Chemistry: Using Formation Constant to calculate an end concentration

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Discussion Overview

The discussion revolves around calculating the equilibrium concentration of copper(I) ions in a solution where copper(I) nitrate is mixed with sodium cyanide, leading to the formation of a complex ion, Cu(CN)3. The context includes a homework problem involving chemical equilibrium and formation constants.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a dilution calculation leading to new concentrations of copper and cyanide ions.
  • Another participant suggests that due to the excess of cyanide, its concentration can be assumed constant after dilution, and that all copper can be assumed to be complexed based on the large formation constant.
  • A participant calculates the concentration of the copper cyanide complex and attempts to use it to find the equilibrium concentration of copper ions, but arrives at a value that differs from the answer booklet.
  • Another participant points out a misunderstanding regarding the stoichiometry of the complex formation, indicating that a concentration higher than 10 M suggests an error in the calculations.

Areas of Agreement / Disagreement

Participants express differing views on the assumptions made regarding the concentrations of reactants and products, particularly concerning the extent of complexation and the validity of calculated concentrations. The discussion remains unresolved regarding the correct equilibrium concentration of copper ions.

Contextual Notes

There are limitations in the assumptions made about the constancy of cyanide concentration and the implications of the formation constant. The calculations presented rely on these assumptions, which may not hold under the conditions described.

Hemolymph
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Homework Statement



50.0 mL sample of 2.0 10–4 M CuNO3 is added to 50.0 mL of 4.0 M NaCN. The formation constant of the complex ion Cu(CN)3 is 1.0 10^9. What is the copper(I) ion concentration in this system at
equilibrium?



The Attempt at a Solution


I did a dilution and got the new concentrations of Cu and CN to be 1x10^-4M for copper
and 2M for CN
Then Cu+3CN→Cu(CN)_3
I (1.0x10^-4)(2)
C
E

Not sure how to proceed
 
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You have a huge excess of cyanide - so you can safely assume its concentration doesn't change (after dilution).

For the same reason - and taking into account quite large formation constant - you can safely assume all copper is complexed.

That gives you approximate, but quite reasonable values for two concentrations. There is only one unknown now.

After you are done, you should check if the assumptions hold.
 
Borek said:
You have a huge excess of cyanide - so you can safely assume its concentration doesn't change (after dilution).

For the same reason - and taking into account quite large formation constant - you can safely assume all copper is complexed.

That gives you approximate, but quite reasonable values for two concentrations. There is only one unknown now.

After you are done, you should check if the assumptions hold.

Ok, so I did that
I got the Concentration of of copper cyanide to be 1.28E7
then did the reverse rxn
CuCN_3_→3CN+Cu
(1.28E7) 3x x

the new K value i got to be 1/1E9=1E-9
so (27x^4/(1.28E7))=1E-9
i got x to be .0147

The answer booklet says 1.3E-14
 
Hemolymph said:
Ok, so I did that
I got the Concentration of of copper cyanide to be 1.28E7

You misunderstood. If all of copper was complexed, concentration of the complex is limited by the stoichiometry.

Whenever you see concentration higher than 10M you can be almost sure something is already wrong.
 
Ah ok thanks
 

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