For a salt of type $\text{AB}$, the dissociation equilibrium in its saturated solution is: $\text{AB}(s) \rightleftharpoons \text{A}^+(aq) + \text{B}^-(aq)$

If $S$ is the molar solubility (molarity of the saturated solution), then the equilibrium concentrations of the ions are: $[\text{A}^+] = S$ $[\text{B}^-] = S$

The solubility product constant ($K_{sp}$) is given by the product of the ion concentrations: $K_{sp} = [\text{A}^+][\text{B}^-] = S \times S = S^2$

Given that $K_{sp} = 4 \times 10^{-8}$: $S^2 = 4 \times 10^{-8}$ $S = \sqrt{4 \times 10^{-8}} = 2 \times 10^{-4} \text{ mol/L}$

Therefore, the molarity of the saturated solution is $2 \times 10^{-4} \text{ mol/L}$.