A) \[{{\left( \frac{K}{4} \right)}^{1/3}}\]
B) \[{{(2K)}^{1/3}}\]
C) \[{{K}^{1/2}}\]
D) \[\frac{K}{2}\]
Correct Answer: A
Solution :
: For lead chloride\[(PbC{{l}_{2}})\] \[{{K}_{sp}}=[P{{b}^{2+}}]{{[C{{l}^{-}}]}^{2}}\] \[{{K}_{sp}}=x.{{(2x)}^{2}}\] \[{{K}_{sp}}=x.4{{x}^{2}}\] \[x={{\left( \frac{K}{4} \right)}^{1/3}}\]You need to login to perform this action.
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