In 1 L saturated solution of AgCl [K_{sp}(Ag | vedclass

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(A) This is a case of simultaneous solubility of salts with a common ion $(Cl^{-})$.Since $K_{sp}(CuCl) \gg K_{sp}(AgCl)$,the concentration of $Cl^{-}

Vedclass · Accepted Answer

$7$

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(A) This is a case of simultaneous solubility of salts with a common ion $(Cl^{-})$.Since $K_{sp}(CuCl) \gg K_{sp}(AgCl)$,the concentration of $Cl^{-}$ ions in the solution is primarily determined by the dissolution of $CuCl$.For $CuCl ightleftharpoons Cu^{+} + Cl^{-}$,let $s$ be the solubility.$K_{sp}(CuCl) = s^2 = 1.0 	imes 10^{-6} \Rightarrow s = 10^{-3} \ M$.Thus,$[Cl^{-}] \approx 10^{-3} \ M$.Now,for $AgCl ightleftharpoons Ag^{+} + Cl^{-}$,the solubility product expression is:$K_{sp}(AgCl) = [Ag^{+}][Cl^{-}] = 1.6 	imes 10^{-10}$.Substituting the value of $[Cl^{-}]$:$[Ag^{+}] 	imes 10^{-3} = 1.6 	imes 10^{-10}$.$[Ag^{+}] = 1.6 	imes 10^{-7} \ M$.Comparing this with $1.6 	imes 10^{-x}$,we get $x = 7$.

In $1 \ L$ saturated solution of $AgCl$ $[K_{sp}(AgCl)=1.6 \times 10^{-10}]$,$0.1 \ mol$ of $CuCl$ $[K_{sp}(CuCl)=1.0 \times 10^{-6}]$ is added. The resultant concentration of $Ag^{+}$ in the solution is $1.6 \times 10^{-x}$. The value of $x$ is

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