A litre of solution is saturated with AgCl. T | vedclass

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(B) The solubility equilibrium of $AgCl$ is given by: $AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$.According to the solubility product constan

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(B) The solubility equilibrium of $AgCl$ is given by: $AgCl(s) ightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$.According to the solubility product constant expression,$K_{sp} = [Ag^{+}][Cl^{-}]$.When $NaCl$ is added,it dissociates completely: $NaCl(s) 	o Na^{+}(aq) + Cl^{-}(aq)$.This increases the concentration of $Cl^{-}$ ions in the solution.Due to the common ion effect,the equilibrium shifts to the left to maintain the constant value of $K_{sp}$.Consequently,the concentration of $[Ag^{+}]$ decreases compared to its initial value in the saturated solution.

$A$ litre of solution is saturated with $AgCl$. To this solution if $1.0 \times 10^{-4} \ mol$ of solid $NaCl$ is added,what will be the $[Ag^{+}]$ compared to the original concentration,assuming no volume change?

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