Pre Lab Questions
1. For the solutions that you will prepare in Step 2 of Part I below, calculate the [FeSCN2+]. Presume that all of the SCN- ions react. In Part I of the experiment, mol of SCN- = mol of FeSCN2+. Record these values in the table below.
Test tube [FeSCN2+]
1 0.00016
2 0.00012
3 0.00008
4 0.00004
2. When 5.00 ml of 0.0200 M Fe(NO3)3 is mixed with 5.00 ml of 0.00200 M NaSCN it is found that at equilibrium FeNCS2+ is produced with a concentration of 7.0!10-4 M. Calculate the equilibrium concentration of all species present and the equilibrium constant Kc for the reaction below?
Fe3+ (aq) + SCN- (aq) → FeNCS2+ (aq)
Initial moles of Fe3+ = volume x concentration of Fe(NO3)3
= 5.00/1000 x 0.0200 = 1 x 10^(-4) mol
Initial moles of SCN- = volume x concentraton of NaSCN
= 5.00/1000 x 0.00200 = 1 x 10^(-5) mol
Total volume = 5 + 5 = 10 mL = 0.01 L
Initial concentration of Fe3+ = moles of Fe3+/total volume
= 1 x 10^(-4)/0.01 = 0.01 M
Initial concentration of SCN- = moles of SCN-/total volume
= 1 x 10^(-5)/0.01 = 0.001 M
At equilibrium:
a = [Fe(SCN)2+] = 7.00 x 10^(-4) M
(1) [Fe3+] = 0.01 - a = 9.30 x 10^(-3) M
(2) [SCN-] = 0.001 - a = 3.00 x 10^(-4) M
(3) Equilibrium constant Kc = [Fe(SCN)2+]/[Fe3+][SCN-]
= 7.00 x 10^(-4)/(9.30 x 10^(-3) x 3.00 x 10^(-4))
= 251