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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?

1 Answer

1 vote

Answer:


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.

User Alan Millirud
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