One millimole of Ni(NO3)2 dissolves in 260.0 mL of a solution that is 0.400 M in...

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 5.5×10^8.

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

Solutions

Expert Solution

Ni^2+(aq) + 6 NH3(aq) <---> Ni(NH3)6^2+(aq)

initial 1/260 = 0.00385 M 0.4 M 0 M

change x M 6x M x M

equilibrium 0.00385-x 0.4-6x x

Kf = [Ni(NH3)6^2+]/[Ni^2+][NH3]^6

at equilibrium, [NH3] = 0.4-6X = 0.4 M

(5.5*10^8) = x/((0.00385-x)(0.4)^6)

x = 0.0038499

at equilibrium, [Ni^2+] = 0.00385-0.0038499 = 1*10^-7 M

queen_honey_blossom answered 1 year ago

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