A) \[6.4\times {{10}^{-15}}\]
B) \[6.4\times {{10}^{-14}}\]
C) \[2.5\times {{10}^{-13}}\]
D) \[6.4\times {{10}^{-13}}\]
Correct Answer: C
Solution :
\[Pb{{I}_{2}}P\underset{s}{\mathop{{{b}^{2+}}}}\,+\underset{2s}{\mathop{2{{I}^{-}}}}\,\] \[{{K}_{sp}}=[P{{b}^{2+}}]{{[{{I}^{-}}]}^{2}}\] \[=(s)\,{{(2s)}^{2}}\] \[=4{{s}^{3}}\] \[=4\times {{(4\times {{10}^{-5}})}^{3}}\] \[=4\times 64\times {{10}^{-15}}\] \[=256\times {{10}^{-15}}\] \[=2.56\times {{10}^{-13}}\]
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