Question

4-Methylphenol, CH3C6H4OH (pKa 10.26), is only slightly soluble in water, but its sodium salt, CH3C6H4O-Na , is quite soluble in water. Describe the solubility of 4-methylphenol in solutions of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3). The pKa values for the conjugate acids of sodium hydroxide, sodium bicarbonate (NaHCO3), and sodium carbonate (Na2CO3) are 15.7, 6.36, and 10.33, respectively.

Answer 1

Explanation:

We know that more is the $pK_{a}$ value, weaker will be the acid. Also, an acid completely dissociates into ions in an aqueous base solution when $pK_{a}$ of conjugate acid of base is greater than acid.

4-methylphenol $(CH_{3}C_{6}H_{4}OH)$ ($pK_{a} = 10.26$) is quite soluble in its sodium salt. In NaOH, the dissociation will be as $Na^{+}$ and $OH^{-}$ ions as NaOH is a strong base.

Therefore, 4-methylphenol will readily dissolve in NaOH solution.

As, $NaHCO_{3}$ is not a strong base but as 4-Methylphenol forms a sodium salt hence, it will have a low solubility as compared to NaOH.

Whereas $Na_{2}CO_{3}$ is not a base but when dissolved in water it shows basic character as it produces NaOH (strong base) and $H_{2}CO_{3}$ (weak acid). As a result, the solution gets basic. Hence, 4-methylphenol will readily dissolve in $Na_{2}CO_{3}$.