Calculate the ionic concentration of a sparin | vedclass

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Question

(A) For a sparingly soluble salt of the type $BA$,the dissociation equilibrium is given by: $BA(s) \rightleftharpoons B^+(aq) + A^-(aq)$.Let the solub

Vedclass · Accepted Answer

$1.643 	imes 10^{-5}$

Vedclass · Answer

(A) For a sparingly soluble salt of the type $BA$,the dissociation equilibrium is given by: $BA(s) ightleftharpoons B^+(aq) + A^-(aq)$.Let the solubility of the salt be $s \ mol \ dm^{-3}$.Then,$[B^+] = s$ and $[A^-] = s$.The solubility product constant $(K_{sp})$ is given by: $K_{sp} = [B^+][A^-] = s 	imes s = s^2$.Given $K_{sp} = 2.7 	imes 10^{-10}$.Therefore,$s^2 = 2.7 	imes 10^{-10}$.$s = \sqrt{2.7 	imes 10^{-10}} = \sqrt{27 	imes 10^{-11}} = \sqrt{2.7} 	imes 10^{-5} \approx 1.643 	imes 10^{-5} \ mol \ dm^{-3}$.Thus,the ionic concentration of each ion is $1.643 	imes 10^{-5} \ mol \ dm^{-3}$.

Calculate the ionic concentration of a sparingly soluble salt $BA$ in $mol \ dm^{-3}$ at $300 \ K$ when equilibrium is attained,if the solubility product of the salt is $2.7 \times 10^{-10}$ at the same temperature.

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