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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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The solubility of $A_2X_3$ is y mol $dm^{-3}$. Its solubility product is

(a) $6y^4$

(b) $64 y^4$

(c) $36 y^5$

(d) $108 y^5$

Can you answer this question?

1 Answer

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Answer: $108 y^5$
$A_2X_3 \rightleftharpoons 2A^{3+} + 3X^{2-}$
At equilibrium [$2A^{3+}$] = 2y and [$3X^{2-}$] = 3y
$\therefore$ solubility product of $A_2X_3, K_{sp} = (2y)^2 (3y)^3 =108 y^5$


answered Nov 27, 2013 by mosymeow_1

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