The solubility product (K_{sp}) of the follow | vedclass

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(D) To determine the most and least soluble compounds,we calculate the molar solubility $(S)$ for each:$1$. For $AgCl$ ($1:1$ type): $K_{sp} = S^2 \Ri

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$Ag_2CO_3$ and $AgI$

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(D) To determine the most and least soluble compounds,we calculate the molar solubility $(S)$ for each:$1$. For $AgCl$ ($1:1$ type): $K_{sp} = S^2 \Rightarrow S = \sqrt{1.1 	imes 10^{-10}} \approx 1.05 	imes 10^{-5} \, M$$2$. For $AgI$ ($1:1$ type): $K_{sp} = S^2 \Rightarrow S = \sqrt{1.0 	imes 10^{-16}} = 1.0 	imes 10^{-8} \, M$$3$. For $PbCrO_4$ ($1:1$ type): $K_{sp} = S^2 \Rightarrow S = \sqrt{4.0 	imes 10^{-14}} = 2.0 	imes 10^{-7} \, M$$4$. For $Ag_2CO_3$ ($2:1$ type): $K_{sp} = 4S^3$ $\Rightarrow S = \sqrt[3]{K_{sp}/4} = \sqrt[3]{8.0 	imes 10^{-12} / 4} = \sqrt[3]{2.0 	imes 10^{-12}} \approx 1.26 	imes 10^{-4} \, M$Comparing the solubilities: $1.26 	imes 10^{-4} > 1.05 	imes 10^{-5} > 2.0 	imes 10^{-7} > 1.0 	imes 10^{-8}$.Thus,$Ag_2CO_3$ is the most soluble and $AgI$ is the least soluble.

Compound	$K_{sp}$
$AgCl$	$1.1 \times 10^{-10}$
$AgI$	$1.0 \times 10^{-16}$
$PbCrO_4$	$4.0 \times 10^{-14}$
$Ag_2CO_3$	$8.0 \times 10^{-12}$

The solubility product $(K_{sp})$ of the following compounds are given at $25\,\text{°C}$:

Compound $K_{sp}$

$AgCl$ $1.1 \times 10^{-10}$

$AgI$ $1.0 \times 10^{-16}$

$PbCrO_4$ $4.0 \times 10^{-14}$

$Ag_2CO_3$ $8.0 \times 10^{-12}$

The most soluble and least soluble compounds are respectively:

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The solubility product $(K_{sp})$ of the following compounds are given at $25\,\text{°C}$: Compound $K_{sp}$ $AgCl$ $1.1 \times 10^{-10}$ $AgI$ $1.0 \times 10^{-16}$ $PbCrO_4$ $4.0 \times 10^{-14}$ $Ag_2CO_3$ $8.0 \times 10^{-12}$ The most soluble and least soluble compounds are respectively: