Question

A standard solution of FeSCN2+ is prepared by combining 9.00 mL of 0.200 M Fe(NO3)3 w/1.00 mL of 0.0020 M KSCN. The equilibrium concentration of FeSCN2+ ([FeSCN2+]std) for this standard solution is assumed to be ____ M.

I tried this problem by adding the moles of Fe(NO3)3 and KSCN and dividing that by 0.01 L. I got 0.1802 but it's wrong.

Answer 1

Write the equation to see what is going on.
Fe(NO3)3 + KSCN ==> FeSCN^+2 + 3NO3^-

mols Fe(NO3)3 = M x L = ??
mols KSCN = M x L = ??
Use the equation to make an educated guess about the FeSCN^+2. Without doing any of the math it is obvious that the KSCN is there is MUCH lesser quantity (said another way the KSCN is the limiting reagent); therefore, mols FeSCN^+2 = mols KSCN. then (FeSCN^+2) = mols FeSCN^+2/liters. Note that the final volume is 10 mL = 0.010 Liters and mols/L = M.

Answer 2

Thanks!!!

Answer 3

actually when you calculate the equilibrium concentrations of each using the 0.010 L you find that the concentration of KSCN is the limiting reagent. so [FeSCN]^2 = mols KSCN/0.010L

Answer 4

actually if you read the previous post again Abi. you'll realize it says the same thing! ;]