The for is . The molar solubility of this compound in pure water is given by: — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

For $Cr(OH)_3$ ($AB_3$ type salt), $K_{sp} = 27s^4$. Therefore, $27s^4 = 1.6 \times 10^{-30}$, which rearranges to molar solubility $s = (1.6 \times 10^{-30} / 27)^{1/4}$.