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:
M2 = 0.0008 M
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+:
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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