Question

Part A

Using (Kf=1.2×109 ) calculate the concentration of Ni2+(aq) and Ni(NH3)2+6 that are present at equilibrium after dissolving 1.22 gNiCl2 in 100.0 mL of 0.20 MNH3(aq).

Answer 1

The initial Ni + 2 concentration is calculated:

[Ni + 2] = g / MM * V = 1.22 g / 130 g / mol * 0.1 L = 0.09 M

It is taken into account that:

[Ni + 2] = 0.09 - X

[NH3] = 0.2 - 2 * X

[Ni (NH3) 2 + 6] = X

Since Kf is very large, it is assumed that all Ni + 2 is consumed and X = 0.09 M, it is calculated:

[NH3] = 0.2 - 2 * 0.09 = 0.02 M

From Kf's expression we clear:

[Ni + 2] = [Ni (NH3) 2 + 6] / Kf * [NH3] ^ 2 = 0.09 / 1.2x10 ^ 9 * (0.02) ^ 2 = 1.9x10 ^ -7 M

If you liked the answer, please rate it in a positive way, you would help me a lot, thank you.