What is the concentration of Cd²⁺(aq)in a mixture of 0.014 mol Cd(NO₃)₂ and 1.0 mol NH₃ diluted to 1.00 L? For Cd(NH₃)₄²⁺,K_f = 1.0 × 10⁷. A) 3.7 × 10^-5 M B) 1.5 × 10^-9 M C) 1.8 × 10^-9 M D) 3.2 × 10^-4 M E) 1.4 × 10^-2 M

Question

Quizplus · Accepted Answer

Firstly, we need to write the chemical equation and calculate the initial concentrations of CdS1U1P12+S1S1P0 and S1S1P0 in the solution before any reaction occurs.

Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0 + S1U1B1U1P0 → CdS1U1P12+S1S1P0 + NHS1U1B13S1U1B0

Initial moles of Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0 = 0.014 mol
Initial moles of NHS1U1B13S1U1B0 = 1.0 mol
Initial moles of CdS1U1P12+S1S1P0 = 0 mol
Initial moles of S1S1P0 = 0 mol

Total volume of solution = 1.00 L

Therefore, the initial concentration of CdS1U1P12+S1S1P0 = 0 M and the initial concentration of S1S1P0 = 0 M.

Next, we need to determine which species is the limiting reagent. This can be done by comparing the initial moles of Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0 and S1U1B1U1P0.

Initial moles of Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0 = 0.014 mol
Initial moles of S1U1B1U1P0 = 1.0 mol

Since there is an excess of S1U1B1U1P0, Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0 is the limiting reagent.

Using stoichiometry, we can calculate the moles of CdS1U1P12+S1S1P0 formed from the reaction between Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0 and S1U1B1U1P0.

Moles of Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0 = 0.014 mol
Moles of CdS1U1P12+S1S1P0 formed = 0.014 mol × 1 mol CdS1U1P12+S1S1P0 / 1 mol Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0 = 0.014 mol

Therefore, the final moles of CdS1U1P12+S1S1P0 = 0.014 mol and the final moles of S1S1P0 = 1.0 mol - 0.014 mol = 0.986 mol.

Using the total volume of the solution, we can calculate the final concentrations of CdS1U1P12+S1S1P0 and S1S1P0.

Concentration of CdS1U1P12+S1S1P0 = 0.014 mol / 1.00 L = 1.4 × 10S1U1P1-2S1S1P0 M
Concentration of S1S1P0 = 0.986 mol / 1.00 L = 9.86 × 10S1U1P0S1S1P0 M

Finally, we can use the equilibrium constant to calculate the concentration of CdS1U1P12+S1S1P0 in the equilibrium mixture.

KS1U1B1fS1U1B0 = [CdS1U1P12+S1S1P0][S1S1P0] / [Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0][S1U1B1U1P0]

[CdS1U1P12+S1S1P0] = KS1U1B1fS1U1B0[Cd(NHS1U1B13S1U1B0)S1U1B12S1U1B0][S1U1B1U1P0] / [S1S1P0]

[CdS1U1P12+S1S1P0] = (1.0 × 10S1U1P17S1S1P0)(0.014 mol / 1.00 L)(9.86 × 10S1U1P0S1S1P0 mol / 1.00 L) / (0.986 mol / 1.00 L)

[CdS1U1P12+S1S1P0] = 1.8 × 10S1U1P1-9S1S1P0 M

Therefore, the best choice is C.

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