If the concentration of \( A \) ion in \( A_{3} B \) is \( 10^{-4} \), so the solubility product is ...........

The solubility product \( K_{sp} \) is determined by the equilibrium concentrations of the ions produced when a compound dissolves. For the compound \( A_{3} B \), which produces 3 \( A^{+} \) ions and 1 \( B^{-} \) ion upon dissociation, we can express \( K_{sp} \) as: \[ K_{sp} = [A^{+}]^3 [B^{-}] \] Given that the concentration of \( A^{+} \) is \( 10^{-4} \), we first note that, at equilibrium, the concentration of \( B^{-} \) will also be related to \( x \) (the solubility), hence we would typically assume it to be \( x \). So, if \( [A^{+}] = 10^{-4} \), then \( K_{sp} = (10^{-4})^3 \times x \). To solve for \( x \) (the actual solubility), we can define \( x \) based on the stoichiometry, and derive accordingly. In a nutshell, calculating \( K_{sp} \) would depend heavily on the actual solubility of \( B \) in this scenario!