The molar solubility of C d O H 2 is 1.84 × 10 - 5 M in water. The expectes solubility of C d O H 2 in a buffer solution of p H = 12 is:

K s p = 4 s 3 = 4 × 1.84 × 10 - 5 3 C d S ' O H 2 ⇌ C d + 2 S ' + 2 O H - 10 - 2 + S ' ≃ 10 - 2 S ' × 10 - 2 2 = 4 × 1.84 × 10 - 5 3 S ' = 4 × 1.84 3 × 10 - 11 ( S ' ) = 2.491 × 10 - 10 M

Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductMedium2 minPYQ_2019

ChemistryMediumstatement

The molar solubility of $C d {(O H)}_{2}$ is $1.84 \times 10^{- 5} M$ in water. The expectes solubility of $C d {(O H)}_{2}$ in a buffer solution of $p H = 12$ is:

Options:

Answer:

Solution:

$K_{s p} = 4 {(s)}^{3}$
$= 4 \times {(1.84 \times 10^{- 5})}^{3}$
$\underset{S^{'}}{C d} {(O H)}_{2} ⇌ \underset{S^{'}}{C d^{+ 2}} + \underset{(10^{- 2} + S^{'}) ≃ 10^{- 2}}{2 O H^{-}}$
$S^{'} \times {(10^{- 2})}^{2} = 4 \times {(1.84 \times 10^{- 5})}^{3}$
$S^{'} = 4 \times {(1.84)}^{3} \times 10^{- 11}$
$(S^{'}) = 2.491 \times 10^{- 10} M$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2019

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