JEE Advanced · Chemistry · 8. Ionic Equilibrium
The of is at . The solubility (in ) of in a solution is
- A
- B
- C
- D
Answer & Solution
Correct Answer
(B)
Step-by-step Solution
Detailed explanation
\(Ag _2 CrO _4(s) \rightleftharpoons \underset{(2 S) \text { molar }}{2 Ag ^{+}( aq )}+\underset{( S ) \text { molar }}{ CrO _4^{2-}}\)
\(AgNO _3( aq ) \longrightarrow \underset{(0.1) \text { molar }}{ Ag ^{+}( aq )}+\underset{(0.1) \text { molar }}{ NO _3^{-}( aq )}\)
\(\left[ Ag ^{+}\right]=(2 S+0.1)\) molar
\(\left[ CrO _4^{2-}\right]=( S )\) molar
Here, S is the solubility of \(Ag _2 CrO _4\) in aqueous \(AgNO _3\).
\(\Rightarrow\left[ Ag ^{+}\right]^2\left[ CrO _4^{2-}\right]=(2 S+0.1)^2(S)\) \(=1.1 \times 10^{-12}\)
\(S \ll 0.1\)
\(\Rightarrow(0.1)^2(S)=1.1 \times 10^{-12}=10^{-2} \times S\)
\(S =1.1 \times 10^{-10} M\)
\(AgNO _3( aq ) \longrightarrow \underset{(0.1) \text { molar }}{ Ag ^{+}( aq )}+\underset{(0.1) \text { molar }}{ NO _3^{-}( aq )}\)
\(\left[ Ag ^{+}\right]=(2 S+0.1)\) molar
\(\left[ CrO _4^{2-}\right]=( S )\) molar
Here, S is the solubility of \(Ag _2 CrO _4\) in aqueous \(AgNO _3\).
\(\Rightarrow\left[ Ag ^{+}\right]^2\left[ CrO _4^{2-}\right]=(2 S+0.1)^2(S)\) \(=1.1 \times 10^{-12}\)
\(S \ll 0.1\)
\(\Rightarrow(0.1)^2(S)=1.1 \times 10^{-12}=10^{-2} \times S\)
\(S =1.1 \times 10^{-10} M\)
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