# 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).

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Formation Constant concept. If you need more Formation Constant practice, you can also practice Formation Constant practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Zelaya's class at Delgado Community College.

What textbook is this problem found in?

Our data indicates that this problem or a close variation was asked in Chemistry - OpenStax 2015th Edition. You can also practice Chemistry - OpenStax 2015th Edition practice problems.