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Q1.24:
Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, \(CH_{3}OH\), \(CH_{3}CN\).

Solution :

Theoretical Foundation: Principles of Solubility

The solubility of a solute in a solvent is governed by the principle of "Like Dissolves Like." This implies that substances with similar intermolecular forces are mutually soluble. n-Octane ($C_{8}H_{18}$) is a non-polar hydrocarbon solvent dominated by weak London dispersion forces. Consequently, solutes with high polarity or ionic characteristics will be poorly soluble, while non-polar or weakly polar solutes will exhibit high solubility [Per the principle of intermolecular attraction vs. lattice energy].

Step 1: Analyzing Solute Polarity and Intermolecular Forces

Solute Nature of Intermolecular Force Interaction with n-Octane
Cyclohexane ($C_{6}H_{12}$) Non-polar (London Dispersion) Highly miscible; similar dispersion forces.
$CH_{3}CN$ (Acetonitrile) Polar (Dipole-Dipole) Low solubility; dipole-induced dipole interactions are weak.
$CH_{3}OH$ (Methanol) Highly Polar (Hydrogen Bonding) Very low solubility; strong H-bonding is incompatible with non-polar solvents.
$KCl$ Ionic Lattice Insoluble; lattice energy is significantly higher than the solvation energy in non-polar media.

Step 2: Visual Representation of Solvent Environment

n-Octane Solvent (Non-polar) Cyclohexane CH₃CN KCl (Insoluble)

Step 3: Determining the Order of Solubility

The solubility increases as the solute becomes more non-polar, effectively minimizing the energy penalty required to create a cavity in the non-polar n-octane solvent [Per the Thermodynamic Model of Solvation].

  • KCl: An ionic solid with a high lattice energy. It cannot interact with non-polar n-octane molecules. (Lowest solubility).
  • $CH_{3}OH$: Contains an $OH$ group capable of strong hydrogen bonding. This results in significant repulsion/non-mixing with non-polar hydrocarbons.
  • $CH_{3}CN$: Possesses a dipole moment, but lacks the ability to form hydrogen bonds. It is slightly more compatible with non-polar solvents than alcohols.
  • Cyclohexane: Being non-polar, it readily mixes with n-octane due to favorable London dispersion forces. (Highest solubility).

Step 4: Final Arrangement

The increasing order of solubility in n-octane is governed by the decrease in polarity and the nature of the intermolecular forces:

$KCl < CH_{3}OH < CH_{3}CN < Cyclohexane$

Final Solution: The increasing order of solubility is KCl < $CH_{3}OH$ < $CH_{3}CN$ < Cyclohexane. This order is dictated by the chemical principle that non-polar solutes dissolve readily in non-polar solvents due to compatible London dispersion forces, whereas ionic and highly polar molecules (like KCl and methanol) remain insoluble due to the mismatch in intermolecular forces.


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