The for is . The molar solubility of this compound in pure water is given by: — Ionic Equilibrium Chemistry Question
Question
The $K_{sp}$ for $Cr(OH)_3$ is $1.6 \times 10^{-30}$. The molar solubility of this compound in pure water is given by:
Answer: A
💡 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 $s = (1.6 \times 10^{-30} / 27)^{1/4}$.
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