Question

What is the molar solubility of nickel(II) sulfide in 0.091 M KCN? For NiS, Ksp = 3.0 × 10 –19; for Ni(CN) 4 2–, Kf = 1.0 × 10 31.

Answer 1

Answer:

The value is  $x = 0.0227 \ M$

Explanation:

From the question we are told that

     The concentration of $KCN \ \ i.e \ \ CN^{-}$ is  $M_1 = 0.091 \ M$

     The solubility product constant for $NiS$ is  $K_{sp} = 3.0 *10^{-19}$

     The stability  constant for $Ni(CN)_4 ^{2-}$ is  $K_f = 1.0 *10^{31}$

Generally the dissociation  reaction for NiS is  

       $Ni S \underset{}{\stackrel{}{\rightleftharpoons}} Ni^{2+} + S^{2-}$

Generally the formation reaction for $Ni(CN)_4 ^{2-}$   is  

      $4CN^- + N_i ^{2+} \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_{4}$

Combining both reaction we have

      $4CN^ - + NiS \ \underset{}{\stackrel{}{\rightleftharpoons}} \ Ni(CN)^{2-}_4 + S^{2-}$

Gnerally the equilibrium constant for this reaction is  

         $K_c = K_{sp} * K_f$

=>       $K_c = 3.0 *10^{-19 } * 1.0 *10^{31}$  

=>       $K_c = 3.0*10^{12}$

Generally the I C E  table for the above reaction is  

                     $4CN^ - \ \ \ + \ \ \ NiS \ \ \ \ \ \ \ \underset{}{\stackrel{}{\rightleftharpoons}} \ \ \ \ \ Ni(CN)^{2-}_4 \ \ \ \ \ \ \ \ \ + \ \ \ \ \ \ \ \ \ S^{2-}$

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 $CN^-$  and gained by   $Ni(CN)_4 ^{2-}$  and  $S^{2-}$

Gnerally the equilibrium constant for this reaction is mathematically represented as

              $K_c = \frac{[Ni (CN)_4^{2-} ] [S^{2-} ] }{ [CN^{-}]^4}$

=>             $3.0*10^{12} = \frac{x * x}{ [0.091 - 4x ]^4}$

=>              $3.0*10^{12}* [0.091 - 4x ]^4 = x^2$

=>              $[0.091 - 4x ]^4 = \frac{x^2}{3.0*10^{12}}$

=>              $[0.091 - 4x ] = \sqrt[4]{ \frac{x^2}{3.0*10^{12}}}$

=>              $[0.091 - 4x ] = \frac{\sqrt{x} }{1316}$

=>              $119.8 - 5264x =\sqrt{x}$

Square both sides

                 $(119.8 - 5264x)^2 =x$

=>               $14352.04 - 1261255 x + 27709696x^2 = 0$

=>                $27709696x^2 - 1261255 x + 14352.04 = 0$

Solving using quadratic equation

   The value of x  is  $x = 0.0227 \ M$

Hence the amount in terms of  molarity (concentration) of  $Ni(CN)_4 ^{2-}$  and  $S^{2-}$ produced at equilibrium is $x = 0.0227 \ M$ it then means that the amount of  NiS (nickel(II) sulfide) lost at equilibrium is  $x = 0.0227 \ M$

So the molar solubility of nickel(II) sulfide at equilibrium is  

        $x = 0.0227 \ M$