# Homework Help: Chemistry: Using Formation Constant to calculate an end concentration

1. Dec 14, 2012

### Hemolymph

1. The problem statement, all variables and given/known data

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?

3. 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

2. Dec 14, 2012

### Staff: Mentor

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.

3. Dec 14, 2012

### Hemolymph

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

4. Dec 15, 2012

### Staff: Mentor

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.

5. Dec 16, 2012

Ah ok thanks