Respuesta :
Answer:
The value is [tex]x = 0.0227 \ M[/tex]
Explanation:
From the question we are told that
The concentration of [tex]KCN \ \ i.e \ \ CN^{-}[/tex] is [tex]M_1 = 0.091 \ M[/tex]
The solubility product constant for [tex]NiS[/tex] is [tex]K_{sp} = 3.0 *10^{-19}[/tex]
The stability constant for [tex]Ni(CN)_4 ^{2-}[/tex] is [tex]K_f = 1.0 *10^{31}[/tex]
Generally the dissociation reaction for NiS is
[tex]Ni S \underset{}{\stackrel{}{\rightleftharpoons}} Ni^{2+} + S^{2-}[/tex]
Generally the formation reaction for [tex]Ni(CN)_4 ^{2-}[/tex] is
[tex]4CN^- + N_i ^{2+} \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_{4}[/tex]
Combining both reaction we have
[tex]4CN^ - + NiS \ \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_4 + S^{2-}[/tex]
Gnerally the equilibrium constant for this reaction is
[tex]K_c = K_{sp} * K_f[/tex]
=> [tex]K_c = 3.0 *10^{-19 } * 1.0 *10^{31}[/tex]
=> [tex]K_c = 3.0*10^{12}[/tex]
Generally the I C E table for the above reaction is
[tex]4CN^ - \ \ \ + \ \ \ NiS \ \ \ \ \ \ \ \underset{}{\stackrel{}{\rightleftharpoons}} \ \ \ \ \ Ni(CN)^{2-}_4 \ \ \ \ \ \ \ \ \ + \ \ \ \ \ \ \ \ \ S^{2-}[/tex]
initial [ I] 0.091 0 0
Change [C] -4x +x + x
Equilibrium [E ] 0.091 - 4x x x
Here is x is the amount in term of concentration that is lost by [tex]CN^-[/tex] and gained by [tex]Ni(CN)_4 ^{2-}[/tex] and [tex]S^{2-}[/tex]
Gnerally the equilibrium constant for this reaction is mathematically represented as
[tex]K_c = \frac{[Ni (CN)_4^{2-} ] [S^{2-} ] }{ [CN^{-}]^4}[/tex]
=> [tex]3.0*10^{12} = \frac{x * x}{ [0.091 - 4x ]^4}[/tex]
=> [tex]3.0*10^{12}* [0.091 - 4x ]^4 = x^2[/tex]
=> [tex][0.091 - 4x ]^4 = \frac{x^2}{3.0*10^{12}}[/tex]
=> [tex][0.091 - 4x ] = \sqrt[4]{ \frac{x^2}{3.0*10^{12}}}[/tex]
=> [tex][0.091 - 4x ] = \frac{\sqrt{x} }{1316}[/tex]
=> [tex]119.8 - 5264x =\sqrt{x}[/tex]
Square both sides
[tex](119.8 - 5264x)^2 =x[/tex]
=> [tex]14352.04 - 1261255 x + 27709696x^2 = 0[/tex]
=> [tex]27709696x^2 - 1261255 x + 14352.04 = 0[/tex]
Solving using quadratic equation
The value of x is [tex]x = 0.0227 \ M[/tex]
Hence the amount in terms of molarity (concentration) of [tex]Ni(CN)_4 ^{2-}[/tex] and [tex]S^{2-}[/tex] produced at equilibrium is [tex]x = 0.0227 \ M[/tex] it then means that the amount of NiS (nickel(II) sulfide) lost at equilibrium is [tex]x = 0.0227 \ M[/tex]
So the molar solubility of nickel(II) sulfide at equilibrium is
[tex]x = 0.0227 \ M[/tex]
