Question

Consider the complex ion \(\left[\mathrm{Ni}(e n)_{3}\right]^{2+}\). 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 \(\left[\mathrm{Ni}(e n)_{3}\right]^{2+} ?\)

Short answer

The concentration of ethylenediamine (en) when 67% of Ni²⁺ is converted to [Ni(en)₃]²⁺ is approximately 2.010000554 M.

Step-by-step solution

Write the formation equation.

The formation of complex ion is given by: $$\mathrm{Ni}^{2+}(aq) + 3e n(aq) \rightleftharpoons \left[\mathrm{Ni}(e n)_{3}\right]^{2+}(aq)$$

Set up the initial and equilibrium concentrations.

Let the initial concentration of \(\mathrm{Ni}^{2+}\) be \(c_0\). At equilibrium, \(67 \%\) of the \(\mathrm{Ni}^{2+}\) is converted to the complex ion. Thus, the equilibrium concentrations are given by: \(\mathrm{Ni}^{2+}\) : \(c_0 \times (1 - 0.67)\) \(e n\) : \(c\) \(\left[\mathrm{Ni}(e n)_{3}\right]^{2+}\) : \(c_0 \times 0.67\) Since the concentration of \(\mathrm{Ni}^{2+}\) decreases by a factor of \(0.67\), the concentration of \(e n\) will decrease by three times that amount (\(0.67 \times 3 = 2.01\)). Therefore, the equilibrium concentration of \(e n\) will be: \(e n\) : \(c - 2.01c_0\)

Write the expression for the formation constant.

The expression for the formation constant \(K_{\mathrm{f}}\) can be written as: $$K_{\mathrm{f}} = \frac{[\left[\mathrm{Ni}(e n)_{3}\right]^{2+}]}{[\mathrm{Ni}^{2+}][e n]^3}$$ Plugging in the equilibrium concentrations: $$2.1 \times 10^{18} = \frac{(c_0 \times 0.67)}{(c_0 \times (1 - 0.67))(c - 2.01c_0)^3}$$

Simplify and solve for the concentration of \(e n\).

Cancelling \(c_0\) in the equation, we obtain: $$2.1 \times 10^{18} = \frac{(0.67)}{(0.33)(c - 2.01c_0)^3}$$ Rearrange to isolate \((c - 2.01c_0)^3\): $$(c - 2.01c_0)^3 = \frac{(0.67)}{(2.1 \times 10^{18} \times 0.33)}$$ Taking the cube root of both sides of the equation: $$c - 2.01c_0 = \left(\frac{(0.67)}{(2.1 \times 10^{18} \times 0.33)}\right)^{\frac{1}{3}}$$ Now, notice that the concentration of \(e n\) that we are trying to find is: $$c = 2.01c_0 + \left(\frac{(0.67)}{(2.1 \times 10^{18} \times 0.33)}\right)^{\frac{1}{3}}$$ Since the value of \(c_0\) can be arbitrary, let's assume \(c_0 = 1\) for simplicity: $$c = 2.01 + \left(\frac{(0.67)}{(2.1 \times 10^{18} \times 0.33)}\right)^{\frac{1}{3}}$$ Evaluating the expression gives: $$c \approx 2.01 + 5.54 \times 10^{-7}$$

Conclusion

The concentration of \(e n\) when \(67 \%\) of the \(\mathrm{Ni}^{2+}\) is converted to \(\left[\mathrm{Ni}(e n)_{3}\right]^{2+}\) is approximately \(2.010000554 \, M\).