A) \[4.1\times {{10}^{5}}M\]
B) \[5.1\times {{10}^{5}}M\]
C) \[8.1\times {{10}^{8}}M\]
D) \[8.1\times {{10}^{7}}M\]
Correct Answer: B
Solution :
\[N{{a}_{2}}C{{O}_{3}}\xrightarrow{{}}2N{{a}^{+}}+CO_{3}^{2-}\]\[1\times {{10}^{-4}}M\] \[BaC{{O}_{3}}\xrightarrow{{}}B{{a}^{2+}}+CO_{3}^{2-}\] \[{{K}_{SP}}=[B{{a}^{2+}}][CO_{3}^{2-}]\] \[5.1\times {{10}^{-9}}=[Ba{{~}^{+2}}][1\times {{10}^{-4}}]\] \[[B{{a}^{2+}}]=\frac{5.1\times {{10}^{-9}}}{1\times {{10}^{-4}}}\] \[=5.1\times {{10}^{-5}}M\]
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