Question

One millimole of Ni(NO3)2 dissolves in 260.0 mL of a solution that is 0.400 M in ammonia. The formation constant of Ni(NH3)62+ is 5.5×108. b)What is the equilibrium concentration of Ni2+(aq ) in the solution?

Answer 1

One millimole = 0.001 moles , 260.0 mL = 0.260 L

0.001 moles Ni(NO₃)₂ / 0.260 L = 3.85 x 10^-3 M Ni(NO₃)₂

Given,

K_f for  [Ni(NH₃)₆]²⁺ is 5.5 x10⁸

The reaction will be--

Ni²⁺ +  6 NH₃ -------> [Ni(NH₃)₆]²⁺

Now, K_f = [Ni(NH₃)₆]²⁺ / [Ni²⁺] [NH₃]⁶

=> 5.5 x 10⁸ = 3.85 x 10^-3 / [Ni²⁺] (0.400)⁶

=> 5.5 x 10⁸ = 3.85 x 10^-3 / [Ni²⁺] (0.004096)

=> [Ni²⁺] = 3.85 x 10^-3 / 5.5 x10⁸ *(0.004096)

=> [Ni²⁺] = 3.85 x 10^-3 / 0.022528 x 10⁸

=> [Ni²⁺] = 170.89 x 10^-11

=> [Ni²⁺] = 1.71 x 10^-9 M