Question

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would precipitate out of solution first?

A. Na3PO4.
B. Ag3PO4.
C. Ca3(PO4)2
When the second cation just starts to precipitate, what percentage of the first cation remains in solution?

Answer 1

C.
`Ca_3(PO_4)_2` will precipitate out first

the percentage of
`Ca^(2+)`remaining = 12.86%

Step-by-step explanation:

Given that:

A solution contains:

`[Ca^(2+)] = 0.0440 \ M`

`[Ag^+] = 0.0940 \ M`

From the list of options , Let find the dissociation of
`Ag_3PO_4`

`Ag_3PO_4 \to Ag^(3+) + PO_4^(3-)`

where;

Solubility product constant Ksp of
`Ag_3PO_4` is
`8.89 * 10^(-17)`

Thus;

`Ksp = [Ag^+]^3[PO_4^(3-)]`

replacing the known values in order to determine the unknown ; we have :

`8.89 * 10 ^(-17) = (0.0940)^3[PO_4^(3-)]`

`(8.89 * 10 ^(-17))/((0.0940)^3) = [PO_4^(3-)]`

`[PO_4^(3-)] =(8.89 * 10 ^(-17))/((0.0940)^3)`

`[PO_4^(3-)] =1.07 * 10^(-13)`

The dissociation of
`Ca_3(PO_4)_2`

The solubility product constant of
`Ca_3(PO_4)_2` is
`2.07 * 10^(-32)`

The dissociation of
`Ca_3(PO_4)_2` is :

`Ca_3(PO_4)_2 \to 3Ca^(2+) + 2 PO_(4)^(3-)`

Thus;

`Ksp = [Ca^(2+)]^3 [PO_4^(3-)]^2`

`2.07 * 10^(-33) = (0.0440)^3 [PO_4^(3-)]^2`

`(2.07 * 10^(-33) )/((0.0440)^3)= [PO_4^(3-)]^2`

`[PO_4^(3-)]^2 = (2.07 * 10^(-33) )/((0.0440)^3)`

`[PO_4^(3-)]^2 = 2.43 * 10^(-29)`

`[PO_4^(3-)] = \sqrt{2.43 * 10^(-29)`

`[PO_4^(3-)] =4.93 * 10^(-15)`

Thus; the phosphate anion needed for precipitation is smaller i.e
`4.93 * 10^(-15)` in
`Ca_3(PO_4)_2` than in
`Ag_3PO_4`
`1.07 * 10^(-13)`

Therefore:

`Ca_3(PO_4)_2` will precipitate out first

To determine the concentration of
`[Ca^+]` when the second cation starts to precipitate ; we have :

`Ksp = [Ca^(2+)]^3 [PO_4^(3-)]^2`

`2.07 * 10^(-33) = [Ca^(2+)]^3 (1.07 * 10^(-13))^2`

`[Ca^(2+)]^3 = (2.07 * 10^(-33) )/((1.07 * 10^(-13))^2)`

`[Ca^(2+)]^3 =1.808 * 10^(-7)`

`[Ca^(2+)] =\sqrt[3]{1.808 * 10^(-7)}`

`[Ca^(2+)] =0.00566`

This implies that when the second cation starts to precipitate ; the concentration of
`[Ca^(2+)]` in the solution is 0.00566

Therefore;

the percentage of
`Ca^(2+)` remaining = concentration remaining/initial concentration × 100%

the percentage of
`Ca^(2+)` remaining = 0.00566/0.0440 × 100%

the percentage of
`Ca^(2+)` remaining = 0.1286 × 100%

the percentage of
`Ca^(2+)`remaining = 12.86%