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6-2.Equilibrium-II (Ionic Equilibrium)
hard
The solubility of $BaSO _4$ in pure water (in $g\, L ^{-1}$ ) is closest to
[Given; $K_{ sp }$ for $BaSO _4$ is $1.0 \times 10^{-10}$ at $25^{\circ} C$. Molecular weight of $BaSO _4$ is $233 \,g\, mol ^{-1}$ ]
A$10 \times 10^{-5}$
B$10 \times 10^{-3}$
C$2.3 \times 10^{-5}$
D$2.3 \times 10^{-3}$
(KVPY-2018)
Solution
(d)
Given, $K_{ sp }=1 \times 10^{-10}$
$BaSO _4 \rightleftharpoons Ba ^{2+}+ SO _4^{2-}$
${}\quad\quad\quad\quad\quad S \quad\quad\quad S$
Let the solubity of $Ba ^{2+}$ and $SO _4^{2-}$ be $S$ $\therefore \quad K_{s p}=S^2$
$1 \times 10^{-10} =S^2$
$S =10^{-5} \,mol / L$
Thus, solubility of $BaSO _4$ in pure water (in $g/L$)
Solubility $($ in mol/L) $\times$ Molecular weight
$=233 \times 10^{-5}=2.3 \times 10^{-3}$
Given, $K_{ sp }=1 \times 10^{-10}$
$BaSO _4 \rightleftharpoons Ba ^{2+}+ SO _4^{2-}$
${}\quad\quad\quad\quad\quad S \quad\quad\quad S$
Let the solubity of $Ba ^{2+}$ and $SO _4^{2-}$ be $S$ $\therefore \quad K_{s p}=S^2$
$1 \times 10^{-10} =S^2$
$S =10^{-5} \,mol / L$
Thus, solubility of $BaSO _4$ in pure water (in $g/L$)
Solubility $($ in mol/L) $\times$ Molecular weight
$=233 \times 10^{-5}=2.3 \times 10^{-3}$
Standard 11
Chemistry
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