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

`3.125* 10^(-19) mol/L` is the concentration of
`Ni^(2+)(aq)` in the solution.

Step-by-step 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+)]=(0.16 mol)/(2 L)=0.08 mol/L`

Concentration of ethylenediamine =
`[en]=(0.80 mol)/(2 L)=0.40 mol/L`

The formation constant of the complex =
`K_f=4.0* 10^(18)`

The expression of formation constant is given as:

`K_f=([[Ni(en)_3]^(2+)])/([Ni^(2+)][en]^3)`

`4.0* 10^(18)=(0.08 mol/L)/(x* (0.40 mol/L)^3)`

`x=(0.08 mol/L)/(4.0* 10^(18)* (0.40 mol/L)^3)`

`x=3.125* 10^(-19) mol/L`

`3.125* 10^(-19) mol/L` is the concentration of
`Ni^(2+)(aq)` in the solution.