In: Chemistry
10.00 mL of 3.48E-3 M Fe(NO3)3 is mixed with 4.00 mL of 2.84E-3 M KSCN and 6.00 mL of water. The equalibrium molarity of FeSCN2+ is found to be 8.70E-5 M. If the reaction proceeds as shown below, what is the equilibrium molarity of Fe3+ and SCN-? Fe3+(aq) + SCN-(aq) FeSCN2+(aq)
Equilibrium Concentration Fe3+ ____________
Equilibrium Concentration SCN- ____________
initial molarity of Fe+3 = 10 x 3.48 x 10^-3 / (10 +4 + 6)
= 1.74 x 10^-3 M
initial molarity of SCN- = 4 x 2.84 x 10^-3 / 20
= 5.68 x 10^-4 M
Fe+3 + SCN- ----------------------------> FeSCN2+
1.74 x 10^-3 5.68 x 10^-4 0 ----------------> initial
1.74 x 10^-3-x 5.68 x 10^-4 -x x -------------------> equilibrium
x = 8.70 x 10^-5 M
equilibrium concentration of Fe+3 = 1.74 x 10^-3-x
= 1.74 x 10^-3 - 8.70 x 10^-5
= 1.65 x 10^-3 M
equilibrium concentration of SCN- = 5.68 x 10^-4 -x
=5.68 x 10^-4 - 8.70 x 10^-5
= 4.81 x 10^-4 M