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(A) Precipitation of $AgCl$ occurs when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.When equal volumes are mixed,the concen

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Precipitation of $AgCl$ occurs when the ionic product $(Q_{sp})$ exceeds the solubility product $(K_{sp})$.When equal volumes are mixed,the concentration of each ion is halved.$a)$ $Q_{sp} = (10^{-4}/2) 	imes (10^{-4}/2) = 0.25 	imes 10^{-8} = 2.5 	imes 10^{-9} \ M^2$. Since $2.5 	imes 10^{-9} > 1.8 	imes 10^{-10}$,precipitation occurs.$b)$ $Q_{sp} = (10^{-5}/2) 	imes (10^{-5}/2) = 2.5 	imes 10^{-11} \ M^2$. Since $2.5 	imes 10^{-11} $c)$ $Q_{sp} = 2.5 	imes 10^{-13} \ M^2 $d)$ $Q_{sp} = 2.5 	imes 10^{-21} \ M^2 Therefore,option $(A)$ is correct.

When equal volumes of the following solutions are mixed,precipitation of $AgCl$ will occur in - ($K_{sp}$ of $AgCl = 1.8 \times 10^{-10}$)

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