A) \[4.18\times {{10}^{-2}}\]
B) \[6.76\times {{10}^{-3}}\]
C) \[3.4\times {{10}^{-4}}\]
D) \[5.44\times {{10}^{-2}}\]
Correct Answer: A
Solution :
Key Idea: First find relationship between solubility and solubility product. \[PbB{{r}_{2}}P{{b}^{2}}+2B{{r}^{-}}\] \[{{K}_{sp}}=[P{{b}^{2+}}]{{[B{{r}^{-}}]}^{2}}\] given \[{{K}_{sp}}\,PbB{{r}_{2}}=10.8\times {{10}^{-5}}\] \[\alpha =\,70%\] \[\text{Solubility}=x\,mol/L\] \[\therefore \] \[[P{{b}^{2+}}]=0.7x,\] \[[B{{r}^{-}}]=2\times 0.7\times x=1.4x\] \[\therefore \] \[{{K}_{sp}}=[P{{b}^{2+}}]{{[B{{r}^{-}}]}^{2}}\] \[=(0.7x)\,{{(1.4x)}^{2}}\] or \[10.8\times {{10}^{-5}}=1.372\,{{x}^{3}}\] or \[x=\sqrt[3]{\frac{10.8\times {{10}^{-5}}}{1.372}}\] \[=4.18\times {{10}^{-2}}\]
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