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

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.