Question

Is octanoic acid more soluble in 1 M HCl, 1 M NaOH, or pure water? Explain. Drugs such as morphine $\left(C_{17}{H}_{19}{O}_{3}N\right)$are often treated with strong acids. The most commonly used form of morphine is morphine hydrochloride (${\mathbf{C}}_{\mathbf{17}}{\mathbf{H}}_{\mathbf{20}}{\mathbf{O}}_{\mathbf{3}}\mathbf{NCl}$). Why is morphine treated in this way? (Hint: Morphine is an amine.)

Short answer

Octanoic acid will be more soluble in the solution with 1 M of NaOH.

Morphine will be more soluble in the solution with 1 M of HCl.

Step-by-step solution

Determine the solubility of octanoic acid  

It is a general rule that species with an overall charge are more soluble in the water. Octanoic acid has all organic acids group -COOH. In the presence of acid (pH below7), octanoic acid will remain in this form. The solubility of octanoic acid in HCl is reduced due to the common ion effect as both are acidic in nature and contain the same types of ions. In the presence of a base (pH above7), octanoic acid will give two species${\text{R - COO}}^{\text{-}}$ and${\text{H}}^{\text{+}}$ . The${\text{H}}^{\text{+}}$ ionwill react with OH^- giving water molecule. As we can observe ${\text{R - COO}}^{\text{-}}$, it has a charge (-1). Therefore, octanoic acid will be more soluble in the solution with 1 M of NaOH.

Determine the solubility of morphine  

Morphine has a different situation because of its amine group$\mathrm{R}-{\mathrm{NH}}_2$_.Morphine will be more soluble in the solution containing acid because the amine group will react with${\text{H}}^{\text{+}}$ giving$\mathrm{R}-{\mathrm{NH}}_3^{+}$ . Treatment of morphine with HCl, results in the generation of an ionic compound which is more soluble in water.