Which gives the more concentrated solution, stearic acid $\left[{\mathbf{CH}}_{\mathbf{3}}{\left({\mathbf{CH}}_{\mathbf{2}}\right)}_{\mathbf{16}}\mathbf{COOH}\right]$ in

a) ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ or

b)${\mathbf{CCl}}_{\mathbf{4}}$ ? Explain.

For a solution to form we have a general rule that like dissolves like it means polar solute can dissolve in polar solvent or vice versa. Also, for a solution to form forces created between solute and solvent must be comparable in strength to the forces destroyed within both the solute and the solvent.

Solute which is more soluble forms a more concentrated solution.

Stearic acid has both polar and non-polar ends. The polar end forms hydrogen bonds with water while the non-polar end forms dipole-induced dipole interactions with water. As the non-polar part of acid increases dipole interactions increases in number and hydrogen bond forming capacity becomes less.

So, acid molecules cannot replace hydrogen bonds between water molecules on increasing n and hence solubility becomes less.

In stearic acid n= 16 which is quite a large number so it is insoluble in water and cannot form a concentrated solution.

Stearic acid will be soluble in${\mathbf{CCl}}_{\mathbf{4}}$because long non-polar chain of the molecule can interact with non-polar solvent I.e.,${\mathbf{CCl}}_{\mathbf{4}}$more effectively by dipole-dipole forces or dispersion forces.

As stearic acid is more soluble in ${\mathbf{CCl}}_{\mathbf{4}}$, it will form a more concentrated solution.