JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
Arrange the following in increasing order of solubility product :
\(\mathrm{Ca}(\mathrm{OH})_2, \mathrm{AgBr}, \mathrm{PbS}, \mathrm{HgS}\)
- A \(\mathrm{HgS} \lt \mathrm{AgBr} \lt \mathrm{PbS} \lt \mathrm{Ca}(\mathrm{OH})_2\)
- B \(\mathrm{Ca}(\mathrm{OH})_2 \lt \mathrm{AgBr} \lt \mathrm{HgS} \lt \mathrm{PbS}\)
- C \(\mathrm{PbS} \lt \mathrm{HgS} \lt \mathrm{Ca}(\mathrm{OH})_2 \lt \mathrm{AgBr}\)
- D \(\mathrm{HgS} \lt \mathrm{PbS} \lt \mathrm{AgBr} \lt \mathrm{Ca}(\mathrm{OH})_2\)
Answer & Solution
Correct Answer
(D) \(\mathrm{HgS} \lt \mathrm{PbS} \lt \mathrm{AgBr} \lt \mathrm{Ca}(\mathrm{OH})_2\)
Step-by-step Solution
Detailed explanation
Based on the Ksp values and salt analysis cation identification, we can say that order of Ksp value is: \(\mathrm{HgS} < \mathrm{PbS} < \mathrm{AgBr} < \mathrm{Ca}(\mathrm{OH})_2\) K sp values…
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