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  • The solubility product of at 298 K is <img alt="6.0\times 10^{-31}" src="http://entrancecorner.

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 :
 

Option: 1

(18\times 10^{-31})^{1/2}


Option: 2

(2.22\times 10^{-31})^{1/4}


Option: 3

(18\times 10^{-31})^{1/4}

 


Option: 4

(4.86\times 10^{-29})^{1/4}


Answers (1)

best_answer

\mathrm{Cr}(\mathrm{OH})_{3} \longrightarrow \mathrm{Cr}^{+3}+3 \mathrm{OH}^{-}

                                   s           3s

\\\mathrm{K}_{\mathrm{sp}}=\mathrm{s} \cdot(3 \mathrm{s})^{3}\\ \\ 6 \times 10^{-31}=27s^{4}\\\\ s=\left(\frac{6}{27} \times 10^{-31}\right)^{1 / 4}\\ \\\left[\mathrm{OH}^{-}\right]=3 \mathrm{s}\\ \left[\mathrm{OH}^{-}\right]=3 \times\left(\frac{6}{27} \times 10^{-31}\right)^{1 / 4}\\\left[\mathrm{OH}^{-}\right]=\left(18 \times 10^{-31}\right)^{1 / 4} \mathrm{M}

Hence, option number (3) is correct.

Posted by

Ritika Jonwal

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