A) \[2.6\times {{10}^{-3}}M\]
B) \[8.8\times {{10}^{-9}}\]
C) \[1.8\times {{10}^{-8}}\]
D) \[3.5\times {{10}^{-8}}\]
Correct Answer: B
Solution :
\[Pb{{I}_{2}}(s)P\underset{S}{\mathop{{{b}^{2+}}}}\,+{{\underset{2S}{\mathop{2I}}\,}^{-}}\] \[2.6\times {{10}^{-3}}M\] Given, \[2s=2.6\times {{10}^{-3}}\] \[s=\frac{2.6\times {{10}^{-3}}}{2}\] \[{{K}_{sp}}=[P{{b}^{2+}}]\,\,{{[{{I}^{-}}]}^{2}}\] \[=(s)\,{{(2s)}^{2}}\] \[={{\left( \frac{2.6\times {{10}^{-3}}}{2} \right)}^{2}}\,{{(2.6\times {{10}^{-3}})}^{2}}\] \[=\frac{2.6\times {{10}^{-3}}\times 6.76\times {{10}^{-6}}}{2}\] \[=8.8\times {{10}^{-9}}\]
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