A) \[1\times {{10}^{-7}}\]
B) \[3.2\times {{10}^{-7}}\]
C) \[1\times {{10}^{-5}}\]
D) \[3.2\times {{10}^{-5}}\]
Correct Answer: D
Solution :
\[PbC{{l}_{2}}\]⇌\[\underset{(S)\,\,}{\mathop{P{{b}^{2+}}}}\,+\underset{{{(2S)}^{2}}}{\mathop{2C{{l}^{-}}}}\,\] \[{{K}_{sp}}=4{{S}^{3}}=4\times {{(2\times {{10}^{-2}})}^{3}}=3.2\times {{10}^{-5}}\]You need to login to perform this action.
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