Two solutions are created by mixing one solution containing lithium nitrate with one containing sodium phosphate.

Solution A is created by mixing 0.500 L of 0.30 M LiNO3 with 0.400 L of 0.20 M Na3PO4.
Solution B is created by mixing 0.400 L of 0.20 M LiNO3 with 0.500 L of 0.30 M Na3PO4.

In which of these solutions will lithium phosphate precipitate? Lithium phosphate has a Ksp = 2.3x10−4 Select the solution(s) that will form a precipitate.

Solution A that will form a precipitate with Ksp = 2.3 x 10−4

Explanation:

                                  Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)

                                                     3S               S

Where S = Solubility(mole/lit) and Ksp = Solubility product

⇒ Ksp = (3S)³ x (S)

⇒ 27S⁴ = 2.3x10−4

⇒ S = 0.05 mol/lit

Concentration of Li₃PO₄ precipitate = 0.05

Solution A

0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole

0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole

                                     3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄

(Mole/Stoichiometry)                    

                                   = 0.05            = 0.24

Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.

So concentration of Li₃PO₄ is equal to 0.05.

1 is the answer to this problem