Question

One millimole of Ni(NO3)2 dissolves in 210.0 mL of a solution that is .300 M in ammonia.

a.) What is the initial concentration of Ni(NO3)2 in the solution?

Answer: 4.76e-3

b.) What is the equilibrium concentration of Ni2+(aq) in the solution?

I keep getting 1.19e-8, but that is not correct.

Answer 1

a) Given that One millimole of Ni(NO₃)₂ dissolves in 210.0 mL of a solution.

moles = 1 millimole = 0.001 mol

volume = 210 mL = 0.210 L

Hence,

[Ni(NO₃)₂] = moles/ volume in lItres

= 0.001 mol/ 0.210 L

= 4.76 x 10^-3 M

Therefore,

Initial concentration of Ni(NO₃)₂ in the solution = 4.76 x 10^-3 M

b)   

Given that [NH₃] = 0.3 M

We know that Kf for [Ni(NH₃)₆]²⁺ = 5.5 x 10⁸

Ni²⁺ + 6NH₃ <--------------> [Ni(NH₃)₆]²⁺

   Kf = [Ni(NH₃)₆]²⁺/ [Ni²⁺] [NH₃]⁶

    5.5 x 10⁸ =  [4.76 x 10^-3] / [Ni²⁺][0.3]⁶

   [Ni²⁺] = [4.76 x 10^-3] /  5.5 x 10⁸ x [0.3]⁶

= 1187 x 10^-11 M

Therefore,

Equilibrium concentration of Ni²⁺(aq) in the solution = 1187 x 10^-11 M