The K_{sp} of AgBr at 25 , ^circ C is 4.9 tim | vedclass

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(B) For $AgBr$,the solubility equilibrium is $AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)$.$K_{sp} = [Ag^+][Br^-] = s^2$,where $s$ is the solubilit

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$14 	imes 10^{-6} 	imes 188$

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(B) For $AgBr$,the solubility equilibrium is $AgBr(s) ightleftharpoons Ag^+(aq) + Br^-(aq)$.$K_{sp} = [Ag^+][Br^-] = s^2$,where $s$ is the solubility in $mol/L$.Given $K_{sp} = 4.9 	imes 10^{-13}$,so $s = \sqrt{4.9 	imes 10^{-13}} = \sqrt{49 	imes 10^{-14}} = 7 	imes 10^{-7} \, mol/L$.For $20 \, L$ of water,the moles of $AgBr$ required $= s 	imes 20 = 7 	imes 10^{-7} 	imes 20 = 140 	imes 10^{-7} = 14 	imes 10^{-6} \, mol$.Mass of $AgBr$ required $= 	ext{moles} 	imes 	ext{molar mass} = 14 	imes 10^{-6} 	imes 188 \, g$.

The $K_{sp}$ of $AgBr$ at $25 \, ^\circ C$ is $4.9 \times 10^{-13}$. If the molar mass of $AgBr$ is $188 \, g/mol$,how much $AgBr$ is required to saturate $20 \, L$ of water?

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