Question

Arrange the following in increasing order of solubility product: \[ {Ca(OH)}_2, {AgBr}, {PbS}, {HgS} \]

• A. HgS<AgBr<PbS<Ca(OH)2
• B. PbS<HgS<Ca(OH)2<AgBr
• C. Ca(OH)2<AgBr<HgS<PbS
• D. HgS<PbS<AgBr<Ca(OH)2

Hint:

For solubility products, compounds with lower Ksp values are less soluble. Compare Ksp values to determine the order of solubility.

Answer 1

The Correct Option is C

The solubility product constant, \(K_{sp}\), is a measure of the solubility of a compound; the smaller the \(K_{sp}\), the less soluble the compound is. To determine the increasing order of solubility product for the given compounds: \(Ca(OH)_2\), \(AgBr\), \(PbS\), and \(HgS\), we compare their \(K_{sp}\) values.

1. \({HgS}\): It has a very low \(K_{sp}\) with a value approximately in the order of \(10^{-54}\), indicating extremely low solubility.

2. \({PbS}\): This compound also has a low \(K_{sp}\) but is slightly more soluble than \(HgS\), with \(K_{sp}\) around \(10^{-28}\).

3. \({AgBr}\): It is more soluble than both \(HgS\) and \(PbS\), with a \(K_{sp}\) around \(10^{-13}\).

4. \({Ca(OH)}_2\): This compound has the highest \(K_{sp}\) among the given compounds, approximately \(10^{-6}\), making it the most soluble.

Based on these \(K_{sp}\) values, the increasing order of solubility product is:

\(HgS<PbS<AgBr<Ca(OH)_2\)