For a concentrated solution of a weak electro | vedclass

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Question

$A _2 B _3 	ext { (aq.) } ightleftharpoons 2 A _{	ext {(aq.) }}^{3+}+3 B _{	ext {(aq) }}^{2-}$

$K _{ eq }=\frac{\left[ A ^{3+}ight]^2\left[

Vedclass · Accepted Answer

$\left(\frac{ K _{ eq }}{108 c ^4}\right)^{\frac{1}{5}}$

Vedclass · Answer

$A _2 B _3 	ext { (aq.) } ightleftharpoons 2 A _{	ext {(aq.) }}^{3+}+3 B _{	ext {(aq) }}^{2-}$

$K _{ eq }=\frac{\left[ A ^{3+}ight]^2\left[ B ^{2-}ight]^3}{\left[ A _2 B _3ight]}=\frac{4 c ^2 \alpha^2 	imes 27 c ^3 \alpha^3}{ c (1-\alpha)}$

$K _{ eq }=\frac{108 c ^5 \alpha^5}{ c } \alpha=\left(\frac{ K _{ eq }}{108 c ^4}ight)^{\frac{1}{5}}$

For a concentrated solution of a weak electrolyte ( $K _{ eq }=$ equilibrium constant) $A _2 B _3$ of concentration ' $c$ ', the degree of dissociation " $\alpha$ ' is

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