Question

EMF of the cell, $Ag|AgN{O}_3\left(0.1M\right)|KBr\left(1N\right),AgBr\left(s\right)|Ag$ is $-0.6\text{V}$ at $298\text{K}$
Calculate the $K_{sp}$ of $AgBr$ at $298\text{K}$.

• A. $K_{sp}=4.0\times {10}^{-12}$
• B. $K_{sp}=3.5\times {10}^{-12}$
• C. $K_{sp}=4.8\times {10}^{-12}$
• D. $K_{sp}=3.8\times {10}^{-12}$

Answer 1

The correct option is C $K_{sp}=4.8\times {10}^{-12}$

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