Solve the following

Question:

The solubility product of $\mathrm{Cr}(\mathrm{OH})_{3}$ at $298 \mathrm{~K}$ is $6.0 \times 10^{-31}$. The concentration of hydroxide ions in a saturated solution of $\mathrm{Cr}(\mathrm{OH})_{3}$ will be:

  1. $\left(2.22 \times 10^{-31}\right)^{1 / 4}$

  2. $\left(18 \times 10^{-31}\right)^{1 / 4}$

  3. $\left(18 \times 10^{-31}\right)^{1 / 2}$

  4. $\left(4.86 \times 10^{-29}\right)^{1 / 4}$

Correct Option: , 2

Solution:

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