A 0.180 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 8 .

1.09 x 10⁻⁴ M

Explanation:

The equation of the reaction in given by

Ni²⁺ (aq) + 6NH₃ (aq) ⇔ Ni(NH₃)₆

At the beginning o the reaction, we have 0.18M concentration of Ni and 1.2M concentration of aqueous NH₃ and zero concentration of the product

As the reaction proceeds towards equilibrium, the concentration of the reactants decrease as the concentration of the product starts to increase

From the equation,

1 mole of Ni²⁺ reacts with 6 moles of aqueous NH₃ to give 1mole of Ni(NH3)

therefore

0.18 M of Ni would react with 1.08M  (6 x 0.18M) aqueous  NH₃ to give 0.18M of Ni(NH₃)6

At equilibrium,

1.08M of NH3 would have reacted to form the product  leaving

(1.2 - 1.08)M = 0.12M of aqueous NH₃ left as reactant.

Therefore, formation constant  K which is the ratio of the concentration of the product to that of the reactant is given by

                       K  = [Ni(NH₃)₆} / [Ni²⁺] 6[NH₃]

   5.5 x 10⁸          =  0.18 M / [Ni²⁺] [0.12]⁶

                   [Ni²⁺]= 0.18 M / (5.5 x 10⁸) (2.986 x 10⁻⁶)

                            =0.18 M / 0.00001642

                           = 1.09 x 10⁻⁴ M

[Ni]²⁺                  =   1.09 x 10⁻⁴ M

Hence the concentration of Ni²⁺ is  1.09 x 10⁻⁴ M

20 drops hope this

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