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(A) The solubility product expression is $K_{sp} = [Ag^+][IO_3^-]$.Since $AgIO_3 \rightleftharpoons Ag^+ + IO_3^-$,let the solubility be $s \ mol \ L^

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The solubility product expression is $K_{sp} = [Ag^+][IO_3^-]$.Since $AgIO_3 ightleftharpoons Ag^+ + IO_3^-$,let the solubility be $s \ mol \ L^{-1}$.Then $K_{sp} = s^2 = 1.0 	imes 10^{-8}$.$s = \sqrt{1.0 	imes 10^{-8}} = 1.0 	imes 10^{-4} \ mol \ L^{-1}$.In $100 \ mL$ $(0.1 \ L)$,the number of moles $n = s 	imes V = 1.0 	imes 10^{-4} \ mol \ L^{-1} 	imes 0.1 \ L = 1.0 	imes 10^{-5} \ mol$.Mass $= n 	imes 	ext{Molar Mass} = 1.0 	imes 10^{-5} \ mol 	imes 283 \ g \ mol^{-1} = 2.83 	imes 10^{-3} \ g$.

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