(B) The dissociation of the sparingly soluble salt $BA$ is given by: $BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$ For a salt of the type

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(B) The dissociation of the sparingly soluble salt $BA$ is given by: $BA_{(s)} \rightleftharpoons B_{(aq)}^{+} + A_{(aq)}^{-}$ For a salt of the type $BA$,the solubility product $K_{sp}$ is related to solubility $S$ by the formula: $K_{sp} = [B^+][A^-] = S \times S = S^2$ Given,solubility $S = 7.2 \times 10^{-7} \ mol \ dm^{-3}$. Substituting the value of $S$ in the formula: $K_{sp} = (7.2 \times 10^{-7})^2$ $K_{sp} = 51.84 \times 10^{-14} = 5.184 \times 10^{-13}$

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Solubility product

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Calculate the solubility product of sparingly soluble salt $BA$ at $25^{\circ} C$ if its solubility is $7.2 \times 10^{-7} \ mol \ dm^{-3}$ at same temperature.

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A
$4.810 \times 10^{-13}$
B
$5.184 \times 10^{-13}$
C
$6.454 \times 10^{-13}$
D
$5.925 \times 10^{-13}$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Calculate the solubility product of sparingly soluble salt $BA$ at $25^{\circ} C$ if its solubility is $7.2 \times 10^{-7} \ mol \ dm^{-3}$ at same temperature.

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