Question

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In a dilute nitric acid solution, Fe³⁺ reacts with thiocyanate ion (SCN⁻) to form a dark red complex:

[Fe(H₂O)₆]³⁺ + SCN⁻ ⇌ H₂O + [Fe(H₂O)₅NCS]²⁺

The equilibrium concentration of [Fe(H₂O)₅NCS]²⁺ may be determined by how dark the colored solution is (measured by a spectrometer). In one such experiment, 1.0 mL of

0.20 M

Fe(NO₃)₃ was mixed with 1.0 mL of 3.8

×

10⁻³M KSCN and 8.0 mL of dilute HNO₃. The color of the solution quantitively indicated that the [Fe(H₂O)₅NCS]²⁺ concentration was 7.3

×

10⁻⁵M. Calculate the formation constant for [Fe(H₂O)₅NCS]²⁺.

Answer 1

Initial Molarity of Fe(NO₃)₃ (M₁)= 0.20 M

Volume (V₁)= 1.0mL

After mixing, total Volume (V₂)= 1mL + 1mL + 8.0 mL = 10.0 mL

Let Molarity of Fe(NO₃)₃ in final solution be M₂

Using Molarity equation, we have :

M₁V₁ = M₂V₂

0.20 M * 1.0 mL = M₂ * 10mL

M₂ = 0.020 M

Similarly Molarity of KSCN in final solution = 1.0 mL * 3.8 *10^-3 M / 10 mL = 3.8 * 10^-4 M

Reaction is:

[Fe(H₂O)₆ ]³⁺ + SCN^- H₂O + [Fe(H₂O)₅NCS]²⁺

I(M) 0.02 3.8 * 10^-4 0

C(M) -7.3 * 10^-5 -7.3 * 10^-5​ +7.3 * 10^-5​

E(M) 0.0199 0.000307 +7.3 * 10^-5​

So, K_f = [Fe(H₂O)₅NCS]²⁺ / (   [Fe(H₂O)₆ ]³⁺ [SCN^- ]

K_f = (7.3 * 10^-5​ ) / (0.000307)(0.0199) = 11.94 12