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:
\(HgS<PbS<AgBr<Ca(OH)_2\)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
| Sample | Van't Haff Factor |
|---|---|
| Sample - 1 (0.1 M) | \(i_1\) |
| Sample - 2 (0.01 M) | \(i_2\) |
| Sample - 3 (0.001 M) | \(i_2\) |
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)