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(A) When equal volumes are mixed,the concentration of each ion is halved. Let the initial concentrations be $C_1$ and $C_2$. The new concentration of

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

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(A) When equal volumes are mixed,the concentration of each ion is halved. Let the initial concentrations be $C_1$ and $C_2$. The new concentration of each ion becomes $C/2$.For precipitation to occur,the ionic product $Q_{sp}$ must be greater than $K_{sp}$ $(Q_{sp} > 1.8 	imes 10^{-10})$.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.

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

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