A 0.130 mole quantity of NiCl 2 is added to a liter of 1.20 M NH 3 solution. What is the concentration of Ni 2 + ions at equilibrium? Assume the formation constant of Ni ( NH 3 ) 2 + 6 is 5.5 × 10

This is an incomplete question, here is a complete question.

A 0.130 mole quantity of NiCl₂ is added to a liter of 1.20 M NH₃ solution. What is the concentration of Ni²⁺ ions at equilibrium? Assume the formation constant of Ni(NH₃)₆²⁺ is 5.5 × 10⁸

Answer : The concentration of ions at equilibrium is,

Explanation :  Given,

Moles of = 0.130 mol

Volume of solution = 1 L

Concentration of = Concentration of = 0.130 M

Concentration of = 1.20 M

The equilibrium reaction will be:

Initial conc.    0.130       1.20                     0

At eqm.             x       [1.20-6(0.130)]      0.130

                                   = 0.42

The expression for equilibrium constant is:

Now put all the given values in this expression, we get:

Thus, the concentration of ions at equilibrium is,

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