The solubility product of silver bromide is 5 | vedclass

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(B) The dissociation of silver bromide is given by: $AgBr(s) \rightleftharpoons Ag^{+}(aq) + Br^{-}(aq)$.For precipitation to start,the ionic product

Vedclass · Accepted Answer

$1.2 	imes 10^{-9} \ g$

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(B) The dissociation of silver bromide is given by: $AgBr(s) ightleftharpoons Ag^{+}(aq) + Br^{-}(aq)$.For precipitation to start,the ionic product must exceed the solubility product $(K_{sp})$.$K_{sp} = [Ag^{+}][Br^{-}] = 5.0 	imes 10^{-13}$.Given $[Ag^{+}] = 0.05 \ M$,the concentration of $Br^{-}$ required is:$[Br^{-}] = \frac{K_{sp}}{[Ag^{+}]} = \frac{5.0 	imes 10^{-13}}{0.05} = 1.0 	imes 10^{-11} \ M$.Since the volume of the solution is $1 \ L$,the number of moles of $Br^{-}$ (or $KBr$) required is $1.0 	imes 10^{-11} \ mol$.The mass of $KBr$ required is: $	ext{Mass} = 	ext{moles} 	imes 	ext{molar mass} = 1.0 	imes 10^{-11} \ mol 	imes 120 \ g \ mol^{-1} = 1.2 	imes 10^{-9} \ g$.

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

The solubility product of silver bromide is $5.0 \times 10^{-13}$. The quantity of potassium bromide (molar mass taken as $120 \ g \ mol^{-1}$) to be added to $1 \ L$ of $0.05 \ M$ solution of silver nitrate to start the precipitation of $AgBr$ is

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