A) \[{{10}^{-3}}M\,(A{{g}^{+}})\] and \[{{10}^{-3}}M(C{{l}^{-}})\]
B) \[{{10}^{-5}}M\,(A{{g}^{+}})\] and \[{{10}^{-5}}M(C{{l}^{-}})\]
C) \[{{10}^{-6}}M\,(A{{g}^{+}})\] and \[{{10}^{-5}}M(C{{l}^{-}})\]
D) \[{{10}^{-4}}M\,(A{{g}^{+}})\] and \[{{10}^{-4}}M(C{{l}^{-}})\]
Correct Answer: D
Solution :
For precipitation, ionic product > solubility product \[[A{{g}^{+}}]=\frac{1}{2}\times {{10}^{-4}}=5\times {{10}^{-5}}M\] \[[C{{l}^{-}}]=\frac{1}{2}\times {{10}^{-4}}=5\times {{10}^{-5}}M\] \[{{K}_{sp}}=[A{{g}^{+}}]\,[C{{l}^{-}}]={{(5\times {{10}^{-5}})}^{2}}=2.5\times {{10}^{-9}}\] In this case ionic product (jp) is greater than the solubility product, hence precipitation will take place. \[C{{H}_{3}}CHO+\underset{\left( Tollens\text{ }reagent \right)}{\mathop{2{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}O{{H}^{\Theta }}}}\,\xrightarrow{{}}\] \[C{{H}_{3}}CO{{O}^{-}}+2Ag+4N{{H}_{3}}+2{{H}_{2}}O\] \[{{H}_{3}}CC{{O}_{2}}{{C}_{2}}{{H}_{5}}\xrightarrow{O{{H}^{-}}(base)}{{H}_{3}}CCO{{O}^{-}}\]\[+{{C}_{2}}{{H}_{5}}OH\] So, is the correct option.
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