Question

AgNO3 is added to a solution containing Cl- and CrO42- in order to separate the ions. If the Cl- and CrO42- concentrations are 0.020 and 0.010 M, respectively, what are the minimum Ag+ concentrations required to precipitate out the anions?

Answer 1

Answer: The minimum
`[Ag^(+)]` concentrations required to precipitate out the anions is
`9 * 10^(-9)` M.

Step-by-step explanation:

We know that,

`K_(sp)` for AgCl is
`1.8 * 10^(-10)`

and,
`K_(sp)` for
`Ag_(2)CrO_(4)` is
`9 * 10^(-12)`

Now, we will calculate the concentration of at which these ions precipitate out are as follows.

For AgCl :

`[Ag^(+)] = (K_(sp))/([Cl^(-)])`

=
`(1.8 * 10^(-10))/(0.02)`

=
`9 * 10^(-9)` M

For
`Ag_(2)CrO_(4)` :

`[Ag^(+)]^(2) = (K_(sp))/(CrO^(2-)_(4))`

=
`(9 * 10^(-12))/(0.01)`

=
`9 * 10^(-10)`

`[Ag^(+)] = \sqrt{(9 * 10^(-9))}`

=
`3 * 10^(-5)` M

This shows that concentration of ions in AgCl is less than the concentration of AgCl will precipitate first.