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In the saturated solution of Sb₂S₃, the concentration of Sb⁺³ is found to be 10^-3M at 300K. The solubility product of Sb₂S₃ at same temperature is

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$\large \left( {\frac{{27}}{8}} \right){10^{ - 15}}{M^5}$

$\left( {\frac{{27}}{8}} \right){10^{ - 12}}{M^5}$

$\left( {\frac{{27}}{2}} \right){10^{ - 15}}{M^5}$

$108 \times {10^{ - 15}}{M^5}$

answer is A.

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Detailed Solution

$\large S{b_2}{S_3}\left( s \right) \rightleftharpoons \mathop {S{b_2}{S_3}\left( {aq} \right)}\limits_S \to \mathop {2S{b^{ + 3}}\left( {aq} \right)}\limits_{2S} + \mathop {3{S^{ - 2}}\left( {aq} \right)}\limits_{3S}$

$\large {K_{sp}} = {\left( {2S} \right)^2} \times {\left( {3S} \right)^3} = 108{S^5}$

Now, 2S = 10^-3 ⇒ S = $\large \frac{{{{10}^{ - 3}}}}{2}$

$\large {K_{sp}} = 108 \times {\left( {\frac{1}{2}} \right)^5} \times {10^{ - 15}}{M^5}$

$\large = \frac{{27}}{8} \times {10^{ - 15}}{M^5}$

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