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JEE Advanced · Chemistry · 8. Ionic Equilibrium

The Ksp of Ag2CrO4 is 1 . 1 × 1 0 - 1 2  at 298 K. The solubility (in mol L-1) of Ag2CrO4 in a 0.1 M AgNO3 solution is

  1. A 1 . 1 × 1 0 - 1 1
  2. B 1 . 1 × 1 0 - 1 0
  3. C 1 . 1 × 1 0 - 1 2
  4. D 1 . 1 × 1 0 - 9
Verified Solution

Answer & Solution

Correct Answer

(B) 1 . 1 × 1 0 - 1 0

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 \(AgNO _3\).
\(\Rightarrow\left[ Ag ^{+}\right]^2\left[ CrO _4^{2-}\right]=(2 S+0.1)^2(S)\) \(=1.1 \times 10^{-12}\)
\(S \ll 0.1\)
\(\Rightarrow(0.1)^2(S)=1.1 \times 10^{-12}=10^{-2} \times S\)
\(S =1.1 \times 10^{-10} M\)
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