The K_{sp} for bismuth sulphide (Bi_{2}S_{3}) | vedclass

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Question

(A) The dissociation of bismuth sulphide is given by: $Bi_{2}S_{3}(s) \rightleftharpoons 2Bi^{3+}(aq) + 3S^{2-}(aq)$.Let the solubility be $s \ mol \

Vedclass · Accepted Answer

$1.0 	imes 10^{-15}$

Vedclass · Answer

(A) The dissociation of bismuth sulphide is given by: $Bi_{2}S_{3}(s) ightleftharpoons 2Bi^{3+}(aq) + 3S^{2-}(aq)$.Let the solubility be $s \ mol \ L^{-1}$. Then the concentrations are $[Bi^{3+}] = 2s$ and $[S^{2-}] = 3s$.The solubility product expression is $K_{sp} = [Bi^{3+}]^{2} [S^{2-}]^{3}$.Substituting the values: $K_{sp} = (2s)^{2} (3s)^{3} = 4s^{2} 	imes 27s^{3} = 108s^{5}$.Given $K_{sp} = 1.08 	imes 10^{-73}$,we have $108s^{5} = 1.08 	imes 10^{-73}$.$s^{5} = \frac{1.08 	imes 10^{-73}}{108} = 0.01 	imes 10^{-73} = 10^{-75}$.Taking the fifth root: $s = (10^{-75})^{1/5} = 10^{-15} \ mol \ L^{-1}$.

The $K_{sp}$ for bismuth sulphide $(Bi_{2}S_{3})$ is $1.08 \times 10^{-73}$. The solubility of $Bi_{2}S_{3}$ in $mol \ L^{-1}$ at $298 \ K$ is ......

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