Formation constants and forming precipitates?

Question

Nickel ion form the complex ion {Ni(NH3)6}2+, Kf=5.6x10^8. intially formed when .002M nickel sulfate is added to .10M ammonia. Predict if nickel (2) hydroxide wil precipitate from the solution when .01M sodium hydroxide is added to the complex ion. The Ksp for Ni(OH)2 is 2.0x10^-15

Answer 1

Ni(2+) + 6NH3 = Ni(NH3)6(2+)
Kf=[Ni(NH3)6(2+)]/[Ni(2+)]*[NH3]^6.
at equilibrium [Ni(NH3)6(2+)]= approximately 0.002M (although it will be a little less), [NH3]=10-0.002=~10M
so [Ni2+]=[Ni(NH3)6(2+)]/(Kf*[NH3]^6) = 0.002/(5.6x10^8*10^6)=3.6x10^-19 (approximately because of the assumptions, if we did it properly, the equation would be: Kf=x/(0.002-x)*(10-x)^6, where x is the concentration of [Ni(NH3)6(2+)]. x would be close to 0.002 because of the huge excess of ammonia used and very high complex formation constant)

now Ni(2+) + 2OH- = Ni(OH)2
Ksp=[Ni2+]*[OH]^2=3.6x10^-19 * (0.01)^2=3.6 x10^-23. The answer is no! Only if the number was higher than the theoretical Ksp you would see the precipitate


The answer below is wrong, as [HAN]/[Ni(2+)]=[NH3]^6 * Kf = 10^6 * 5.6x10^8 = 5.6x10^15 not 560!

Answer 2

The hexammoniumnickel complex [HAN] formation is given by the equilibrium
[HAN]/[Ni+2][NH3]^6 = 5.6x10^8
an approximation is [HAN]/[Ni+2] = 560
and [HAN]+[Ni+2] = 0.002 Roughly, [HAN]=0.001995 and [Ni+2]= 0.000005 or 5x10-6 M.

When [OH-] ion is added, the Ksp problem comes into play. Here [Ni+2][OH-]^2= Ksp. The OH- content would limit [Ni+2] in solution to 2x10-11M, which is about 5 orders of magnitude less than what is there. So, there should be a ppt.

Answer 3

Lancenigo di Villorba (television), Italy Scientists say that "answer is a chemical gadget the place various of species dissolved in a fluid performing like solvent : ensuing gadget has HOMOGENEOUS character, e.g. each and every factor of this methodology has to define all gadget's mass". in addition to, AQUEOUS answer is "a chemical gadget the place various of chemical species are dissolved in water". those suggestions may be UNSATURATED (e.g. suggestions attainable to dissolve another volume of solutes) SATURATED (e.g. suggestions which will not be able to not greater get carry of solutes) or SUPERSATURATED (e.g. suggestions ran faraway from Equilibrium, each and every disturb would get surprising precipitations). end as quickly as an Aqueous answer ran Precipitation, you will desire to attend to roughly SATURATED answer. i wish this permits you.