Multiple Choice
A solution contains 2.2 × 10^-3 M in Cu2+ and 0.33 M in LiCN. If the formation constant (Kf) for the complex ion Cu(CN)4^2- is 1.0 × 10^25, how much copper ion (Cu2+) remains at equilibrium?
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18. Aqueous Equilibrium
Complex Ions: Formation Constant
Multiple Choice
If your equilibrium constant K is equal to the product of Ksp and Kf, find the solubility of AgCl in 2.0 M NH3. Ksp of AgCl = 1.77 × 10−10; Kf of Ag(NH3)2+ = 1.7 × 107.
A
1.04 × 10−17 M
B
9.78 × 10−3 M
C
9.89 × 10−2 M
D
3.01 × 10−3 M
Verified step by step guidance1
Identify the relevant chemical reactions: The dissolution of AgCl in water and the formation of the complex ion Ag(NH3)2+. The reactions are: AgCl(s) ⇌ Ag+(aq) + Cl−(aq) and Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq).
Write the expression for the equilibrium constant K for the overall reaction: K = Ksp × Kf, where Ksp is the solubility product of AgCl and Kf is the formation constant of Ag(NH3)2+.
Substitute the given values into the expression: Ksp = 1.77 × 10^−10 and Kf = 1.7 × 10^7. Calculate K = (1.77 × 10^−10) × (1.7 × 10^7).
Set up the expression for the solubility of AgCl in terms of the concentration of Ag(NH3)2+ formed. Assume that the initial concentration of NH3 is 2.0 M and that the change in concentration due to complex formation is negligible.
Solve for the solubility of AgCl by equating the concentration of Ag(NH3)2+ to the solubility of AgCl, using the calculated value of K. This will give you the solubility of AgCl in 2.0 M NH3.
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