The moles of Ag^{+} which must be added to de | vedclass

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Question

(D) $AgCl_{(s)} \leftrightarrow Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$Initially: $[Cl^{-}] = 4 	imes 10^{-5} \ M$Final: $[Cl^{-}] = 10^{-5} \ M$Decrease in $

Vedclass · Accepted Answer

$4 	imes 10^{-6} \ mol$

Vedclass · Answer

(D) $AgCl_{(s)} \leftrightarrow Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$Initially: $[Cl^{-}] = 4 	imes 10^{-5} \ M$Final: $[Cl^{-}] = 10^{-5} \ M$Decrease in $[Cl^{-}] = 3 	imes 10^{-5} \ M$Moles of $Cl^{-}$ precipitated $= 3 	imes 10^{-5} \ M 	imes 0.1 \ L = 3 	imes 10^{-6} \ mol$Since $AgCl$ precipitates in $1:1$ ratio,moles of $Ag^{+}$ used for precipitation $= 3 	imes 10^{-6} \ mol$Remaining $Ag^{+}$ in solution to maintain equilibrium $= K_{sp} / [Cl^{-}]_{final} = 10^{-10} / 10^{-5} = 10^{-5} \ M$Moles of $Ag^{+}$ remaining in solution $= 10^{-5} \ M 	imes 0.1 \ L = 10^{-6} \ mol$Total moles of $Ag^{+}$ to be added $= 3 	imes 10^{-6} + 10^{-6} = 4 	imes 10^{-6} \ mol$

The moles of $Ag^{+}$ which must be added to decrease the concentration of $Cl^{-}$ from $4 \times 10^{-5} \ M$ to $10^{-5} \ M$ in $100 \ mL$ solution,if $K_{sp}$ for $AgCl$ is $10^{-10} \ M^2$ at $25 \ ^oC$.

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