Using your average Keq value (.3948), please calculate what the equilibrium concentrations of Fe3+ , SCN-...

Using your average K_eq value (.3948), please calculate what the equilibrium concentrations of Fe³⁺ , SCN^- , and FeSCN²⁺ in a solution made by combining 10.00mL of 0.0020M Fe(NO₃)₃ and 10.00mL of 0.0020M KSCN.

Expert Solution

initial molarity of Fe+3 = 10 x 0.002 / 10 + 10 = 0.001 M

initial molarity of SCN- = 10 x 0.002 / 10 + 10 = 0.001 M

Fe+3 + SCN- ----------------------------------> FeSCN+2

0.001 0.001 0 -------------------------> initial

-x -x +x -----------------------------> changed

0.001-x 0.001-x x -------------------------> equilibrium

Keq = [FeSCN²⁺] [Fe+3] [SCN-]

0.3948 = x / (0.001-x)^2

0.3948 = x / x^2 + 10^-6 - 0.002 x

0.3948 x^2 - 0.999 x + 3.95 x 10^-7 = 0

x = 3.95 x 10^-7

equilibrium concentrations :

[Fe+3] = 0.001 - x = 9.996 x 10^-4 M

[SCN-] = 0.001 - x = 9.996 x 10^-4 M

[FeSCN+2] = x = 3.95 x 10^-7 M

queen_honey_blossom answered 2 years ago

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