A 0.160-mole quantity of NiCl2 is added to a liter of 1.20 M NH3 solution. What is the concentration of Ni2+ ions at equilibrium? Assume the formation constant* of Ni(NH3)6^2 is 5.5 × 10^8
Initial Moles of NH3 = 1.20 mol/L x 1 L = 1.20 Moles
Initial Moles of NiCl2 = 0.160 Moles
The problem reaction can be restated as : Ni2+ + 6NH3 --> [Ni(NH3)6]2+
Since the formation Constant is very large( of te order 108), it can be assumed that the entire Ni2+ forms the complex, without any residual Ni2+ in the product. Thus , it can be asserted that the amount of [Ni(NH3)6]2+ formed is equal to the Initial amount of NiCl2,i.e. 0.160 moles.
By stoichiometry, the amount of NH3 unconsumed = 1.20 mol - (6x 0.160) = 1.20 - (0.96) = 0.24 moles
Since volume of the solution is constant at 1 L , Concentration will be numerically equal to the number of moles.
Kf = [ [Ni(NH3)6]2+ ]/ [ Ni2+][NH3]6
Using values from above,
5.5x108= 0.160/ [ Ni2+](0.246)
Solving the Above, we get :
[ Ni2+] = [ 5.5x108 x (0.246) / 0.160 ] -1
[ Ni2+] = 1.522263 x 10-6 M at Equillibrium.
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