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6-2.Equilibrium-II (Ionic Equilibrium)
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The solubility product of $BaSO _4$ is $1 \times 10^{-10}$ at $298\,K$. The solubility of $BaSO _4$ in $0.1\,M\,K _2 SO _4( aq )$ solution is $.........\times 10^{-9}\,g\,L ^{-1}$ (nearest integer). Given : Molar mass of $BaSO _4$ is $233\,g\,mol ^{-1}$
A
$233$
B
$232$
C
$231$
D
$234$
(JEE MAIN-2023)
Solution
$K _2 SO _4 \longrightarrow 2 K ^{+}+ SO _4{ }^{2-}$
$0.1 M \quad\quad 0.2 M \quad0.1 M$
$BaSO _4 \rightleftharpoons Ba ^{+2}+ SO _4{ }^{2-}$
$a – S \quad S \quad S +0.1 \approx 0.1$
$K _{ SP }= S \times 10^{-1}$
$\Rightarrow 1 \times 10^{-10}= S \times 10^{-1}$
$\Rightarrow S =10^{-9} mol L^{-1 }$
So, $S =10^{-9} \times 233\,g\,L ^{-1}$
So, Answer : $233$
Standard 11
Chemistry
