Answer:
It's slightly soluble in an aqueous solution of \(NaHCO_{3}\), 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 \(H_{3}O^{+}\) and \(C_{7}H_{5}O_{2} ^{-}\)
It also implies, according to the Le Châtelier's principle, that the ion \(C_{7}H_{5}O_{2}^{-}\) tends to form the acid \(C_{7}H_{6} O_{2}\) more easily. It can be seen in the following equation:
\(C_{7}H_{6} O_{2}\) ⇔ \(C_{7}H_{5}O_{2}^{-}\) + \(H_{3}O^{+}\)
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 \(H_{3}O^{+}\) 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 \(NaHCO_{3}\), both chemical species producing the \(OH^{-}\) ions in aqueous solution, would make the equilibrium shift to the right because it would be a higher need of \(H_{3}O^{+}\) ions to offset the presence of \(OH^{-}\).
However, the effect of NaOH is not the same due to \(NaHCO_{3}\), 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 \(NaHCO_{3}\) and NaOH.
Solubility in water > solubility in \(NaHCO_{3}\)> solubility in NaOH.
