The solubility product of $Cr(OH)_3$ at $298 \ K$ is $6.0 \times 10^{-31}$. The concentration of hydroxide ions in a saturated solution of $Cr(OH)_3$ will be

  • A
    $(18 \times 10^{-31})^{1/4}$
  • B
    $(2.22 \times 10^{-31})^{1/4}$
  • C
    $(4.86 \times 10^{-29})^{1/4}$
  • D
    $(18 \times 10^{-31})^{1/2}$

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