Question

8. A saturated solution of Ag Croq has a silver-ion concentration of 1.3 x 10-4M. Which is the Ksp of Ag CrO 4?

O 1.3 x 10-4
O 3.9 x 10-12
O 6.5 x 10-5
O 1.1 x 10-12

Answer 1

Answer: The
`K_(sp)` of
`AgCrO_(4)` is
`1.1 * 10^(-12)`.

Step-by-step explanation:

Given:
`[Ag^(+)] = 1.3 * 10^(-4) M`

The reaction equation will be written as follows.

`Ag_(2)CrO_(4) \rightleftharpoons 2Ag^(+) + CrO^(2-)_(4)`

This shows that the concentration of
`CrO^(2-)_(4)` is half the concentration of
`Ag^(+)` ion. So,

`[CrO^(2-)_(4)] = (1.3 * 10^(-4))/(2)\\= 0.65 * 10^(-4) M`

The expression for
`K_(sp)` of this reaction is as follows.

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

Substitute values into the above expression as follows.

`K_(sp) = [Ag^(+)]^(2)[CrO^(2-)_(4)]\\= (1.3 * 10^(-4))^(2) * 0.65 * 10^(-4)\\= 1.1 * 10^(-12)`

Thus, we can conclude that the
`K_(sp)` of
`AgCrO_(4)` is
`1.1 * 10^(-12)`.