The solubility product of silver bromide (AgB | vedclass

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(B) The precipitation of $AgBr$ begins when the ionic product exceeds the solubility product $(K_{sp})$.$K_{sp} = [Ag^+][Br^-] = 5.0 	imes 10^{-13}$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The precipitation of $AgBr$ begins when the ionic product exceeds the solubility product $(K_{sp})$.$K_{sp} = [Ag^+][Br^-] = 5.0 	imes 10^{-13}$.Given $[Ag^+] = 0.05 \, M$,we find the concentration of $Br^-$ required:$[Br^-] = \frac{K_{sp}}{[Ag^+]} = \frac{5.0 	imes 10^{-13}}{0.05} = 1.0 	imes 10^{-11} \, M$.Since the volume is $1 \, L$,the number of moles of $KBr$ required 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 $5.0 \times 10^{-13}$. Calculate the mass of potassium bromide $(KBr)$ (molar mass = $120 \, g \, mol^{-1}$) that must be added to $1 \, L$ of a $0.05 \, M$ silver nitrate $(AgNO_3)$ solution to initiate the precipitation of $AgBr$.

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