A) \[5.1\times {{10}^{-5}}M\]
B) \[7.1\times {{10}^{-8}}M\]
C) \[4.1\times {{10}^{-5}}M\]
D) \[8.1\times {{10}^{-7}}M\]
Correct Answer: A
Solution :
[a] Given \[N{{a}_{2}}C{{O}_{3}}=1.0\times {{10}^{-4}}M\] \[\therefore \text{ }\left[ CO_{3}^{2-} \right]=1.0\times {{10}^{-4}}M\] i.e. \[s=1.0\times {{10}^{-4}}M\] At equilibrium \[[B{{a}^{2+}}][CO_{3}^{2-}]={{K}_{sp}}\] of \[BaC{{O}_{3}}\] \[[B{{a}^{2+}}]=\frac{{{K}_{sp}}}{[CO_{3}^{2-}]}=\frac{5.1\times {{10}^{-9}}}{1.0\times {{10}^{-4}}}\] \[=5.1\times {{10}^{-5}}M\]You need to login to perform this action.
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