The Solubility Product Constant for nickel(II) sulfide is 3.0x10^-21. The molar solubility of nickel(II) sulfide in a 0.184 M ammonium sulfide solution is ___ M.

Answer 1

The dissociation equation for ammonium sulfide (NH4)2S is (NH4)2S(s) → 2NH4+(aq) + S2-(aq). 

NH4)2S completely dissociates in water. Therefore in this solution there is 0.184M [S]2- from ammonium sulfide.

Assume the molar solubility of NiS in this solution is x, if there is no addition S2- from (NH4)2S, the dissociation equation will be:

NiS (s)⇌Ni 2+ (aq)+S2− (aq)

1-x x x

Now, consider the S2- from (NH4)2S, this equation become

NiS (s)⇌Ni 2+ (aq)+S2− (aq)

1-x x x+0.184x

Therefore, put [Ni]2+ and [S]2- into Ksp equation

We have Ksp=[Ni]2+[S]2- =x(x+0.184)= x^2+ 0.184x =3.0x10^-21

Because the small Ksp value, x^2 is negligible compared to 0.184x, the above equation could be simplified to 0.184x = 3.0x10^-21

Now, solving x, we have: x= 1.6 x 10^-20