Question

You mix a 125.0 mL sample of a solution that is 0.0100 M in NiCl2 with a 166.0 mL sample of a solution that is 0.253 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)

Answer 1

0.0109 moles/liter NiCl2 x 0.1265 liters = 0.00138 moles NiCl2

0.246 moles NH3/1liter x 0.1835 liters = 0.0451 moles NH3

I am assuming the formula of the complex ion is: [Ni(NH3)6]+2

NiCl2 + 6NH3 ==⇒ [Ni(NH3)6]+2 + 2Cl-
Since the Kf is very large, assume all the NiCl2 reacts:

0.00138 moles NiCl2 x 6 moles NH3/1mole NiCl2 = 0.00828 moles of NH3 are consumed.

Moles of NH3 that remain = 0.0451 – 0.00828 = 0.0368

Moles of [Ni(NH3)6]+2 that are formed =
0.00138 moles NiCl2 x 1mole [Ni(NH3)6]+2/ 1 mole NiCl2 = 0.00138 moles ]Ni(NH3)6]+2.

Final volume = 126.5 mL + 183.5 mL = 310 mL = 0.310 Liters

[NH3] = 0.0368 moles/0.310 liters = 0.119 M
[Ni(NH3)6]+2 = 0.00138 moles/0.310 liters = 0.00445 M

[Ni(NH3)6{+2}]/ [Ni+2][NH3]^6 = Kf = 2.0 x 10^8

(0.00445)/[Ni+2](0.119)^6 = 2.0 x 10^8

[Ni+2] = 0.00445/(0.119)^6(2.0 x 10^8)

[Ni+2] = 7.84 x 10^-6 M