The solubility product of AgCl is 1.8 times 1 | vedclass

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(A) Precipitation occurs when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.Given $K_{sp} = 1.8 	imes 10^{-10}$.When equal v

Vedclass · Accepted Answer

$10^{-4} \ M \ Ag^{+}$ and $10^{-4} \ M \ Cl^{-}$

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(A) Precipitation occurs when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.Given $K_{sp} = 1.8 	imes 10^{-10}$.When equal volumes are mixed,the concentration of each ion is halved: $[Ag^{+}]_{new} = [Ag^{+}]_{initial} / 2$ and $[Cl^{-}]_{new} = [Cl^{-}]_{initial} / 2$.For option $A$: $[Ag^{+}] = 10^{-4} / 2 = 5 	imes 10^{-5} \ M$ and $[Cl^{-}] = 10^{-4} / 2 = 5 	imes 10^{-5} \ M$.$Q_{sp} = [Ag^{+}][Cl^{-}] = (5 	imes 10^{-5}) 	imes (5 	imes 10^{-5}) = 25 	imes 10^{-10} = 2.5 	imes 10^{-9}$.Since $2.5 	imes 10^{-9} > 1.8 	imes 10^{-10}$,precipitation will occur.

The solubility product of $AgCl$ is $1.8 \times 10^{-10}$. Mixing equal volumes of which of the following solutions will result in the precipitation of $AgCl$?

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