A precipitate of AgCl is formed when equal vo | vedclass

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Question

(C) For the precipitation of $AgCl$ to occur,the ionic product $(Q_{sp})$ must exceed the solubility product $(K_{sp} = 10^{-10})$.When equal volumes

Vedclass · Accepted Answer

$10^{-5} \ M \ AgNO_3$ and $10^{-4} \ M \ HCl$

Vedclass · Answer

(C) For the precipitation of $AgCl$ to occur,the ionic product $(Q_{sp})$ must exceed the solubility product $(K_{sp} = 10^{-10})$.When equal volumes are mixed,the concentration of each ion is halved.For option $(C)$: $[Ag^+] = \frac{10^{-5}}{2} \ M$ and $[Cl^-] = \frac{10^{-4}}{2} \ M$.$Q_{sp} = [Ag^+][Cl^-] = (\frac{10^{-5}}{2}) 	imes (\frac{10^{-4}}{2}) = \frac{10^{-9}}{4} = 0.25 	imes 10^{-9} = 2.5 	imes 10^{-10}$.Since $2.5 	imes 10^{-10} > 10^{-10}$,the ionic product exceeds $K_{sp}$,and precipitation occurs.

$A$ precipitate of $AgCl$ is formed when equal volumes of the following are mixed. [$K_{sp}$ for $AgCl = 10^{-10}$]

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