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JEE Advanced2019ChemistryIonic EquilibriumActual

For the following reaction, the equilibrium constant K e at 298   K is 1.6 × 10 17 . F e 2 + a q + S 2 - a q   ⇌ F e S s When equal volumes of 0.06   M   F e 2 + a q and 0.2   M   S 2 - a q solutions are mixed, the equilibrium concentration of F e 2 + a q is found to be Y × 10 - 17 M . The value of Y is __________

Correct answer

0

Step-by-step solution

Fe 2+ + S 2- ⇌ Fe  S ( s ) ..... K c = 1.6 × 10 17 t = 0 : 0.03  M  0.1  M t = t;   0.03 - H   0.1  H Since K c > > 1 , n ≃ 0.03 ∴   [ S 2- ] = 0.1 - 0.031 = 0.07  M K c = Fe 2 + S 2- = 1 1.6 × 10 17 Fe 2 + = 1 1.6 × 0.07 × 10 - 17 y = 1 1.6 × 0.07 ≃ 8.93

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