A) at constant temperature, the product \[[B{{a}^{2+}}]\,\,.\,C{{l}^{-2}}{{]}^{2}}\] remains constant in a saturated solution
B) ionic product \[[B{{a}^{2+}}]\,\,.\,\,[C{{l}^{-}}]\] remains constant in a saturated collection
C) the ionic product \[[B{{a}^{2+}}]\,{{[C{{l}^{-2}}]}^{2}}\] is greater than solubility product
D) it follows from Le-Chateliers principle
Correct Answer: C
Solution :
Key Idea For electrolyte AB, we have \[AB{{A}^{+}}+{{B}^{-}}\] \[{{K}_{sp}}=[{{A}^{+}}]\,\,[{{B}^{-}}]\] for saturated solution If \[[{{A}^{+}}]\,[{{B}^{-}}]\,\,>{{K}_{sp}}\] precipitation occur. If \[[{{A}^{+}}]\,[{{B}^{-}}]\,\,<{{K}_{sp}}\] substance will dissolve. In the saturated solution of \[BaC{{l}_{2}}\] \[BaC{{l}_{2}}B{{a}^{2+}}+2C{{l}^{-}}\] \[{{K}_{sp}}=[B{{a}^{2+}}]\,\,{{[C{{l}^{-}}]}^{2}}\] If cone. \[HCl\] is added through the saturated solution of \[BaC{{l}_{2}},\,\,C{{l}^{-}}\] ion concentration will increase. \[HCl{{H}^{+}}+C{{l}^{-}}\] So, that \[{{K}_{sp}}<[B{{a}^{2+}}]\,\,{{[C{{l}^{-}}]}^{2}}\]and precipitation of \[BaC{{l}_{2}}\] occur. Therefore, addition of cone. \[HCl\] to saturated \[BaC{{l}_{2}}\] solution, precipitates \[BaC{{l}_{2}}\]because the ionic product of \[[B{{a}^{2+}}]\,\,{{[C{{l}^{-}}]}^{2}}\] is greater than solubility product.
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