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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 empirical rule: "Like dissolves like." This principle relies on the nature of intermolecular forces. Solutes that exhibit similar polarity and magnitude of intermolecular forces as the solvent will dissolve readily [Per the Principle of Intermolecular Interaction].

  • Solvent: $n\text{-octane} (CH_{3}(CH_{2})_{6}CH_{3})$ is a non-polar hydrocarbon.
  • Solute-Solvent Interaction: Solutes with non-polar characteristics will exhibit strong dispersion (London) forces with $n\text{-octane}$, facilitating solubility. Polar or ionic solutes will have negligible interaction with the non-polar solvent, leading to poor solubility.

Step 1: Characterization of Solute Polarity

Solute Chemical Nature Polarity
Cyclohexane ($C_{6}H_{12}$) Non-polar hydrocarbon Non-polar
$CH_{3}CN$ (Acetonitrile) Polar molecule (dipole moment $\approx 3.92\text{ D}$) Moderately Polar
$CH_{3}OH$ (Methanol) Highly polar (Hydrogen bonding capability) Highly Polar
$KCl$ (Potassium Chloride) Ionic compound (Electrostatic lattice) Ionic

Step 2: Analyzing Interactions with $n\text{-octane}$

Solubility increases as the solute becomes more non-polar and capable of van der Waals interactions with the non-polar $n\text{-octane}$ chain.

  1. Cyclohexane: Since both are non-polar hydrocarbons, they exhibit strong induced-dipole/induced-dipole interactions [London Dispersion Forces]. This results in maximum solubility.
  2. $CH_{3}CN$: Contains a polar $-CN$ group, which disrupts the homogeneity of the non-polar solvent. Solubility is low.
  3. $CH_{3}OH$: Features strong intermolecular hydrogen bonding. Energy required to disrupt these hydrogen bonds is not compensated by the non-polar solvent interaction [Enthalpic penalty], leading to very low solubility.
  4. $KCl$: As an ionic solid, the lattice energy ($\Delta H_{lattice}$) is significantly higher than the energy released by solvation with non-polar molecules ($\Delta H_{solvation}$). It is effectively insoluble in $n\text{-octane}$.

Visual Representation of Polarity Gradient

KCl CH3OH CH3CN Cyclohexane Increasing Solubility in n-Octane

Step 3: Final Ordering

Based on the progressive decrease in solute-solvent compatibility, the order of increasing solubility is:

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

Final Solution: The order of increasing solubility in n-octane is $KCl < CH_{3}OH < CH_{3}CN < \text{Cyclohexane}$. Solubility increases as the solute transitions from ionic to non-polar, maximizing interaction via London dispersion forces with the non-polar solvent.


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