A) \[4\times {{10}^{-3}}\]
B) \[4\times {{10}^{-9}}\]
C) \[1\times {{10}^{-3}}\]
D) \[1\times {{10}^{-9}}\]
Correct Answer: C
Solution :
Key Idea: Find relationship between solubility product and solubility of \[BaC{{l}_{2}}\] and then solve problem (solubility product of \[BaC{{l}_{2}}\] is \[4\times {{10}^{-9}}\]) \[BaC{{l}_{2}}\xrightarrow{{}}B{{a}^{2+}}+2C{{l}^{-}}\] Let the solubility of \[BaC{{l}_{2}}\,\,x\,\,mol/L\] \[\therefore \] \[{{K}_{sp}}=[B{{a}^{2+}}]\,{{[C{{l}^{-}}]}^{2}}\] \[=(x)\times {{(2x)}^{2}}\] \[=x\times 4{{x}^{2}}=4{{x}^{3}}\] or solubility of \[BaC{{l}_{2}}\] \[=\frac{{{(solubility\text{ }product\text{ }of\,BaC{{l}_{2}})}^{\frac{1}{3}}}}{4}\] \[={{\left( \frac{4\times {{10}^{-9}}}{4} \right)}^{1/3}}\] \[={{10}^{-3}}mol/L\]
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