Arrange the following in increasing order of solubility product:
\[
{Ca(OH)}_2, {AgBr}, {PbS}, {HgS}
\]
Show Hint
- HgS<AgBr<PbS<Ca(OH)2
- PbS<HgS<Ca(OH)2<AgBr
- Ca(OH)2<AgBr<HgS<PbS
- HgS<PbS<AgBr<Ca(OH)2
The Correct Option is C
Solution and Explanation
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:
\(HgS<PbS<AgBr<Ca(OH)_2\)
Top Questions on Solutions
- Elements A and B form two non-volatile compounds (AB$_2$ and AB$_4$). When 1 g of AB$_2$ is added to 20 g of C$_6$H$_6$ (molar mass =78 gmol$^{-1}$) the freezing point of C$_6$H$_6$ is lowered by 2.3 K. When 1 g of AB$_4$ is added to 20 g of C$_6$H$_6$, the freezing point of C$_6$H$_6$ was lowered by 1.3 K. The atomic masses of A and B are respectively (K$_f$(C$_6$H$_6$)=5.1 K kg mol$^{-1}$)
- Mole fraction of H$_2$SO$_4$ in aqueous solution is 0.9. Find mass % of H$_2$SO$_4$. (H=1, S=32, O=16 u)
- Give reasons:
(a) Cooking is faster in a pressure cooker than in an open pan.
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y? Observe the following data given in the table. (\(K_H\) = Henry's law constant)
Gas CO₂ Ar HCHO CH₄ \(K_H\) (k bar at 298 K) 1.67 40.3 \(1.83 \times 10^{-5}\) 0.413 The correct order of their solubility in water is
- Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is:
Questions Asked in JEE Main exam
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
- JEE Main - 2025
- Determinants
- An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density 1000 kg/m\(^3\). If the pressure inside the bubble is 2100 N/m\(^2\) greater than the atmospheric pressure, then the surface tension of the liquid in SI units is (use \(g = 10 \, {m/s}^2\)).
- JEE Main - 2025
- Surface Tension
- A force of 49 N acts tangentially at the highest point of a sphere (solid of mass 20 kg) kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
- JEE Main - 2025
- Quantum Mechanics
- Let a circle C pass through the points (4, 2) and (0, 2), and its centre lie on \(3x + 2y + 2 = 0\). Then the length of the chord of the circle C, whose midpoint is (1, 2), is:
- JEE Main - 2025
- Coordinate Geometry
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is:
- JEE Main - 2025
- Gauss's Law