Final answer:
In a solution with a pH of 3.8, dissolved acetic acid with a pKa of 4.8 exists primarily in its protonated form as CH3COOH, because the pH is lower than the pKa, leading to limited ionization and equilibrium favoring the undissociated acid.
Step-by-step explanation:
When acetic acid (CH₃COOH) with a pKa of 4.8 is dissolved in a solution with a pH of 3.8, the solution has a higher concentration of hydronium ions (H₃O+) compared to a neutral pH 7 solution. In this scenario, the pH of the solution is lower than the pKa of acetic acid, which implies that the acid is in its protonated form, predominately as CH₃COOH, rather than its conjugate base, acetate ion (CH₃COO−). The pH is within 1 unit below the pKa, which means some ionization does occur but acetic acid is not fully ionized, leading to a solution that contains a larger amount of the undissociated acid form.
According to the Henderson-Hasselbalch equation, a solution with a pH lower than the pKa of the acid in question will favor the formation of the un-ionized form of that acid. An equilibrium exists between the acetic acid and its ionized form, the acetate ion in water, represented by the equation CH₃COOH (aq) = H+ (aq) + CH₃COO− (aq). Since only a small percentage of the acetic acid molecules ionize to form hydronium (H₃O+) and acetate ions, the equilibrium lies far to the left, indicating that most of the acetic acid remains in its molecular form.