Question #26133

.160-mole quantity of NiCl2 is added to a liter of 1.20M NH3 solution. what is the concentration of Ni2+ ions at equilibrium? assume the formation constant of Ni(NH3)6^2+ is 5.5*10^8

Expert's answer

The starting concentration of NH_{3} solutionis 1.20 M.

The starting concentration of NiCl_{2} is 0.160 mole in 1 L = 0.160 M

The starting concentration of Ni(NH_{3})_{6}^{2+} is 0

Formation constant is: K = [Ni(NH_{3})_{6}^{2+}] / [Ni^{2+}][NH_{3}]^{6}

Equilibrium concentrations:

If equilibrium concentrations of Ni(NH_{3})_{6}^{2+} is x, equilibriumconcentrations of Ni^{2+} is 0.160-x, and equilibrium concentrations of NH_{3} is 1.20^{-6}x.

K = [x]/[0.160 - x][1.20 - 6x]^{6}

x = 0.159 M

So concentration of Ni^{2+} is 0.160 - 0.159 = 0.001 M

The starting concentration of NiCl

The starting concentration of Ni(NH

Formation constant is: K = [Ni(NH

Equilibrium concentrations:

If equilibrium concentrations of Ni(NH

K = [x]/[0.160 - x][1.20 - 6x]

x = 0.159 M

So concentration of Ni

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