A) \[3.4\times {{10}^{-26}}\]
B) \[3.4\times {{10}^{-30}}\]
C) \[1\times {{10}^{-2}}\]
D) \[3.4\times {{10}^{-22}}\]
Correct Answer: A
Solution :
\[PbSP{{b}^{2+}}+{{S}^{2-}}\] For precipitation, the ionic product of PbS should be greater than solubility product. Hence the concentration of\[{{S}^{2-}}\]ions will be minimum. Concentration of \[{{S}^{2-}}>\frac{Solubility\text{ }product}{concentration\text{ }of\text{ }P{{b}^{2+}}}\] \[>\frac{3.4\times {{10}^{-28}}}{1\times {{10}^{-2}}}\] \[>3.4\times {{10}^{-26}}mol\]
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