JEE Chemistry 2025 Question with Solution

Given: The compounds are $\text{Ca(OH)}_2$, $\text{AgBr}$, $\text{PbS}$ and $\text{HgS}$.

Find: The increasing order of solubility product $K_{sp}$.

The solution states that smaller $K_{sp}$ means lower solubility product value.

Approximate values given are:

• $K_{sp}[\text{Ca(OH)}_2] \approx 5.5 \times 10^{-6}$
• $K_{sp}[\text{AgBr}] \approx 5 \times 10^{-13}$
• $K_{sp}[\text{HgS}] \approx 1 \times 10^{-52}$
• $K_{sp}[\text{PbS}] \approx 3 \times 10^{-28}$

Comparing these values in increasing order:

However, the solution also explicitly states The Correct Option is C and its second approach concludes:

There is a clear discrepancy in the solution. Since the page explicitly marks C as the correct option, the extracted answer is C.

Therefore, the correct option is C.

Given: Four sparingly soluble salts are to be arranged by increasing $K_{sp}$.

Find: Which option matches the order.

The dissociation equations shown in the solution are:

1. $\text{Ca(OH)}_2 \rightarrow \text{Ca}^{2+} + 2\text{OH}^-$
2. $\text{AgBr} \rightarrow \text{Ag}^+ + \text{Br}^-$
3. $\text{PbS} \rightarrow \text{Pb}^{2+} + \text{S}^{2-}$
4. $\text{HgS} \rightarrow \text{Hg}^{2+} + \text{S}^{2-}$

The page explains that a smaller $K_{sp}$ corresponds to a less soluble compound.

Using the approximate numerical values listed on the page, the ascending numerical order should be read by comparing powers of $10$. The first approach and the second approach contradict each other, but the page's declared answer is C.

Hence, following the solution's authority for answer extraction, the correct option is C.