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Which among the following is the least soluble? (a) \(\operatorname{MnS}\left(\mathrm{K}_{s p}=7 \times 10^{-16}\right)\) (b) \(\mathrm{FeS}\left(\mathrm{K}_{\mathrm{sp}}^{\mathrm{sp}}=4 \times 10^{-19}\right)\) (c) \(\operatorname{PtS}\left(\mathrm{K}_{\mathrm{sp}}^{\mathrm{sp}}=8 \times 10^{-73}\right)\) (d) \(\operatorname{NiS}\left(K_{s p}^{5 p}=3 \times 10^{-12}\right)\)

Short Answer

Expert verified
PtS is the least soluble, with the smallest Ksp of \(8 \times 10^{-73}\).

Step by step solution

01

Understand Solubility Product (Ksp)

The solubility product constant, \(K_{sp}\), is an indication of the solubility of a compound. A smaller \(K_{sp}\) value means that the compound is less soluble in solution. Therefore, the compound with the smallest \(K_{sp}\) value is likely the least soluble.
02

Compare Ksp Values

List the given solubility product constants:- \(K_{sp}\) for \(\text{MnS} = 7 \times 10^{-16}\)- \(K_{sp}\) for \(\text{FeS} = 4 \times 10^{-19}\)- \(K_{sp}\) for \(\text{PtS} = 8 \times 10^{-73}\)- \(K_{sp}\) for \(\text{NiS} = 3 \times 10^{-12}\).Compare these values to determine which is the smallest.
03

Identify the Least Soluble Compound

Determine which compound has the smallest \(K_{sp}\) value. In this case, \(K_{sp}\) for \(\text{PtS} = 8 \times 10^{-73}\) is the smallest among the given values, indicating that \(\text{PtS}\) is the least soluble compound.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility
Solubility refers to the ability of a substance to dissolve in a solvent, creating a homogeneous solution. It is a key characteristic that describes how much of the substance, or solute, can be dissolved before the solution becomes saturated. Solubility is influenced by several factors, including temperature, pressure, and the nature of the solvent.
  • Temperature: Generally, the solubility of solids in liquids increases with temperature.
  • Pressure: Pressure significantly affects the solubility of gases; higher pressure increases gas solubility.
  • Nature of the solvent: The chemical nature of the solvent and solute can either facilitate or hinder solubility, with like-dissolving-like being a common rule of thumb.
For ionic compounds in a liquid solvent, solubility is often quantified using the solubility product constant, or Ksp, especially when dealing with sparingly soluble salts.
Chemical Equilibrium
Chemical equilibrium is a state where the rate of the forward chemical reaction is equal to the rate of the reverse reaction. In this state, the concentrations of the reactants and products remain constant over time, although the individual molecules continue to react with one another.
This concept is crucial in understanding how solubility is assessed for ionic compounds. When a sparingly soluble salt dissolves, it reaches a point where an equilibrium is established between the undissolved salt and the ions in solution. The equilibrium can be represented by a balanced chemical equation.
  • Forward Reaction: The dissolution of solid solute into ions.
  • Reverse Reaction: The precipitation of ions back into solid form.
The equilibrium constant expressed as Ksp (solubility product) further quantifies this balance and helps predict how much of the salt can dissolve without additional factors altering the equilibrium.
Ionic Compounds
Ionic compounds consist of positive and negative ions held together by strong electrostatic forces. These compounds typically form solid crystal structures, where ions are arranged in a repeating three-dimensional pattern. Ionic compounds are sharp in defining properties such as high melting and boiling points, and their ability to conduct electricity when molten or in solution.
In terms of solubility, the polar nature of water makes it an excellent solvent for ionic compounds. When an ionic compound dissolves in water, it dissociates into its constituent ions. Thus, solubility product (Ksp) helps in determining how well an ionic compound dissolves.
  • Water as a solvent promotes the separation of ions from the solid structure.
  • Dissolved ions result in an electrolytic solution capable of conducting electricity.
  • When equilibrium is reached, no further ions can dissolve and excess undissolved salt remains in solid form.
The extent to which an ionic compound dissolves in a solvent is part of its intrinsic property and is often represented through its Ksp value.
Sparingly Soluble Salts
Sparingly soluble salts are salts that dissolve to a small extent in a given solvent, such as water. They are associated with very low solubility product constants, meaning only a tiny amount of the salt dissolves before equilibrium is reached. These salts tend to remain mostly undissolved in their solid state when added to a solvent.
The Ksp value is critical in identifying and quantifying the solubility of such salts. A smaller Ksp value indicates lesser solubility.
  • Examples: Silver chloride (AgCl) and calcium sulfate (CaSO₄) are classic examples of sparingly soluble salts.
  • Application: The concept is widely applied in various chemical processes to predict precipitation and dissolution reactions.
In practical terms, knowing the Ksp helps in industries such as pharmaceuticals and water purification to control the formation of precipitates and manage dissolution processes efficiently.

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Most popular questions from this chapter

For the following three reactions \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), equilibrium constants are given: (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{1}\) (b) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{2}\) (c) \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{3}\) Which of the following relation is correct? (a) \(\mathrm{K}_{1} \sqrt{\mathrm{K}_{2}}=\mathrm{K}_{2}\) (b) \(\mathrm{K}_{2} \mathrm{~K}_{3}=\mathrm{K}_{1}\) (c) \(\mathrm{K}_{3}=\mathrm{K}_{1} \mathrm{~K}_{2}\) (d) \(\mathrm{K}_{3} \cdot \mathrm{K}_{2}^{3}=\mathrm{K}_{1}^{2}\)

One mole of \(\mathrm{A}(\mathrm{g})\) is heated to \(300^{\circ} \mathrm{C}\) in a closed one litre vessel till the following equilibrium is reached. \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\) The equilibrium constant of this reaction at \(300^{\circ} \mathrm{C}\) is 4\. What is the concentration of \(\mathrm{B}\) (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) at equilibrium: (a) \(0.2\) (b) \(0.6\) (c) \(0.75\) (d) \(0.1\)

For a gaseous reaction \(2 \mathrm{~A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\), the par- tial pressures of \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) at equilibrium are \(0.5\), \(0.8,0.7\) and \(1.2 \mathrm{~atm}\). The value of \(\mathrm{K}_{\mathrm{p}}\) for this reaction is: (a) \(2.4 \mathrm{~atm}\) (b) \(6.2 \mathrm{arm}^{-2}\) (c) \(4.2 \mathrm{~atm}^{-1}\) (d) \(8.4 \mathrm{~atm}^{-3}\)

A saturated solution of non-radioactive sugar was taken and a little radioactive sugar was added to it. A small amount of it gets dissolved in solution and an equal amount of sugar was precipitated. This proves: (a) The equilibrium has been established in the solution (b) Radioactive sugar can displace non-radioactive sugar from its solution.

For the reaction \(2 \mathrm{SO}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{SO}_{3}\) \(\mathrm{g}\) \(\mathrm{g}\) If we start with 2 mol. \(\mathrm{SO}_{2}\) and \(1 \mathrm{~mol} . \mathrm{O}_{2}\) in \(1 \mathrm{~L}\) flask, the mixture needs \(0.4 \mathrm{~mol} \mathrm{MnO}_{4}^{-}\) in acidic medium for the complete oxidation of \(\mathrm{SO}_{2}\). The value of \(\mathrm{K}_{\mathrm{c}}\) is: (a) \(1 / 2\) (b) 2 (c) 1 (d) \(0.6\)

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