A) \[1.8\times {{10}^{-7}}\]
B) \[3.6\times {{10}^{-10}}\]
C) 0.01
D) \[2\times {{10}^{-4}}\]
Correct Answer: C
Solution :
[c] \[{{K}_{sp}}(AgBr)<{{K}_{sp}}(AgCl)\] Therefore, \[AgBr\] will precipitate first and at that time all \[C{{l}^{-}}\] will be present.You need to login to perform this action.
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