Nickel (ii) ions form a complex ion in the presence of ammonia with a formation constant (kf) of 2.0×10^8:
ni2+ + 6nh3 ⇌ [ni(nh3)6]2+
calculate the molar solubility of nis in 3.1 m nh3. g

The molar solubility of NiS is 7.7 * 10⁻⁷ M

Explanation:

To answer this question, we need to keep in mind two equilibriums.

First, we have the solubilization of NiS:

NiS ⇄ Ni²⁺ + S²⁻   ksp= 3.0 * 10⁻²¹  (we know this from standard tables)

Second, we have the formation of the complex:

Ni²⁺ + 6NH₃ ⇄ [Ni(NH₃)₆]²⁺   kf=2.0 * 10⁻⁸

Combine the two equilibriums and we have

NiS + 6NH₃ ⇄ [Ni(NH₃)₆]²⁺ + S²⁻  K= ksp * kf =6.0* 10⁻¹³=

The molar solubility s is equal to both [Ni(NH₃)₆²⁺] and [S²]

At equilibrium, [NH₃]= 3,1 M - 6s

Thus, if we replace those terms in the formula for K, we're left with:

Using an approximation we can ignore the denominator and we have