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

Explanation:

 From the question we are told that

      The number of moles of is

       The volume  of is =

       The concentration of

     The for   is

The number of moles of is mathematically evaluated as

substituting values

  The equation for this reaction is  

      ⇄    

From this equation we can mathematically evaluate the limiting reactant as

For  

For  

    Hence the Limiting reactant is  

At the initial the number of moles of   produced is mathematically evaluated as

The remaining amount of is  

        Now the reaction is complete

The number of moles of   = 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    after disassociation be  z

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

The 6  is from the reaction equation

    The the number of moles of  after disassociation be

Now the equilibrium constant for this reaction is

The powers of these concentration is  are their moles  

               substituting value

Since the back  reaction is  very little so we can neglecting subtracting it from the moles of    and  adding  it to the  number of moles of  

Because it won't affect the value of z obtained  

So the number of moles  of     at equilibrium is  

   The  concentration of   is mathematically represented as

The moles of   is  =  

The  concentration of   is mathematically represented as