Question

what is the concentration of Cd2+ in a 0.010 M Cd(NO3)2 solution that is also 1.0 M NH3? For Cd(NH3)4^2+, Kf=1.0x10^7

Answer 1

[Cd2+] = 0.010 M

[NH3] = 1.0 M

Kf for Cd(NH3)42+ = 1.0 x 107

Assume the volume of the solution is 1 L.

Determine the moles of Cd2+ and NH3 as follows:

Molarity = moles of solute / Volume(L)

Rearrange the formula for moles of solute as follows:

Moles of solute = Molarity x Volume(L)

Moles of Cd2+ = 0.010 M x 1L = 0.010 mol

Moles of NH3 = 1.0 M x 1 L = 1 mol

The formation reaction of Cd(NH3)42+ is as follows:

Cd2+(aq) + 4NH3a(q) Cd(NH3)42+(aq)

Draw an ICE chart as follows:

	Cd2+(aq)	4NH3a(q)	Cd(NH3)42+(aq)
I(moles)	0.010	1	0
C(moles)	-0.010	-4x(0.010)	0.010
E(moles)	0	0.96	0.010

Now,

[NH3] = 0.96 mol / 1 L = 0.96 M

[Cd(NH3)42+] = 0.010 mol / 1 L = 0.010 M

The dissociation equilibrium for Cd(NH3)42+ is as follows:

Cd(NH3)42+(aq) Cd2+(aq) + 4NH3(aq)

Draw an ICE chart for the equilibrium as follows:

	Cd(NH3)42+(aq)	Cd2+(aq)	4NH3(aq)
I(M)	0.010	0	0.96
C(M)	-x	+x	+4x
E(M)	0.010-x	x	0.96+4x

Determine the Kd value using given Kf value as follows:

Kd = 1/ Kf

Kd = 1/(1.0 x 107)

Kd = 1.0 x 10-7

The Kd expression is as follows:

Kd = [Cd2+][NH3]4 / [Cd(NH3)42+]

1.0 x 10-7 = [x][0.96+4x]4 / [0.010-x]

1.0 x 10-7 = [x][0.96]4 / [0.010] ---------Here, [ 0.96+4x] 0.96 and [0.010-x] 0.010, since x << 0.96 and 0.010

x = 1.18 x 10-9

As per the ICE chart,

x = [ Cd2+] = 1.2 x 10-9 M [2 S.F]