Respuesta :
Answer: The minimum [tex][Ag^{+}][/tex] concentrations required to precipitate out the anions is [tex]9 \times 10^{-9}[/tex] M.
Explanation:
We know that,
[tex]K_{sp}[/tex] for AgCl is [tex]1.8 \times 10^{-10}[/tex]
and, [tex]K_{sp}[/tex] for [tex]Ag_{2}CrO_{4}[/tex] is [tex]9 \times 10^{-12}[/tex]
Now, we will calculate the concentration of at which these ions precipitate out are as follows.
For AgCl :
[tex][Ag^{+}] = \frac{K_{sp}}{[Cl^{-}]}[/tex]
= [tex]\frac{1.8 \times 10^{-10}}{0.02}[/tex]
= [tex]9 \times 10^{-9}[/tex] M
For [tex]Ag_{2}CrO_{4}[/tex] :
[tex][Ag^{+}]^{2} = \frac{K_{sp}}{CrO^{2-}_{4}}[/tex]
= [tex]\frac{9 \times 10^{-12}}{0.01}[/tex]
= [tex]9 \times 10^{-10}[/tex]
[tex][Ag^{+}] = \sqrt{(9 \times 10^{-9})}[/tex]
= [tex]3 \times 10^{-5}[/tex] M
This shows that concentration of ions in AgCl is less than the concentration of AgCl will precipitate first.
