Question

The molar solubility of nickel(II) sulfide in a 0.182 M ammonium sulfide solution is

Answer 1

NiS ---> Ni+2 + S-2

let the molar solubility be s

then

[Ni+2] = s

[S-2] = s

now

the solubility product is given by

Ksp = [Ni+2] [S-2]

now

(NH4)2S --> 2 NH4+ + S-2

we know that

(NH4)2S is completely soluble

So

[S-2] = 0.182

now

using common ion effect we get

[S-2= 0.182

also for NiS , Ksp = 1 x 10-22

so

Ksp = [s] [0.182]

1 x 10-22 = [s] [0.182]

[s] = 5.5 x 10-22

so

the molar solubility of NiS is 5.5 x 10-22 M