Question

Using (Kf=1.2×109 ) calculate the concentration of Ni2+(aq) and Ni(NH3)2+6 that are present at equilibrium after dissolving 1.48 gNiCl2 in 100.0 mL of 0.20 MNH3(aq).

Express your answer using three significant figures.

Answer 1

Ni²⁺(aq) + 6NH₃(aq) Ni(NH₃)₆²⁺

Molar mass of NiCl₂ = 129.6 g/mol

Concentration of Ni²⁺ = (Mass/Molar mass) x (1000/100)

                                          = (1.48/129.6) x 10 = 0.114197 M

We have,

kf = [Ni(NH₃)₆²⁺]/[NH₃]6[Ni²⁺]

1.2 x 10⁹ = x/(0.2)⁶(0.114197 -x)

(1.2 x 10⁹) x 0.2⁶ x (0.114197 –x) = x

8770.37037 = 76801x

x = 0.114196

[Ni²⁺] = 0.114197 -0.114196 = 1 x 10^-6 M

[Ni(NH₃)₆²⁺] = x = 0.114 M