Question

Ethylenediamine (en) forms an octahedral complex with Ni2+(aq) with the formula [Ni(en)3]2+. Ni2+(aq) + 3 en ⇌ [Ni(en)3]2+(aq) Kf = 4.0 x 1018 If there are 0.16 mol [Ni(en)3]2+ and 0.80 mol ethylenediamine at equilibrium in a 2-L solution, what is the concentration of Ni2+(aq) in the solution?

Answer 1

Answer:

$3.125\times 10^{-19} mol/L$ is the concentration of $Ni^{2+}(aq)$ in the solution.

Explanation:

$Ni^{2+}(aq) + 3 en\rightleftharpoons [Ni(en)_3]^{2+}(aq)$

Concentration of nickel ion = $[Ni^{2+}]=x$

Concentration of nickel complex= $[[Ni(en)_3]^{2+}]=\frac{0.16 mol}{2 L}=0.08 mol/L$

Concentration of ethylenediamine = $[en]=\frac{0.80 mol}{2 L}=0.40 mol/L$

The formation constant of the complex = $K_f=4.0\times 10^{18}$

The expression of formation constant is given as:

$K_f=\frac{[[Ni(en)_3]^{2+}]}{[Ni^{2+}][en]^3}$

$4.0\times 10^{18}=\frac{0.08 mol/L}{x\times (0.40 mol/L)^3}$

$x=\frac{0.08 mol/L}{4.0\times 10^{18}\times (0.40 mol/L)^3}$

$x=3.125\times 10^{-19} mol/L$

$3.125\times 10^{-19} mol/L$ is the concentration of $Ni^{2+}(aq)$ in the solution.