The solubility product of C r ( O H ) 3 at 298 K is 6.0 × 1 0 - 31 . The concentration of hydroxide ions in a saturated solution of C r ( O H ) 3 will be

Cr OH 3 ⇌ Cr 3 + s + 3 OH − 3 s 6 × 1 0 - 31 = S × ( 3 S ) 3 6 × 1 0 - 31 = 27 S 4 S = 6 27 × 1 0 - 31 1 / 4 [ O H - ] = 3 S = 3 6 27 × 1 0 - 31 1 / 4 = ( 18 × 1 0 - 31 ) 1 / 4 M

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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductEasy2 minPYQ_2020

ChemistryEasysingle choice

The solubility product of $C r (O H)_{3}$ at 298 K is $6.0 \times 1 0^{- 31}$ . The concentration of hydroxide ions in a saturated solution of $C r (O H)_{3}$ will be

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Solution:

$Cr {(OH)}_{3} \underset{}{⇌} \underset{s}{{Cr}^{3 +}} + \underset{3 s}{3 {OH}^{-}}$
$6 \times 1 0^{- 31} = S \times (3 S)^{3}$
$6 \times 1 0^{- 31} = 27 S^{4}$
$S = {(\frac{6}{27} \times 1 0^{- 31})}^{1 / 4}$
$[O H^{-}] = 3 S = 3 {(\frac{6}{27} \times 1 0^{- 31})}^{1 / 4} = (18 \times 1 0^{- 31})^{1 / 4} M$

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2020

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