Molar solubility of Bismuth iodide is related to its solubility product by equation: S=sqrt [4](K_(s)^0/27) S=sqrt [3](K_(z)^0/3)

The correct equation for the molar solubility of Bismuth iodide is:<br /><br />$S=\sqrt [4]{K_{s}^{0}/27}$<br /><br />This equation is derived from the solubility product constant (Ks) of Bismuth iodide, which is the product of the concentrations of the ions raised to the power of their stoichiometric coefficients. In this case, Bismuth iodide dissociates into one Bismuth ion (Bi) and two iodide ions (I-), so the equation is:<br /><br />$S=\sqrt [4]{K_{s}^{0}/27}$<br /><br />where S is the molar solubility of Bismuth iodide and Ks is the solubility product constant.