1. millimols = mL x M = 5 mL x 0.002M = 0.01 mmols or 1E-5 mols.
mols SCN = same
2.
...........Fe^3+ + 2SCN^- ==> Fe(SCN)2^+
I..........1E-5....1E-5........0
C...........-x......-2x........x
E.........1E-5-x...1E-5-2x......x
x from the problem is 3E-4M and that x 0.01L (10 mL) = 3E-6 mols. So
Fe = 1E-5-3E-6 = ? mols
SCN = 1E-5 - 2*3E-6) = ?mols
complex = 3E-6 mols
3. Fe = mols/L
SCN = mols/L
complex = 3E-4 M from the problem.
4. Substitute the E values (of concentration--not mols) into Keq expression and solve for Keq.
Note: Your teacher, trying to make things easier for you by going through steps 1,2 has actually made it more complicated and easier to make a mistake. The easy way to do this is to calculate (Fe^2+), (SCN^-) from the beginning and use that with the concn of the complex already given and solve for Keq.
Suppose that the reaction of Fe3+ and SCN- produces Fe(SCN)2 (+). 5.00mL of 2.0mM Fe3+ is mixed with 5.00 mL of 2.0mM SCN-. The student finds the equilibrium concentrations of Fe(SCN)2(+) to be 0.3mM.
1.) What is the initial number of moles of each species present?
2.) What is the equilibrium number of moles of each species present?
3.) What is the equilibrium concentration of each species present
4.) What is the value of the equilibrium constant?
1 answer