Textbook Question
Calculate the molar solubility of Cr(OH)3 in 0.50 M NaOH; Kf for Cr(OH)4- is 8 x 10^29.
18. Aqueous Equilibrium
Complex Ions: Formation Constant
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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?
A
0.33 M
B
1.1 × 10^-3 M
C
2.2 × 10^-3 M
D
2.2 × 10^-28 M
Verified step by step guidance1
Identify the initial concentrations of the ions involved: Cu2+ is 2.2 × 10^-3 M and LiCN is 0.33 M.
Write the balanced equation for the formation of the complex ion: Cu2+ + 4CN- ⇌ Cu(CN)4^2-.
Use the formation constant (Kf) expression for the complex ion: Kf = [Cu(CN)4^2-] / ([Cu2+][CN-]^4).
Assume that x is the change in concentration of Cu2+ that forms the complex ion. At equilibrium, [Cu2+] = (2.2 × 10^-3 - x) M and [CN-] = (0.33 - 4x) M.
Substitute the equilibrium concentrations into the Kf expression and solve for x, which represents the amount of Cu2+ that reacts. The remaining Cu2+ concentration will be (2.2 × 10^-3 - x) M.
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