Question

What Is The Molar Solubility Of Nickel(II) Sulfide In 0.091 M CN-? For Nis, Ksp = 3.0 10-19; For Ni(CN)4²-, Kf,= 1.0x 10³¹. Show All Equilibrium Equations Involved Here.

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Answer 1

Given :

For NiS, Ksp = 3.0*10^-19

For [Ni(CN)₄]^2-, Kf = 1.0*10³¹

............... Ksp = 3.0*10^-19

..... Kf = 1.0*10³¹

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........K_overall = Ksp*Kf = (3.0*10^-19)*(1.0*10³¹) = 3.0*10¹²

Now, K_overall = 3.0*10¹² = [Ni(CN)₄^2-] [S^2-] / [CN^-]⁴

Set up ICE table for the overall reaction to calculate the equilibrium concentrations:

	4CN-	NiS(s)	Ni(CN)42-	S2-
Initial(M)	0.091	--	0	0
Change(M)	-4s	--	+s	+s
Equilibrium(M)	0.091-4s	--	s	s

Substitute the equilibrium values in the K_overall expression.

K_overall = 3.0*10¹² = s*s / (0.091-4s)⁴ s² / (0.091)⁴ = s² / 6.86*10^-5

s² = 6.86*10^-5 * 3.0*10¹² = 20.58*10⁷ = 2.058*10⁸

Take square root of both sides.

s = 1.4*10⁴ M

So, the molar solubility of NiS is 1.4*10⁴ M.