Answer:
It's slightly soluble in an aqueous solution of [tex]NaHCO_{3}[/tex], and almost insoluble in an aqueous solution of NaOH.
Explanation:
Sodium benzoate comes from benzoic acid, which is a weak acid. It means that in an aqueous solution benzoic acid does not ionize easily to form the ions [tex]H_{3}O^{+}[/tex] and [tex]C_{7}H_{5}O_{2} ^{-}[/tex]
It also implies, according to the Le Châtelier's principle, that the ion [tex]C_{7}H_{5}O_{2}^{-}[/tex] tends to form the acid [tex]C_{7}H_{6} O_{2}[/tex] more easily. It can be seen in the following equation:
[tex]C_{7}H_{6} O_{2}[/tex] ⇔ [tex]C_{7}H_{5}O_{2}^{-}[/tex] + [tex]H_{3}O^{+}[/tex]
In an aqueous solution, the equilibrium shifts to the left, thus letting water dissolve sodium benzoate. But why? Because water in that case would produce enough [tex]H_{3}O^{+}[/tex] ions to facilitate the disolution of sodium benzoate. It's shown by its solubility in water at 15°C (62.78g/100mL, according to Wikipedia).
In contrast, the presence of NaOH or [tex]NaHCO_{3}[/tex], both chemical species producing the [tex]OH^{-}[/tex] ions in aqueous solution, would make the equilibrium shift to the right because it would be a higher need of [tex]H_{3}O^{+}[/tex] ions to offset the presence of [tex]OH^{-}[/tex].
However, the effect of NaOH is not the same due to [tex]NaHCO_{3}[/tex], because the first is a strong base and the other is a weak one. Thereby it is reasonable to think that solubility of sodium benzoate is greater in water than in [tex]NaHCO_{3}[/tex] and NaOH.
Solubility in water > solubility in [tex]NaHCO_{3}[/tex]> solubility in NaOH.
