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:

M1V1V2=M2V2V2M2=M1V1V2


[Cd2+]:

M2=(0.0100 M)(0.100 L)(1.250 L)

M2 = 0.0008 M


[NH3]:

M2=(0.100 M)(1.150 L)(1.250 L)

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+:

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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).

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