- Home
- Standard 11
- Chemistry
When equal volume of following solutions are mixed, precipitation of $AgCl$ will occur in -$(k_{sp}$ of $AgCl = 1.8\times10^{-10})$
$10^{-4}\, M\, Ag^+$ and $10^{-4}\, M\, Cl^-$
$10^{-5}\, M\, Ag^+$ and $10^{-5}\, M\, Cl^-$
$10^{-6}\, M\, Ag^+$ and $10^{-6}\, M\, Cl^-$
$10^{-10}\, M\, Ag^+$ and $10^{-10}\, M\, Cl^-$
Solution
Precipitation of $AgCl$ will occur only when ionic product exceeds the solubility product.
$a)$ The ionic product of the solution obtained by mixing $10^{-4}\, M\, Ag ^{+}$ and $10^{-4} \,M\, Cl ^{-}$ is $10^{-8}\, M ^2$
$b)$ The ionic product of the solution obtained by mixing $10^{-5} \,M\, Ag ^{+}$ and $10^{-5} \,M \,Cl ^{-}$ is $10^{-10} \,M ^2$
$C)$ The ionic product of the solution obtained by mixing $10^{-6}\, M \,Ag ^{+}$ and $10^{-6}\, M\, Cl ^{-}$ is $10^{-12} \,M ^2$
$d)$ The ionic product of the solution obtained by mixing $10^{-10} \,M\, Ag ^{+}$ and $10^{-10}\, M \,Cl ^{-}$ is $10^{-20}\, M ^2$
Since $10^{-8} \,M ^2\,>\,1.8 \times 10^{-10}$, hence option $(a)$ is correct.
