JEE Main 2025 — Ionic Equilibrium Question with Solution

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 ${\mathrm{K}}_{\mathrm{SP}}$.