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Ionic Equilibrium question

2025 · 2 Apr · Shift 1 · Q3
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  5. /2025 · 2 Apr · Shift 1 · Q3

Ionic Equilibrium question

2025 · 2 Apr · Shift 1 · Q3

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
If equal volumes of AB2A B_2AB2​ and XYX YXY(both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of AY2\mathrm{AY}_2AY2​ at 300 K ? (Given Ksp(\mathrm{K}_{\mathrm{sp}}\left(\right.Ksp​( at 300 K ) for AY2=5.2×10−7\mathrm{AY}_2=5.2 \times 10^{-7}AY2​=5.2×10−7 )
  1. A
    2.0×10−4MAB2,0.8×10−3MXY2.0 \times 10^{-4} \mathrm{M} \mathrm{AB}_2, 0.8 \times 10^{-3} \mathrm{M} \mathrm{XY}2.0×10−4MAB2​,0.8×10−3MXY
  2. B
    2.0×10−2MAB2,2.0×10−2MXY2.0 \times 10^{-2} \mathrm{M} \mathrm{AB}_2, 2.0 \times 10^{-2} \mathrm{M} \mathrm{XY}2.0×10−2MAB2​,2.0×10−2MXY
  3. C
    1.5×10−4MAB2,1.5×10−3MXY1.5 \times 10^{-4} \mathrm{M} \mathrm{AB}_2, 1.5 \times 10^{-3} \mathrm{M} \mathrm{XY}1.5×10−4MAB2​,1.5×10−3MXY
  4. D
    3.6×10−3MAB2,5.0×10−4MXY3.6 \times 10^{-3} \mathrm{M} \mathrm{AB}_2, 5.0 \times 10^{-4} \mathrm{M} \mathrm{XY}3.6×10−3MAB2​,5.0×10−4MXY
View written solutionFree

Correct answer: B

  1. Identify the ions formed on mixing

Given salts:

  • AB2→A2++2B−AB_2 \to A^{2+} + 2B^-AB2​→A2++2B−
  • XY→X++Y−XY \to X^+ + Y^-XY→X++Y−

The possible precipitate is AY2AY_2AY2​, so its dissolution equilibrium is: AY2(s)⇌A2++2Y−AY_2(s) \rightleftharpoons A^{2+} + 2Y^-AY2​(s)⇌A2++2Y−

Hence, Ksp=[A2+][Y−]2K_{sp} = [A^{2+}][Y^-]^2Ksp​=[A2+][Y−]2

A precipitate forms if the ionic product Q=[A2+][Y−]2>Ksp=5.2×10−7Q = [A^{2+}][Y^-]^2 > K_{sp} = 5.2 \times 10^{-7}Q=[A2+][Y−]2>Ksp​=5.2×10−7

  1. Effect of mixing equal volumes

When equal volumes of the two solutions are mixed, each solution gets diluted to half its original concentration.

So after mixing:

  • [A2+]=C(AB2)2[A^{2+}] = \dfrac{C(AB_2)}{2}[A2+]=2C(AB2​)​
  • [Y−]=C(XY)2[Y^-] = \dfrac{C(XY)}{2}[Y−]=2C(XY)​

Thus,

= \frac{C_{AB_2}C_{XY}^2}{8}$$ We now check each option. --- 3. **Option A** $$C_{AB_2} = 2.0 \times 10^{-4}, \quad C_{XY} = 0.8 \times 10^{-3} = 8.0 \times 10^{-4}$$ After mixing: $$[A^{2+}] = 1.0 \times 10^{-4}$$ $$[Y^-] = 4.0 \times 10^{-4}$$ So, $$Q = (1.0 \times 10^{-4})(4.0 \times 10^{-4})^2$$ $$= (1.0 \times 10^{-4})(16 \times 10^{-8})$$ $$= 1.6 \times 10^{-11}$$ Since $$1.6 \times 10^{-11} < 5.2 \times 10^{-7}$$ no precipitate forms. --- 4. **Option B** $$C_{AB_2} = 2.0 \times 10^{-2}, \quad C_{XY} = 2.0 \times 10^{-2}$$ After mixing: $$[A^{2+}] = 1.0 \times 10^{-2}$$ $$[Y^-] = 1.0 \times 10^{-2}$$ So, $$Q = (1.0 \times 10^{-2})(1.0 \times 10^{-2})^2$$ $$= 1.0 \times 10^{-6}$$ Since $$1.0 \times 10^{-6} > 5.2 \times 10^{-7}$$ precipitate forms. --- 5. **Option C** $$C_{AB_2} = 1.5 \times 10^{-4}, \quad C_{XY} = 1.5 \times 10^{-3}$$ After mixing: $$[A^{2+}] = 7.5 \times 10^{-5}$$ $$[Y^-] = 7.5 \times 10^{-4}$$ So, $$Q = (7.5 \times 10^{-5})(7.5 \times 10^{-4})^2$$ $$= (7.5 \times 10^{-5})(56.25 \times 10^{-8})$$ $$= 4.21875 \times 10^{-11}$$ Since $$4.22 \times 10^{-11} < 5.2 \times 10^{-7}$$ no precipitate forms. --- 6. **Option D** $$C_{AB_2} = 3.6 \times 10^{-3}, \quad C_{XY} = 5.0 \times 10^{-4}$$ After mixing: $$[A^{2+}] = 1.8 \times 10^{-3}$$ $$[Y^-] = 2.5 \times 10^{-4}$$ So, $$Q = (1.8 \times 10^{-3})(2.5 \times 10^{-4})^2$$ $$= (1.8 \times 10^{-3})(6.25 \times 10^{-8})$$ $$= 1.125 \times 10^{-10}$$ Since $$1.125 \times 10^{-10} < 5.2 \times 10^{-7}$$ no precipitate forms. --- 7. **Conclusion** Only **Option B** gives $$Q > K_{sp}$$ and hence only this mixture will produce a precipitate of $AY_2$. Therefore, the correct answer is: $$\boxed{\text{B}}$$ --- 8. **Comparison with stored correct answer** Stored correct answer: **B** My derived answer: **B** They match.
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