# Problem: Calculate the cadmium ion concentration, [Cd2+], in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO3)2 with 1.150 L of 0.100 NH3(aq).

###### FREE Expert Solution

Cd(NO3)2 → all nitrates are soluble

Cd(NO3)2(aq)Cd2+(aq) + 2 NO3-(aq)

Cd2+(aq) + 4 NH3(aq) → [Cd(NH3)4]2+                      Kf = 1.3×107

[Cd(NO3)2] = [Cd2+] = 0.0100 M

[NH3] = 0.100 M

Vsol’n = 0.100 L + 1.150 L

Vsol’n = 1.250 L

Calculate the concentration of Cd2+ and NH3 in the solution:

$\frac{{\mathbf{M}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{V}}_{\mathbf{2}}}\mathbf{=}\frac{{\mathbf{M}}_{\mathbf{2}}\overline{){\mathbf{V}}_{\mathbf{2}}}}{\overline{){\mathbf{V}}_{\mathbf{2}}}}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{M}}_{\mathbf{2}}\mathbf{=}\frac{{\mathbf{M}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{V}}_{\mathbf{2}}}}$

[Cd2+]:

M2 = 0.0008 M

[NH3]:

M2 = 0.0920 M

Cd2+ → Limiting reactant

1 mol Cd2+ react with 4 moles NH3:

amount of NH3 that will react = 0.0008 M × 4 = 0.0032 M

amount of NH3 left = 0.0920 M – 0.0032 M = 0.0888 M

1 mol of Cd2+ form 1 mol [Cd(NH3)4]2+

amount of [Cd(NH3)4]2+ formed = 0.0008 M

Construct an ICE Chart for the dissociation of [Cd(NH3)4]2+: ###### Problem Details

Calculate the cadmium ion concentration, [Cd2+], in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO3)2 with 1.150 L of 0.100 NH3(aq).