Homework Help: Chemistry

Posted by Sara on Thursday, February 3, 2011 at 5:24pm.

Hi
Please help
Initial [Fe3+] = 0.00050 M
Initial [SCN-] = 0.0010 M
Equilibrium [Fe(SCN)2+] = 5.7 x 10-5

Find [Fe3+]E and [SCN-]E (in mol/L)
Find K.

Balanced equation is:
Fe3+ + SCN- -> Fe(SCN)2+ (reversible)

I know I can get K if I have the equilibrium concentrations, but not sure how to calculate these with so little info.

Thanks.

Chemistry - DrBob222, Thursday, February 3, 2011 at 5:54pm
...........Fe^+3 + SCN^- ==> FeSCN^+2
begin......5E-4....1E-3......0
change
equilib......................5.7E-5
Wouldn't you think that if the equilibrium concn FeSCN^+2 = 5.7E-5 and it was zero initially, then 5.7E-5 M must have been used from the initial SCN^- of 1E-3 and 5.7E-5 must have been used from the initial 5E-4 of Fe^+3.
So equilibrium concn Fe^+3 is 5E-4 - 5.7E-5 and equilibrium concn SCN^- is 1E-3 - 5.7E-5. Check for typos

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