Question

You mix a 126.5 mL sample of a solution that is 0.0109 M in NiCl2 with a 183.5 mL sample of a solution that is 0.246 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)

Answer 1

the reaction taking place will be

Ni²⁺ + 4 NH₃ ----> Ni(NH₃)₄²⁺

0.01090 0.246 0 initial concentration

-0.01090 0.0436 (0.01090 x4) change (since 1 mol of Ni2+ reqires 4 moles of NH₃)

0 0.2024 0.01090 at equilibrium

once the equiblrium is established the complexe will dissociate to give it's reactants at equiblirium

Ni²⁺ + 4 NH₃    Ni(NH₃)₄²⁺

0 0.2024 0.01090   initial concentration
let the change be x , so
+x +4x 0.01090-4x

2.0 x 10⁸ = 0.01090-4x / [Ni²⁺] ( 0.2024+4x)⁴
= 0.01090 / [Ni2+] ( 0.2024)^4

= 6.49 / [Ni2+]

[Ni²⁺] =3.24 x 10^-8 M