A_{3}B_{2} is a sparingly soluble salt of mol | vedclass

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(D) The dissociation of the salt is given by: $A_{3}B_{2(s)} \rightleftharpoons 3A^{2+}_{(aq)} + 2B^{3-}_{(aq)}$Let the molar solubility be $s \ mol \

Vedclass · Accepted Answer

$108$

Vedclass · Answer

(D) The dissociation of the salt is given by: $A_{3}B_{2(s)} \rightleftharpoons 3A^{2+}_{(aq)} + 2B^{3-}_{(aq)}$Let the molar solubility be $s \ mol \ L^{-1}$.At equilibrium,the concentrations are $[A^{2+}] = 3s$ and $[B^{3-}] = 2s$.The solubility product expression is $K_{sp} = [A^{2+}]^{3} [B^{3-}]^{2}$.Substituting the values: $K_{sp} = (3s)^{3} (2s)^{2} = (27s^{3}) (4s^{2}) = 108s^{5}$.Given that molar solubility $s = \frac{x}{M}$,where $x$ is solubility in $g \ L^{-1}$ and $M$ is molar mass in $g \ mol^{-1}$.Thus,$K_{sp} = 108(\frac{x}{M})^{5}$.Comparing this with the given expression $K_{sp} = a(\frac{x}{M})^{5}$,we get $a = 108$.

$A_{3}B_{2}$ is a sparingly soluble salt of molar mass $M \ g \ mol^{-1}$ and solubility $x \ g \ L^{-1}$. The solubility product satisfies $K_{sp} = a(\frac{x}{M})^{5}$. The value of $a$ is ...... . (Integer answer)

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