What is the equilibrium constant for the following reaction?

Question

What is the equilibrium constant for the following reaction?
NiCO3(s) + 6NH3(aq) Ni(NH3)62+(aq) + CO32-(aq)
Ksp for NiCO3 = 1.3 x 10-7 and Kf for Ni(NH3)62+(aq) = 1.2 x 10 9
1.2 x 109
1.56 x 102
9.23 x 1015
1.08 x 10-16
6.41 x 10-3
8.33 x 10-10

Answer 1

NiCO3(s) + 6NH3 <=> Ni(NH3)6(+2) + CO3(-2)

solids are not included in equilibrium constants, thus

Kc= [Ni(NH3)6(+2)]* [CO3(-2)] / [NH3]^6 (1)

NiCO3 <=> Ni+2 +CO3(-2)

Ksp= [Ni+2][CO3(-2)]=. [Ni2+]= Ksp/[CO3(-2)] (2)

Ni+2 + 6NH3 <=> Ni(NH3)6(+2)

Kf= [Ni(NH3)6(+2)] / ([Ni+2]*[NH3]^6), substitute [Ni+2] from (2)and you get
Kf=[CO3(-2)]*[Ni(NH3)6(+2)] / Ksp*[NH3]^6 =>
Kf*Ksp = [Ni(NH3)6(+2)] *[CO3(-2)]/ [NH3]^6 =Kc according to (1)

So
Kc= (1.2*10^9)*(1.3*10^-7) =156= 1.56*10^2