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

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(B) The solubility product expression for $AgBr$ is $K_{sp} = [Ag^+][Br^-] = S^2$.Given $K_{sp} = 4.9 	imes 10^{-13} = 49 	imes 10^{-14}$.Thus,solub

Vedclass · Accepted Answer

$14 	imes 10^{-6} 	imes 188 \, g$

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(B) The solubility product expression for $AgBr$ is $K_{sp} = [Ag^+][Br^-] = S^2$.Given $K_{sp} = 4.9 	imes 10^{-13} = 49 	imes 10^{-14}$.Thus,solubility $S = \sqrt{49 	imes 10^{-14}} = 7 	imes 10^{-7} \, mol/L$.To find the mass dissolved in $20 \, L$,we use the formula: $	ext{Mass} = S 	imes 	ext{Volume} 	imes 	ext{Molar Mass}$.$	ext{Mass} = (7 	imes 10^{-7} \, mol/L) 	imes (20 \, L) 	imes (188 \, g/mol)$.$	ext{Mass} = 140 	imes 10^{-7} 	imes 188 \, g = 14 	imes 10^{-6} 	imes 188 \, g$.

The $K_{sp}$ for $AgBr$ at $25 \, ^\circ C$ is $4.9 \times 10^{-13}$. How much $AgBr$ (in grams) will dissolve in $20 \, L$ of its saturated solution? (Molecular weight of $AgBr = 188$)

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