JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
\(M_3 A_2\) is a sparingly soluble salt of molar mass \(y\) g mol\(^{-1}\) and solubility \(x\) g L\(^{-1}\). The ratio of the molar concentration of the anion (\(A^{3-}\)) to the solubility product of the salt is
- A \(\dfrac{1}{54} \cdot \dfrac{y^4}{x^4}\)
- B \(\dfrac{y^5}{108 x^4}\)
- C \(108 \cdot \dfrac{x^5}{y^5}\)
- D \(\dfrac{1}{108} \cdot \dfrac{y^4}{x^4}\)
Answer & Solution
Correct Answer
(A) \(\dfrac{1}{54} \cdot \dfrac{y^4}{x^4}\)
Step-by-step Solution
Detailed explanation
The molar solubility \(S\) of the salt is given by \(S = \dfrac{x}{y}\) mol L\(^{-1}\). The dissociation of the sparingly soluble salt \(M_3A_2\) is: \(M_3A_2 \rightleftharpoons 3M^{2+} + 2A^{3-}\) The molar concentration of the anion \(A^{3-}\) is: \([A^{3-}] = 2S\) The…
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