The solubility product of silver bromide (AgB | vedclass

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(C) The solubility product expression for $AgBr$ is $K_{sp} = [Ag^+][Br^-]$.Given $[Ag^+] = [AgNO_3] = 0.05 \, M$.Substituting the values into the $K_

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$1.2 	imes 10^{-9} \, g$

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(C) The solubility product expression for $AgBr$ is $K_{sp} = [Ag^+][Br^-]$.Given $[Ag^+] = [AgNO_3] = 0.05 \, M$.Substituting the values into the $K_{sp}$ expression:$5.0 	imes 10^{-13} = (0.05) 	imes [Br^-]$.$[Br^-] = \frac{5.0 	imes 10^{-13}}{0.05} = 1.0 	imes 10^{-11} \, M$.Since $[Br^-] = [KBr]$,the concentration of $KBr$ required is $1.0 	imes 10^{-11} \, mol \, L^{-1}$.For $1 \, L$ of solution,the number of moles of $KBr$ is $1.0 	imes 10^{-11} \, mol$.Mass of $KBr = 	ext{moles} 	imes 	ext{molar mass} = (1.0 	imes 10^{-11} \, mol) 	imes (120 \, g \, mol^{-1}) = 1.2 	imes 10^{-9} \, g$.

The solubility product of silver bromide $(AgBr)$ is $5.0 \times 10^{-13}$. How much potassium bromide $(KBr)$ (molar mass $120 \, g \, mol^{-1}$) should be added to $1 \, L$ of $0.05 \, M$ silver nitrate $(AgNO_3)$ solution to initiate the precipitation of $AgBr$?

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