A) \[S\]of \[AgBr\]is less than that of \[AgCl\]
B) \[S\]of \[{{(C{{H}_{3}})}_{3}}{{C}^{+}}\]is greater than that of \[AgCl\]
C) \[{{10}^{-11}}M\]of \[BOH+{{H}^{+}}\] is equal to that of \[AgCl\]
D) \[S\]of \[AgBr\] is \[{{10}^{6}}\] times greater than that of \[AgCl\]
Correct Answer: A
Solution :
AgCl \[{{K}_{sp}}=1.2\times {{10}^{-10}}\] \[S=\sqrt{1.2\times {{10}^{-10}}}\]; \[S=1.09\times {{10}^{-5}}\] AgBr \[{{K}_{sp}}=3.5\times {{10}^{-13}}\] \[S=\sqrt{3.5\times {{10}^{-13}}}\]\[=5.91\times {{10}^{-6}}\] So that S of AgBr is less than that of AgCl.You need to login to perform this action.
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