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JEE MAIN - Chemistry (2025 - 2nd April Morning Shift - No. 3)

If equal volumes of $A B_2$ and $X Y$ (both are salts) aqueous solutions are mixed, which of the following combination will give a precipitate of $\mathrm{AY}_2$ at 300 K ? (Given $\mathrm{K}_{\mathrm{sp}}\left(\right.$ at 300 K ) for $\mathrm{AY}_2=5.2 \times 10^{-7}$ )
$2.0 \times 10^{-4} \mathrm{M} \mathrm{AB}_2, 0.8 \times 10^{-3} \mathrm{M} \mathrm{XY}$
$2.0 \times 10^{-2} \mathrm{M} \mathrm{AB}_2, 2.0 \times 10^{-2} \mathrm{M} \mathrm{XY}$
$1.5 \times 10^{-4} \mathrm{M} \mathrm{AB}_2, 1.5 \times 10^{-3} \mathrm{M} \mathrm{XY}$
$3.6 \times 10^{-3} \mathrm{M} \mathrm{AB}_2, 5.0 \times 10^{-4} \mathrm{M} \mathrm{XY}$

Explanation

When equal volumes of solutions are mixed, the molarity of each solution halves. To determine if a precipitate will form, calculate the ionic product $\mathrm{Q}_{\mathrm{SP}}$ using the formula:

$ \mathrm{Q}_{\mathrm{SP}} = \left[\mathrm{A}^{2+}\right]\left[\mathrm{Y}^{-}\right]^2 $

A precipitate of $\mathrm{AY}_2$ will form if $\mathrm{Q}_{\mathrm{SP}} > \mathrm{K}_{\mathrm{SP}}$, where $\mathrm{K}_{\mathrm{SP}}$ is given as $5.2 \times 10^{-7}$.

Option 1 Calculation:

$ \mathrm{Q}_{\mathrm{SP}} = \left(1.8 \times 10^{-3}\right)\left(\frac{5}{2} \times 10^{-4}\right)^2 = \text{Calculated value} < \mathrm{K}_{\mathrm{SP}} $

Option 2 Calculation:

$ \mathrm{Q}_{\mathrm{SP}} = \left(10^{-4}\right)\left(0.4 \times 10^{-3}\right)^2 = \text{Calculated value} < \mathrm{K}_{\mathrm{SP}} $

Option 3 Calculation:

$ \mathrm{Q}_{\mathrm{SP}} = \left(10^{-2}\right)\left(10^{-2}\right)^2 = \text{Calculated value} > \mathrm{K}_{\mathrm{SP}} $

Option 4 Calculation:

$ \mathrm{Q}_{\mathrm{SP}} = \left(\frac{1.5}{2} \times 10^{-4}\right)\left(\frac{1.5}{2} \times 10^{-3}\right)^2 = \text{Calculated value} < \mathrm{K}_{\mathrm{SP}} $

From the calculations, the solution in Option 3 will form a precipitate since the ionic product $\mathrm{Q}_{\mathrm{SP}}$ is greater than the solubility product $\mathrm{K}_{\mathrm{SP}}$.

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