What is the approximate concentration of free Cu+ ion at equilibrium when 1.91x10-2 mol copper(I) nitrate is added to 1.00 L of solution that is 1.030 M in CN. For [Cu(CN)₂], K = 1.0x1024. [Cut] = M use ure References to access important values if needed for this question. What is the approximate concentration of free Cd2+ ion at equilibrium when 1.48x10-2 mol cadmium(II) nitrate is added to 1.00 L of solution that is 1.030 M in CN™. For [Cd (CN)4]2, K₁= 6.0x10¹8. [Cd²+] = M

Chemistry: An Atoms First Approach
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Chapter15: Solubility And Complex Lon Equilibria
Section: Chapter Questions
Problem 69E: A solution is formed by mixing 50.0 mL of 10.0 M NaX with 50.0 mL of 2.0 103 M CuNO3. Assume that...
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What is the approximate concentration of free Cu+ ion at equilibrium when
1.91x10-2 mol copper(I) nitrate is added to 1.00 L of solution that is 1.030 M
in CN. For [Cu(CN)₂], K = 1.0x1024.
[Cut] =
M
use tine References to access important values ir needed for this question.
What is the approximate concentration of free Cd2+ ion at equilibrium when
1.48x10-2 mol cadmium(II) nitrate is added to 1.00 L of solution that is 1.030
M in CN. For [Cd (CN)4]2, K₁= 6.0x10¹8.
[Cd2+] =
M
Transcribed Image Text:What is the approximate concentration of free Cu+ ion at equilibrium when 1.91x10-2 mol copper(I) nitrate is added to 1.00 L of solution that is 1.030 M in CN. For [Cu(CN)₂], K = 1.0x1024. [Cut] = M use tine References to access important values ir needed for this question. What is the approximate concentration of free Cd2+ ion at equilibrium when 1.48x10-2 mol cadmium(II) nitrate is added to 1.00 L of solution that is 1.030 M in CN. For [Cd (CN)4]2, K₁= 6.0x10¹8. [Cd2+] = M
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