The K_{sp} of AgCl at 18, ^oC is 1.8 times 10 | vedclass

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Question

(A) $AgCl(s) ightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$The solubility product expression is $K_{sp} = [Ag^{+}][Cl^{-}]$.Given $K_{sp} = 1.8 	imes 10

Vedclass · Accepted Answer

$4.5 	imes 10^{-8}\ mol/L$

Vedclass · Answer

(A) $AgCl(s) ightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$The solubility product expression is $K_{sp} = [Ag^{+}][Cl^{-}]$.Given $K_{sp} = 1.8 	imes 10^{-10}$ and $[Ag^{+}] = 4 	imes 10^{-3}\ M$.For precipitation to occur,the ionic product must exceed the solubility product $(Q > K_{sp})$.$1.8 	imes 10^{-10} = (4 	imes 10^{-3}) 	imes [Cl^{-}]$$[Cl^{-}] = \frac{1.8 	imes 10^{-10}}{4 	imes 10^{-3}} = 0.45 	imes 10^{-7} = 4.5 	imes 10^{-8}\ M$.Therefore,the concentration of $Cl^{-}$ must be greater than $4.5 	imes 10^{-8}\ M$ for $AgCl$ to precipitate.

The $K_{sp}$ of $AgCl$ at $18\, ^oC$ is $1.8 \times 10^{-10}$. If the concentration of $Ag^{+}$ is $4 \times 10^{-3}\ mol/L$,what is the minimum concentration of $Cl^{-}$ required for $AgCl$ precipitation?

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