Question

The reaction between the compound [Fe (H2O).) (NO3)2 and Na (NSC) produces a red ferric thiocyanate complex, [Fe (H2O):(NSC)] (NO3)2. Initial concentration of [Fe (H2O).) (NO3)3 and Na (NSC) were determined to be 0.025M and 0.050M. Equilibrium concentration of [Fe (H2O)s(NSC)) (NO3)2 was determined to be 0.016M. Calculate the equilibrium, Kc. [Fe (H20).] (NO3)2 (aq) + Na (NSC) (aq) → [Fe (H2O):(NSC)] (NO3)2 (aq) +NaNO3(aq) + H20(1)

Answer 1

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ICE table for the given reaction:

[Fe (H20).] (NO3)2 (aq) + Na (NSC) (aq) →[Fe (H2O)(NSC)] (NO3)2 (aq) +NaNO3(aq) + H20(0) Initial: 0.025 0.05 Change: +x Eqbm: 0.025-x 0.05-x -X +x

Given that, Final concentration of [Fe(H₂O)₅(NSC)](NO₃)₂ is 0.016 M.

Therefore, change, x = 0.016 M.

Therefore:

[Fe(H2O).(NO3)41 = 0.025 - I = 0.025 -0.016 M = 0.009 M

Na(NSC) = 0.05 - I = 0.05 – 0.016 M = 0.034 M

[[Fe(H20)(NSC)(NO3)2) = 0.016 M

NaNO3) = 0.016 M

H2O = 0.016 M

Therefore, equilibrium constant is given by,

[[Fe(H20)(NSC)](NO3)2] - [H20].(NaNO3). Kc = [[Fe(H20)6)(NO3)4] - [Na(NSC)]

:. Kc = 0.016 M X 0.016 M x 0.016 M 0.009 M X 0.034 M

Kc = 0.01339

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