the rate of the forward reaction equals the rate of the reverse reaction. This balance results in no observable changes in the concentrations of reactants and products over time, even though both reactions are still occurring.

For the complex ion \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\), the equilibrium is established between the nickel ion \(\mathrm{Ni}^{2+}\), ethylenediamine \(\mathrm{en}\), and the complex ion. At equilibrium, the rate at which the complex forms from the nickel ion and ethylenediamine is the same as the rate at which it dissociates back into these components. Students must remember that equilibrium does not mean the concentrations are equal, but that they are constant.

Formation Constant

The formation constant, \(K_\text{f}\), is an essential concept in understanding complex ion formation. It quantifies the stability of the complex in a given solution. More specifically, it measures the tendency of a metal ion to bind with ligands to form a complex.

Mathematically, the formation constant is the ratio of the concentration of the complex ion to the concentrations of the free ions, each raised to the power of its stoichiometric coefficient in the balanced equilibrium equation. In the example given, the equation is \(K_\mathrm{f} = \frac{[\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}]}{[\mathrm{Ni}^{2+}][\mathrm{en}]^3}\).

A large value of \(K_\mathrm{f}\) indicates a strong tendency to form the complex ion, signifying a high stability and low concentration of free metal and ligand ions in the solution. Conversely, a small formation constant suggests a weakly bound complex, which is prone to dissociate.

Equilibrium Concentration

The term 'equilibrium concentration' refers to the concentration of each species in a reaction mixture when the chemical system has reached equilibrium. Importantly, these concentrations remain constant over time as long as the system isn't disturbed.

In the problem we are examining, the Ni(II) ions bind with ethylenediamine (en) to form \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}\). When 67% of \(\mathrm{Ni}^{2+}\) ions have formed the complex, it means that 67% of the initial \(\mathrm{Ni}^{2+}\) concentration, denoted as \(\left[\mathrm{Ni}^{2+}\right]_0\), is now in the form of the complex ion.

Using the formation constant and the given percentage of conversion, we can calculate the equilibrium concentration of the unbound ethylenediamine. This concentration is a crucial figure in calculating how much ligand is needed in the solution to form the desired amount of complex ion. It assures that students can apply the concept of equilibrium to find out not only the existence but also the extent of a reaction under given conditions.