For a sparingly soluble strong electrolyte Ag | vedclass

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Question

(B) The dissociation equilibrium is: $AgIO_3(s) \rightleftharpoons Ag^+(aq) + IO_3^-(aq)$.Let the solubility be $S \, mol/L$. Then $K_{sp} = [Ag^+][IO

Vedclass · Accepted Answer

$2.83 	imes 10^{-3} \, g$

Vedclass · Answer

(B) The dissociation equilibrium is: $AgIO_3(s) ightleftharpoons Ag^+(aq) + IO_3^-(aq)$.Let the solubility be $S \, mol/L$. Then $K_{sp} = [Ag^+][IO_3^-] = S 	imes S = S^2$.Given $K_{sp} = 1.0 	imes 10^{-8}$,so $S = \sqrt{1.0 	imes 10^{-8}} = 10^{-4} \, mol/L$.The molar mass of $AgIO_3$ is $283 \, g/mol$.The mass of $AgIO_3$ in $100 \, mL$ $(0.1 \, L)$ is calculated as:$	ext{Mass} = 	ext{Molarity} 	imes 	ext{Molar Mass} 	imes 	ext{Volume in Liters}$$	ext{Mass} = 10^{-4} \, mol/L 	imes 283 \, g/mol 	imes 0.1 \, L = 2.83 	imes 10^{-3} \, g$.

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