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6-2.Equilibrium-II (Ionic Equilibrium)
hard
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
A
$\left(18 \times 10^{-31}\right)^{1 / 4}$
B
$\left(2.22 \times 10^{-31}\right)^{1 / 4}$
C
$\left(4.86 \times 10^{-29}\right)^{1 / 4}$
D
$\left(18 \times 10^{-31}\right)^{1 / 2}$
(JEE MAIN-2020)
Solution
$\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Cr}^{3+}(\mathrm{aq} .)+3 \mathrm{OH}^{-}(\mathrm{aq} .)$
$\mathrm{k}_{\mathrm{sp}}=27(\mathrm{s})^{4}=6 \times 10^{-31}$
$\Rightarrow[3(\mathrm{s})]^{4}=18 \times 10^{-31}$
$\left[\mathrm{OH}^{-}\right]=3(\mathrm{s})=\left[18 \times 10^{-31}\right]^{1 / 4}$
Standard 11
Chemistry
