What is the answer to the Equilibrium JEE 2019 FeS question?
The answer to the Equilibrium JEE 2019 FeS question is 8.93, the dimensionless coefficient, not the iron concentration itself. This is a numerical-answer question from JEE Advanced 2019, Paper 1, Chemistry, not a multiple-choice question.
At 298 K, the supplied precipitation reaction and equilibrium constant are:
Equal volumes of 0.06 M iron(II) solution and 0.20 M sulphide solution are mixed. Find the coefficient in:
Why is the precipitation equilibrium constant the reciprocal of the ion product?
For precipitation, solid FeS belongs in the numerator, but its activity is unity. The supplied equilibrium constant is therefore the reciprocal of the dissolved-ion concentration product, not the product itself. Write the expression from the reaction’s direction before inserting numbers.
For pure solid FeS:
This does not mean the solid has a concentration of 1 M. Its activity is fixed while the pure solid phase is present.
For the reaction as written:
Consequently:
The reverse reaction is dissolution, so its solubility product satisfies:
Use this relationship to check the direction. For the main calculation, retain the supplied precipitation expression, as the official solution does.
How do dilution and precipitation change the two concentrations?
Equal-volume mixing first halves both concentrations. Precipitation then removes equal amounts of iron and sulphide because the reaction has 1:1 stoichiometry. Halving the concentrations accounts only for dilution, not precipitation.
Let each starting volume be:
The combined volume is:
Before accounting for precipitation:
These are mixed concentrations, not equilibrium concentrations. Iron is the limiting ion because less of it is present and the reaction consumes the ions in equal amounts.
The very large precipitation constant means almost all the iron precipitates. Approximately 0.03 M of each ion is therefore removed from solution.
Keep the stages separate in a concentration ledger:
- After dilution: iron is 0.03 M; sulphide is 0.10 M.
- Change through precipitation: iron decreases by approximately 0.03 M; sulphide decreases by approximately 0.03 M.
- At equilibrium: iron has a tiny positive concentration; sulphide is approximately 0.07 M.
The official subtraction is:
Iron is almost exhausted, not exactly zero. Use equilibrium to find its residual concentration.
How does the residual iron concentration give 8.93?
Substituting the remaining sulphide concentration, approximately 0.07 M, gives an iron concentration whose coefficient rounds to 8.93. The concentration retains the power of ten and the molarity unit; the requested coefficient does not. Neither original stock concentration belongs in this substitution.
Set:
Following the official approximation:
Rearrange and calculate:
Compare this with the requested form:
Therefore, after rounding:
Check the approximation using the residual iron concentration. The concentration removed by precipitation is:
The exact sulphide bookkeeping is then:
Using the calculated residual:
The correction is negligible compared with the excess sulphide concentration, so replacing the exact expression by 0.07 M is justified. This is an approximation check, not a reason to replace the official dilution–stoichiometry–equilibrium method with a quadratic calculation.
Why does using 0.10 M sulphide give the wrong answer of 6.25?
That substitution handles dilution correctly but ignores sulphide consumed during precipitation. It treats the post-mixing concentration as the equilibrium concentration, even though approximately 0.03 M sulphide has been removed into solid FeS.
This numerical-answer question has no supplied options. 6.25 is a derived wrong result, not an official distractor.
The faulty calculation is:
It incorrectly gives:
Being in excess means sulphide remains after the limiting ion has almost completely precipitated. It does not mean sulphide remains unchanged: approximately 0.03 M is consumed, leaving approximately 0.07 M.
Before substituting into an equilibrium expression, label each concentration as stock, mixed or equilibrium. Use only the equilibrium values in the ion product.
How can you test this method on two related equilibrium questions?
Test one mixture with excess sulphide and another with equal reactant amounts. The first uses an approximately fixed excess-ion concentration; the second requires equal residual concentrations. Both questions below are original practice variations from the Equilibrium chapter, not verified additional JEE PYQs.
Use the same FeS precipitation reaction at 298 K for both:
Question 1: Mix equal volumes of 0.04 M iron(II) and 0.10 M sulphide solutions. Find the equilibrium iron concentration.
After dilution:
Iron is limiting. Subtract the sulphide consumed before applying equilibrium:
Question 2: Mix equal volumes of 0.06 M iron(II) and 0.06 M sulphide solutions. Find the equilibrium concentration of each ion.
Both mixed concentrations are 0.03 M. Neither ion has a stoichiometric excess, and equal amounts precipitate, so the dissolved concentrations remain equal:
Zero stoichiometric excess does not mean zero dissolved ions. Before substituting, check which case you have: use the excess-ion concentration in the original question, but equal residual concentrations in the equal-reactant variation.
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Frequently asked questions
What is the answer to the Equilibrium JEE 2019 FeS question?
The answer is 8.93, the dimensionless coefficient Y in [Fe²⁺] = Y × 10⁻¹⁷ M. The equilibrium iron concentration is approximately 8.93 × 10⁻¹⁷ M. This was a numerical-answer question in JEE Advanced 2019, Paper 1, Chemistry.
Why is Ksp the reciprocal of Kc in the FeS question?
The supplied reaction forms solid FeS from dissolved Fe²⁺ and S²⁻, so its precipitation constant is Kc = 1/([Fe²⁺][S²⁻]). Pure solid FeS has unit activity, not a concentration of 1 M. Dissolution reverses this reaction, giving Ksp = 1/Kc = 6.25 × 10⁻¹⁸.
Why is the equilibrium sulphide concentration 0.07 M?
Equal-volume mixing first reduces the iron and sulphide concentrations to 0.03 M and 0.10 M, respectively. FeS precipitation consumes the ions in a 1:1 ratio, removing approximately 0.03 M of each. The remaining sulphide concentration is therefore approximately 0.10 − 0.03 = 0.07 M.
Why do I get 6.25 instead of 8.93 for the FeS question?
Using 0.10 M sulphide in the equilibrium expression gives 6.25 because it accounts for dilution but ignores sulphide consumed during precipitation. Use approximately 0.07 M instead to obtain [Fe²⁺] = 8.93 × 10⁻¹⁷ M and Y = 8.93. The value 6.25 is a derived wrong result, not an official answer option.
What if equal volumes of 0.06 M iron and 0.06 M sulphide are mixed?
In this practice variation, both concentrations become 0.03 M after mixing, and neither ion has a stoichiometric excess. Their equilibrium concentrations are equal, so x² = 1/Kc = 6.25 × 10⁻¹⁸. Each dissolved-ion concentration is therefore 2.50 × 10⁻⁹ M, not zero.