A) \[1\times {{10}^{-3}}\]
B) \[1\times {{10}^{-9}}\]
C) \[4\times {{10}^{-27}}\]
D) \[1\times {{10}^{-27}}\]
Correct Answer: A
Solution :
\[BaC{{l}_{2}}\underset{s}{\overset{2+}{\mathop{Ba}}}\,+\underset{2s}{\mathop{2C{{l}^{-}}}}\,{{K}_{sp}}=4{{s}^{3}}\], \[{{s}^{3}}=\frac{4\times {{10}^{-9}}}{4}={{10}^{-3}}\] \[s={{10}^{-3}}M\]
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