JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
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 \(234\)
Answer & Solution
Correct Answer
(A) \(233\)
Step-by-step Solution
Detailed explanation
\(K _2 SO _4 \longrightarrow 2 K ^{+}+ SO _4{ }^{2-}\) \(0.1 M \quad\quad 0.2 M \quad0.1 M\) \(BaSO _4 \rightleftharpoons Ba ^{+2}+ SO _4{ }^{2-}\) \(a - S \quad S \quad S +0.1 \approx 0.1\) \(K _{ SP }= S \times 10^{-1}\) \(\Rightarrow 1 \times 10^{-10}= S \times 10^{-1}\)…
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