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(B) The precipitation order depends on the $K_{sp}$ values. Since $[Cl^-] = [Br^-] = [I^-] = 0.1 \ M$,the ion requiring the lowest $[Ag^+]$ will preci

Vedclass · Accepted Answer

$4.5 	imes 10^{-8}$

Vedclass · Answer

(B) The precipitation order depends on the $K_{sp}$ values. Since $[Cl^-] = [Br^-] = [I^-] = 0.1 \ M$,the ion requiring the lowest $[Ag^+]$ will precipitate first.$1$. For $AgI$: $[Ag^+] = \frac{K_{sp}(AgI)}{[I^-]} = \frac{9 	imes 10^{-17}}{0.1} = 9 	imes 10^{-16} \ M$.$2$. For $AgBr$: $[Ag^+] = \frac{K_{sp}(AgBr)}{[Br^-]} = \frac{5 	imes 10^{-13}}{0.1} = 5 	imes 10^{-12} \ M$.$3$. For $AgCl$: $[Ag^+] = \frac{K_{sp}(AgCl)}{[Cl^-]} = \frac{2 	imes 10^{-10}}{0.1} = 2 	imes 10^{-9} \ M$.Thus,the order of precipitation is $I^- > Br^- > Cl^-$. The third ion to precipitate is $Cl^-$. When $Cl^-$ starts precipitating,$[Ag^+] = 2 	imes 10^{-9} \ M$.At this point,the concentration of the first ion $(I^-)$ is:$[I^-] = \frac{K_{sp}(AgI)}{[Ag^+]} = \frac{9 	imes 10^{-17}}{2 	imes 10^{-9}} = 4.5 	imes 10^{-8} \ M$.

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Salt	$K_{sp}$
$AgCl$	$2 \times 10^{-10}$
$AgBr$	$5 \times 10^{-13}$
$AgI$	$9 \times 10^{-17}$