A) \[A{{g}_{2}}S{{O}_{4}}\]
B) \[BaS{{O}_{4}}\]
C) \[CaS{{O}_{4}}\]
D) All of these
Correct Answer: B
Solution :
[b] The species having minimum value of \[{{K}_{sp}}\] will get precipitated first of all because ionic product will exceed the solubility product of such a species. The \[{{K}_{sp}}\] value is minimum for \[BaS{{O}_{4}}({{10}^{-11}}),\] so, \[BaS{{O}_{4}}\] will get precipitated first of all.You need to login to perform this action.
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