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Chemistry Question on Solutions

The solubility product of Cr(OH)3Cr(OH)_3 at 298 K298 \,K is 6.0×10−31.6.0 \times 10^{-31}. The concentration of hydroxide ions in a saturated solution of Cr(OH)3Cr(OH)_3 will be :

A

(18×10−31)1/2\left(18\times10^{-31}\right)^{1/ 2}

B

(2.22×10−31)1/4\left(2.22\times10^{-31}\right)^{1/ 4}

C

(18×10−31)1/4\left(18\times10^{-31}\right)^{1/ 4}

D

(4.86×10−29)1/4\left(4.86\times10^{-29}\right)^{1/ 4}

Answer

(18×10−31)1/4\left(18\times10^{-31}\right)^{1/ 4}

Explanation

Solution

Ksp=27(s)4=6×10−31K_{sp}=27\left(s\right)^{4}=6\times10^{-31} ⇒[3(s)]4=18×10−31\Rightarrow\left[3\left(s\right)\right]^{4}=18\times10^{-31} [OH−]=3(s)=[18×10−31]1/4\left[OH-\right]=3\left(s\right)=\left[18\times10^{-31}\right]^{1 /4}