A) \[1\times {{10}^{-6}}\]
B) \[1\times {{10}^{-3}}\]
C) \[1.1\times {{10}^{-6}}\]
D) \[1.1\times {{10}^{-5}}\]
Correct Answer: B
Solution :
\[PbC{{l}_{2}}\]completely ionised in the solution as\[PbC{{l}_{2}}\xrightarrow{{}}P{{b}^{2+}}+2C{{l}^{-}}\]ie, 1 mole of \[PbC{{l}_{2}}\]in the solution gives 1 mole of\[P{{b}^{2+}}\]ion and 2 moles of\[C{{l}^{-}}\]ions. Now, as the solubility of\[PbC{{l}_{2}}\] \[=6.3\times {{10}^{-2}}mol/L\] \[\therefore \] \[[P{{b}^{2+}}]=6.3\times {{10}^{-2}}mol/L\] And \[[C{{l}^{-}}]=2\times 6.3\times {{10}^{-2}}\] \[=12.6\times {{10}^{-2}}mol/L\] \[\therefore \]\[{{K}_{sp}}\]for\[PbC{{l}_{2}}=[P{{b}^{2+}}]{{[C{{l}^{-}}]}^{2}}\] \[=(6.3\times {{10}^{-2}})\times {{(12.6\times {{10}^{-2}})}^{2}}\] \[=1\times {{10}^{-3}}\]
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