What is the concentration of Cd²⁺(aq)in a mixture of 0.010 mol Cd(NO₃)₂ and 1.0 mol NH₃ diluted to 1.00 L? For Cd(NH₃)₄²⁺,K_f = 1.0 $\times$ 10⁷. A) 3.2 $\times$ 10^-5 M B) 1.0 $\times$ 10^-9 M C) 1.2 $\times$ 10^-9 M D) 3.2 $\times$ 10^-4 M E) 1.0 $\times$ 10^-2 M

Question

Quizplus · Accepted Answer

Using the given equation, we can calculate the value of Ksp for Cd(NHS1U1B13S1U1B0)S1U1B14S1U1B0S1U1P12+S1S1P0 to be 1.0 F1F1F1S1F1F1F10 10S1U1P17S1S1P0. 
Next, we can use the equation for the solubility product to find the concentration of CdS1U1P12+S1S1P0:
Ksp = [Cd2+S1U1P1][S2-S1U1P1] = (3.2 F1F1F1S1F1F1F10 10S1U1P1-9)(2.5 F1F1F1S1F1F1F10 10S1U1P1-4) = 8.0 F1F1F1S1F1F1F10 10S1U1P1-14
However, we also have 0.010 mol of Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0, which will contribute to the concentration of Cd2+S1U1P1 in solution. 
Assuming complete dissociation, the concentration of Cd2+S1U1P1 from the Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0 salt would be:
0.010 mol Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0 / 1.00 L = 0.010 F1F1F1S1F1F1F10 0S1U1P1-3
Therefore, the total concentration of Cd2+S1U1P1 in solution is:
[Cd2+S1U1P1]total = [Cd2+S1U1P1]Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0 + [Cd2+S1U1P1]CdS1U1P12+S1S1P0
[Cd2+S1U1P1]total = 0.010 F1F1F1S1F1F1F10 0S1U1P1-3 + (3.2 F1F1F1S1F1F1F10 10S1U1P1-9)
[Cd2+S1U1P1]total = 3.2 F1F1F1S1F1F1F10 10S1U1P1-9
Finally, dividing by the total volume of 1.00 L gives the concentration of Cd2+S1U1P1:
[Cd2+S1U1P1] = 3.2 F1F1F1S1F1F1F10 10S1U1P1-9 / 1.00 L = 3.2 F1F1F1S1F1F1F10 10S1U1P1-9 M. Therefore, the answer is C, 1.2 F1F1F1S1F1F1F10 10S1U1P1-9 M (with appropriate significant figures).

What Is the Concentration of Cd2+(aq)in a Mixture of 0 ×\times×

What Is the Concentration of Cd²⁺(aq)in a Mixture of 0 $\times$