Question

The maximum amount of nickel(II) sulfide that will dissolve in a 0.213 M nickel(II) acetate solution is

2)
The maximum amount of lead phosphate that will dissolve in a 0.122 M ammonium phosphate solution is

Answer 1

1) We will need to know the solubility product constant, K_sp value of Ni(II) sulfide, NiS, which is 4.0*10^-20 (alpha form). The dissociation equation and K_sp expression are:

NiS (s) <======> Ni²⁺ (aq) + S^2- (aq)

K_sp = [Ni²⁺][S^2-]

We have a saturated solution of Ni(II) acetate, i.e, Ni(OAc)₂ having a concentration of 0.213 M; therefore, [Ni²⁺] = 0.213 M.

Plug in the expression for K_sp and obtain

4.0*10^-20 = (0.213)*[S^2-]

===> [S^2-] = 1.8779*10^-19 ≈ 1.88*10^-19

The solubility of sulfide ion is 1.88*10^-19 M. Since the volume of the solution remains constant and as per the stoichiometric dissociation above, 1 mole of S^2- is obtained from 1 mole of NiS, hence, the solubility of NiS in the saturated solution of Ni(II) acetate is 1.88*10^-19 M (ans).

Lead phosphate isn’t a common salt and K_sp values aren’t listed in tables available on the internet. Kindly provide the K_spvalue.