Question

calculate the concentration of nickel (II) ion in the solution after the addition of 25.0 mL of 0.200 M NaCN to 60.0 mL of 0.0100 M Ni(NO3)2 (Kf is 2.0x10^31 for [Ni(CN)4]^2-)

Answer 1

The overall reaction:

Ni(NO₃)₂ + NaCN -----------> [Ni(CN)₄]^2-

Total Nickel = (60 x 0.01)/1000 = 0.0006 moles

total CN^- = (25 x 0.2)/1000 = 0.0005 moles

CN^- is in excess compared to the stoichiometry of 1 : 4 required to form [Ni(CN)₄]^2-

The equation for formation of [Ni(CN)₄]^2- :

Ni²⁺ + 4 CN^- [Ni(CN)₄]^2- Kf = [Ni(CN)₄]^2-/[Ni²⁺][CN^-]⁴

Since Kf is too big, the Nickel goes into forming [Ni(CN)₄]^2- and the amount of CN^- consumed in forming [Ni(CN)₄]^2- is 0.0006 moles x 4 = 0.0024 moles (0.0005 - 0.0024 mmoles = 0.0026 moles of CN^- is left in solution. This then comes into equilibrium.

The equilibrium table for this is:

	Ni2+	CN-	[Ni(CN)4]2-
Initial	0	0.0026	0.0006
change	+x	+4x	-x
final	+x	0.0026+4x	0.0006-x

2 x 10³¹ = 0.0006-x/[x][0.0026+4x]⁴ since x is small we can ignore the higher powers of x as well as the x in 0.6-x

2 x 10³¹ = 0.0006/45.7x (0.0026⁴ = 4.57x10^-11)

x = 0.0006/4.57x10^-11 x 2x10³¹

x = 6.56 x 10^-25 moles

The concentration would be dividing this moles between the total volume which is 85 mL. Hope this helps.