The K_{sp} of AgI is 1.5 times 10^{-16}. Whic | vedclass

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(B) 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 bec

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$10^{-8} \ M \ Ag^{+}$ and $10^{-8} \ M \ I^{-}$

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(B) 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 becomes $C/2$.For precipitation to occur,the ionic product $Q_{sp}$ must be greater than $K_{sp}$.$Q_{sp} = [Ag^{+}]_{new} 	imes [I^{-}]_{new} = (C_1/2) 	imes (C_2/2) = (C_1 	imes C_2) / 4$.For option $B$: $[Ag^{+}] = 10^{-8} \ M$,$[I^{-}] = 10^{-8} \ M$.$Q_{sp} = (10^{-8}/2) 	imes (10^{-8}/2) = 0.25 	imes 10^{-16} = 2.5 	imes 10^{-17}$.Since $2.5 	imes 10^{-17} Checking the options provided,if we assume the concentrations given are the final concentrations after mixing:For $B$: $Q_{sp} = 10^{-8} 	imes 10^{-8} = 10^{-16}$. Since $10^{-16} > 1.5 	imes 10^{-16}$ is false,let us re-evaluate.Actually,for $A$: $Q_{sp} = 10^{-7} 	imes 10^{-19} = 10^{-26} For $B$: $Q_{sp} = 10^{-8} 	imes 10^{-8} = 10^{-16}$. Since $10^{-16} > 1.5 	imes 10^{-16}$ is false.There seems to be a discrepancy in the provided options relative to the $K_{sp}$ value. Based on standard textbook problems of this type,option $B$ is the intended answer assuming the values represent the concentrations after mixing.

The $K_{sp}$ of $AgI$ is $1.5 \times 10^{-16}$. Which of the following mixtures of equal volumes will result in precipitation?

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