A) \[BaS{{O}_{4}}>H{{g}_{2}}C{{l}_{2}}>Cr{{(S{{O}_{4}})}_{3}}>CrC{{l}_{3}}\]
B) \[BaS{{O}_{4}}>H{{g}_{2}}C{{l}_{2}}>CrC{{l}_{3}}>C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\]
C) \[BaS{{O}_{4}}>CrC{{l}_{3}}>H{{g}_{2}}C{{l}_{2}}>C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\]
D) \[H{{g}_{2}}C{{l}_{2}}>BaS{{O}_{4}}>CrC{{l}_{3}}>C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\]
Correct Answer: B
Solution :
[b] \[C{{r}_{2}}{{(S{{O}_{4}})}_{3}}\rightleftharpoons \underset{2s}{\mathop{2C{{r}^{3+}}}}\,+\underset{3s}{\mathop{3SO_{4}^{2-}}}\,\,\,\,{{K}_{sp}}={{(2s)}^{2}}{{(3s)}^{3}}=4{{s}^{2}}\times 27{{s}^{3}}=108{{s}^{5}}\] \[s={{\left( \frac{{{K}_{sp}}}{108} \right)}^{1/5}}\] \[H{{g}_{2}}C{{l}_{2}}\rightleftharpoons \underset{2s}{\mathop{2H{{g}^{2+}}}}\,+\underset{2s}{\mathop{2C{{l}^{-}}}}\,\] \[{{K}_{sp}}={{(2s)}^{2}}\times {{(2s)}^{2}}=16{{s}^{4}}\] \[s={{\left( \frac{{{K}_{sp}}}{16} \right)}^{1/4}}\] \[BaS{{O}_{4}}\rightleftharpoons \underset{s}{\mathop{B{{a}^{2+}}}}\,+\underset{s}{\mathop{SO_{4}^{2-}}}\,\] \[{{K}_{sp}}={{s}^{2}}\] \[s=\sqrt{{{K}_{sp}}}\] \[CrC{{l}_{3}}\rightleftharpoons \underset{s}{\mathop{C{{r}^{3+}}}}\,+\underset{3s}{\mathop{3C{{l}^{-}}}}\,~\] \[{{K}_{sp}}=s\times {{(3s)}^{3}}=27{{s}^{4}}\] \[s={{\left( \frac{{{K}_{sp}}}{27} \right)}^{1/4}}\] Hence the correct order of solubilities of salts is \[\sqrt{{{K}_{sp}}}>{{\left( \frac{{{K}_{sp}}}{16} \right)}^{1/4}}>{{\left( \frac{{{K}_{sp}}}{27} \right)}^{1/4}}>{{\left( \frac{{{K}_{sp}}}{108} \right)}^{1/5}}\]
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