A) \[1\times {{10}^{-5}}\]
B) \[1\times {{10}^{-10}}\]
C) \[1.435\times {{10}^{-5}}\]
D) \[108\times {{10}^{-3}}\]
Correct Answer: B
Solution :
\[AgClA{{g}^{+}}+C{{l}^{-}}\] \[{{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]\] (S = solubility in mol/L) \[S=\frac{1.435\times {{10}^{-3}}g/L}{143.5}=1\times {{10}^{-10}}\text{mol/L}\]You need to login to perform this action.
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