Observe the following solutions:(i) 1 L of 1 | vedclass

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(C) precipitate of $AgCl$ forms when the ionic product $[Ag^{+}][Cl^{-}]$ exceeds the solubility product constant $K_{sp} = 1 	imes 10^{-10} \ M^2$.W

Vedclass · Accepted Answer

$iv, v$

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(C) precipitate of $AgCl$ forms when the ionic product $[Ag^{+}][Cl^{-}]$ exceeds the solubility product constant $K_{sp} = 1 	imes 10^{-10} \ M^2$.When mixing $1 \ L$ of each solution,the final volume becomes $2 \ L$,so the concentrations are halved.For $(iv)$ and $(v)$:$[Ag^{+}] = \frac{10^{-3} \ M}{2} = 0.5 	imes 10^{-3} \ M$$[Cl^{-}] = \frac{10^{-5} \ M}{2} = 0.5 	imes 10^{-5} \ M$Ionic product = $(0.5 	imes 10^{-3}) 	imes (0.5 	imes 10^{-5}) = 0.25 	imes 10^{-8} = 2.5 	imes 10^{-9}$.Since $2.5 	imes 10^{-9} > 1 	imes 10^{-10}$,a precipitate will form.

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