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Use the following data to calculate the KspK_{sp} value for each solid. The solubility of BiI3 is 1.32×105mol/L.\mathrm { BiI } _ { 3 } \text { is } 1.32 \times 10 ^ { - 5 } \mathrm { mol } / \mathrm { L }.

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The reaction is

BiI3Bi3++3IBiI_3 \rightleftharpoons Bi^{3+}+3I^-

From the reaction, the following relationship can be obtained:

[BiI3]=[Bi3+][BiI_3]=[Bi^{3+}] [I]=3[BiI3][I^-]=3[BiI_3]

Therefore,

Ksp=[Bi3+][I]3=[BiI3](3[BiI3])3=27[BiI3]4=27(1.32×105)4K_{sp}=[Bi^{3+}][I^-]^3=[BiI_3](3[BiI_3])^3=27[BiI_3]^4=27(1.32\times10^{-5})^4

Ksp=8.20×1019K_{sp}=8.20\times10^{-19}

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