One millimole of Ni(NO3)2 dissolves in 230.0mL of a solution that is 0.400M in a

Question

One millimole of Ni(NO3)2 dissolves in 230.0mL of a solution that is 0.400M in ammonia. The formation constant of Ni(NH3)6^2+ is Kr= 5.5x10^8.

I figured out how to get the first value, but cannot figure out the 2nd part. Please help.

Question 9 of 21 (1 point) x Incorrect X Incorrect X Incorrect X Incorrect X Incorrect X Incorrect 1 One millimole of Ni(NO3)2 dissolves in 230.0 mL of a solution that is 0.400 Min ammonia The formation constant of Ni(NH306 is Kr 5.5 x 10 (a) What is the initial concentration of Ni(NO3)2 in the solution? Number One millimole is 10-3 moles. 0.0043 (b) What is the equilibrium concentration of Ni2 (aq) in the solution? The N forms a hexammonia complex: Ni(NH3)6 Kr 5.5x108 Number M Since complex ion formation constants are very large, it is almost a 1.93 x 10 limiting reagent problem. It is easier to start by converting all of the Ni (the limiting reagent) into the ammonia complex and then letting a little of the complex (x M) turn back into Ni? Very little will turn back: 6x

Explanation / Answer

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

As Ni^2+ is the limiting reagent , amount of complex formed = 0.0043 M

amount of excess NH3 = (0.4 M *0.230L) - 6*0.001 mol = 0.086 moles

Keq of the reverse reaction is :

Ni(NH3)6^2+ (aq) <==> Ni^2+ (aq) + 6NH3 (aq) Keq = 1/Kf = 1.82*10^-9

Keq = (0.086+6x)*x/(0.001-x)

or, 1.82*10^-9 = (0.086+6x)*x/(0.001-x)

as x is very small, 0.086 + 6x ~0.086 and 0.001-x ~0.001

or, 1.82*10^-9 =x* 0.086/0.001

or, x = 0.021*10^-9 moles

[Ni^2+] = 0.021*10^-9 moles

molarity = 0.021*10^-9 moles/0.23 L = 0.092 *10^-9 M = 9.2*10^-11 M

concentration of Ni^2+ (aq) = 9.2*10^-11 M

or, x =

Ni^2+ (aq) 6NH3 Ni(NH3)6^2+ initial 0 0.086 0.001 change x 6x -x equilibrium x 0.086+6x 0.001-x

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