In: Chemistry
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
1) We will need to know the solubility product constant, Ksp value of Ni(II) sulfide, NiS, which is 4.0*10-20 (alpha form). The dissociation equation and Ksp expression are:
NiS (s) <======> Ni2+ (aq) + S2- (aq)
Ksp = [Ni2+][S2-]
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 Ksp and obtain
4.0*10-20 = (0.213)*[S2-]
===> [S2-] = 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 S2- 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 Ksp values aren’t listed in tables available on the internet. Kindly provide the Kspvalue.