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# Solve this problem - Solutions - JEE Main-3

The molar solubility of $Cd(OH)_{2}$ is $1.84\times 10^{-5}M$ in water. The expected solubility of $Cd(OH)_{2}$ in a buffer solution of pH = 12 is :

• Option 1)

$1.84\times 10^{-9}M$

• Option 2)

$\frac{2.49}{1.84}\times 10^{-9}M$

• Option 3)

$6.23\times 10^{-11}M$

• Option 4)

$2.49\times 10^{-10}M$

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$Ksp$ of $C\alpha (OH)^{3}=4S^{3}$

$Ksp = 45^{3}=4\times (1.84\times 10^{-5})$

$Cd(OH)_{2}\rightleftharpoons Cd^{2+}+ 2(OH)^{-}$

S               2S+ $10^{-2}$

2S - > this OH comes from dissociation of $Cd(OH)_{2}$

$10^{-2}$ - > this OH comes from a buffer

Given $P^{H}=12$

$P^{OH}=2$

$[OH] = 10^{-2}$

$Ksp= Kea$

$45^{3}=s[25+10^{-2}]^{2}$

$4\times (1.84\times 10^{-5})^{3}=s\times 10^{-4}$

$S=2.49\times 10^{-10}M$

Option 1)

$1.84\times 10^{-9}M$

Option 2)

$\frac{2.49}{1.84}\times 10^{-9}M$

Option 3)

$6.23\times 10^{-11}M$

Option 4)

$2.49\times 10^{-10}M$

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