Question

A basic solution contains the iodide and phosphate ions that are to be separated via selective precipitation. the i– concentration, which is 9.00×10-5 m, is 10,000 times less than that of the po43– ion at 0.900 m . a solution containing the silver(i) ion is slowly added. answer the questions below. ksp of agi is 8.30×10-17 and of ag3po4, 8.90×10-17.

Answer 1

The given question is incomplete. The complete question is as follows.

A basic solution contains the iodide and phosphate ions that are to be separated via selective precipitation. the i– concentration, which is 9.00×10-5 m, is 10,000 times less than that of the
`PO^(3-)_(4)` ion at 0.900 m . a solution containing the silver(i) ion is slowly added. answer the questions below. ksp of agi is 8.30×10-17 and of
`Ag_(3)PO_(4)`,
`8.90 * 10^(-17)`.

Calculate the minimum
`Ag^(+)` concentration required to cause precipitation of AgI.

Step-by-step explanation:

It is known that at the stage of precipitation, the
`K_(sp)` is equal to the ionic product.

Therefore, expression for
`K_(sp)` for
`Q_(1)` is as follows.

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

In the given case,

`[I^(-)] = 7.7 * 10^(-5)`

`K_(sp) = 8.3 * 10^(-17)`

Hence,
`[Ag^(+)] = (k_(sp))/(I^(-))`

=
`(8.3 * 10^(-17))/(9 * 10^(-5))`

=
`0.92 * 10^(-12) M`

Now, for
`Q_(2)` the expression for
`K_(sp)` will be as follows.

`K_(sp) = [Ag^(+)]^(3)[PO^(3-)_(4)]`

or,
`[Ag^(+)] = ((K_(sp))/([PO^(3-)_(4)]))^{(1)/(3)}`

=
`((8.90 * 10^(-17))/(0.9 * 10^(-5)))^{(1)/(3)}`

=
`9.72 * 10^(-5)` M