A) \[{{S}_{AgBr{{O}_{3}}}}>{{S}_{A{{g}_{2}}S{{O}_{4}}}}\]
B) \[{{S}_{AgBr{{O}_{3}}}}<{{S}_{A{{g}_{2}}S{{O}_{4}}}}\]
C) \[{{S}_{AgBr{{O}_{3}}}}={{S}_{A{{g}_{2}}S{{O}_{4}}}}\]
D) \[{{S}_{AgBr{{O}_{3}}}}\approx {{S}_{A{{g}_{2}}S{{O}_{4}}}}\]
Correct Answer: B
Solution :
\[A{{g}_{2}}S{{O}_{4}}\] ⇌ \[\underset{4{{S}^{2}}}{\mathop{2A{{g}^{+}}}}\,+\underset{S\,\,\,\,\,\,\,\,}{\mathop{SO_{4}^{-\,-}}}\,\] \[{{K}_{sp}}=4{{S}^{3}};\,\,{{K}_{sp}}=2\times {{10}^{-5}}\] \[S=\sqrt[3]{\frac{2\times {{10}^{-5}}}{4}}=0.017\,m/l\]\[=1.7\times {{10}^{-2}}\] Ag\[Br{{O}_{3}}\] ⇌ \[\underset{S}{\mathop{A{{g}^{+}}}}\,+\underset{S}{\mathop{BrO_{3}^{-}}}\,\] \[{{K}_{sp}}={{S}^{2}};\,\,{{K}_{sp}}=5.5\times {{10}^{-5}}\] \[S=\sqrt{5.5\times {{10}^{-5}}}=7.4\times {{10}^{-3}}\,m/l.\]
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