Question

A solution is prepared by adding 0.10 mole of Ni(NH3)6Cl2 to 0.50 L of 3.0 M NH3. Calculate [Ni(NH3)62 ] and [Ni2 ] in this solution. Koverall for Ni(NH3)62 is 5.5 x 108. That is, 5.5 x 108

Answer 1

Answer:

$[Ni^{2+}]= 2.327*10^{-8} M$

$[Ni(NH_3)_6]^{2+} _{(aq)}]= 0.3 M$  

Explanation:

 From the question we are told that

      The number of moles of $Ni(NH_3)_6Cl_2$ is $n = 0.10 \ moles$

       The volume  of $NH_3$ is = $V = 0.50L$

       The concentration of $NH_3$ $C = 3.0M$

     The $k_{overall}$ for  $[Ni(NH_3) _6^{2+}]$ is $k_{overall} = 5.5*10^{8}$

The number of moles of $NH_3$ is mathematically evaluated as

        $No\ of mole = concentration \ * \ volume$

substituting values

       $Moles = 0.50* 3.0$

                  $=1.5 \ moles$

  The equation for this reaction is  

    $Ni^{2+}_{(aq)} + 6NH_3_{(aq)}$   ⇄     $[Ni(NH_3)_6]^{2+} _{(aq)}$

From this equation we can mathematically evaluate the limiting reactant as

For   $Ni^{2+}$

                  $0.10moles [Ni ^{2+}] * \frac{6moles NH_3}{1 mole [N_i^{2+}]}$

                            $= 0.6 moles [Ni^{2+}]$

For   $NH_3$

                                 $1.5 moles [NH_3] * \frac{6moles NH_3}{1 mole [N_i^{2+}]}$

                            $= 9 moles [NH_3]$

    Hence the Limiting reactant is  $Ni^{2+}$

At the initial the number of moles of  $[Ni(NH_3)_6]^{2+} _{(aq)}$ produced is mathematically evaluated as

           $0.10 [Ni^{2+}] *\frac{1 moles [Ni(NH_3)_6]^{2+}}{1 mol [Ni^{2+}]}$

                          $= 0.10 \ moles$

The remaining amount of $NH_3$ is  $1.5 - 0.6 = 0.9 \ moles$

        Now the reaction is complete

The number of moles of  $Ni^{2+}$ = 0 moles  

 For this kind of reaction after the complex solution(which is at equilibrium) has been formed they disassociate  again ( which is also at equilibrium )

   Let the number of moles of  $Ni^{2+}$  after disassociation be  z

    The the number of moles of $NH_3$ after disassociation be   0.9 + 6z

The 6  is from the reaction equation

    The the number of moles of $[Ni(NH_3)_6]^{2+} _{(aq)}$  after disassociation be $0.10 -z$

Now the equilibrium constant for this reaction is

                $K_{overall} = \frac{[Ni (NH_3) _6]^{2+}}{[Ni^{2+} [NH_3] ^6]}$

The powers of these concentration is  are their moles  

               substituting value

                 $5.5*10^{8} = \frac{0.10 - z}{[z] [0.9 + 6z]^6}$

Since the back  reaction is  very little so we can neglecting subtracting it from the moles of  $[Ni(NH_3)_6]^{2+} _{(aq)}$  and  adding  it to the  number of moles of  $NH_3$

Because it won't affect the value of z obtained  

                 $5.5*10^{8} = \frac{0.10}{[z] [0.5]^6}$

                            $z = \frac{0.10}{5.5*10^8 * (0.9)}$

                                $= 1.16*10^{-8} \ mol$

So the number of moles  of  $Ni^{2+}$   at equilibrium is  $= 1.16*10^{-8} \ mol$

   The  concentration of  $Ni^{2+}$ is mathematically represented as

                            $concentration = \frac{mole}{volume}$

                                                    $= \frac{1.1636*10^{-8}}{0.50}$

                                                    $[Ni^{2+}]= 2.327*10^{-8} M$

The moles of  $[Ni(NH_3)_6]^{2+} _{(aq)}$ is  =   $0.10 - 2.327*10^{-8}$

                                                       $= 0.10 \ moles$

The  concentration of  $[Ni(NH_3)_6]^{2+} _{(aq)}$ is mathematically represented as

                            $concentration = \frac{mole}{volume}$

                                                    $= \frac{0.10}{0.50}$

                                                    $[Ni(NH_3)_6]^{2+} _{(aq)}]= 0.3 M$