Solubility product of BaC{l₂} is 4 × {10^{ - 9}} . Its solubility in moles/litre would be

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6-2.Equilibrium-II (Ionic Equilibrium)

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Solubility product of $BaC{l_2}$ is $4 \times {10^{ - 9}}$. Its solubility in moles/litre would be

A

$1 \times {10^{ - 3}}$

B

$1 \times {10^{ - 9}}$

C

$4 \times {10^{ - 27}}$

D

$1 \times {10^{ - 27}}$

Solution

(a) ${K_{sp}} = 4{S^3},$

${S^3} = \frac{{4 \times {{10}^{ – 9}}}}{4} = {10^{ – 9}}$

$\therefore \,S = {10^{ – 3}}M$.

Standard 11

Chemistry

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