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

a. Solution A is created by mixing 0.500 L of 0.30 M LiNO3 with 0.400 L of 0.20 M Na3PO4.
b. Solution B is created by mixing 0.400 L of 0.20 M LiNO3 with 0.500 L of 0.30 M Na3PO4.
c. In which of these solutions will lithium phosphate precipitate? Lithium phosphate has a Ksp = 2.3x10−4
d. Select the solution(s) that will form a precipitate.

Solution A

Explanation:

For this question we have to compare Qsp, the solubility product reaction quotient, with Ksp.

If Qsp > Ksp we know a precipitate of Li₃PO₄ will form.

Therefore, what we need to do is calculate the concentrations of Li⁺ and PO₄³⁻ from the data given.

Q which has the same form as Ksp

Q = [Li⁺]³ x [PO₄³⁻]

a. [Li⁺] = mol Li⁺ / Vol soln (L)

mol Li⁺ = 0.500 L x 0.30 mol/L = 0.15 mol

Vol solution = 0.500 L + 0.400 L = 0.900 L

[Li⁺] = 0.15 mol / 0.900 L = 0.167 M

mol PO₄³⁻ = 0.400 L x 0.20 mol/L = 0.08 mol

[PO₄³⁻] = 0.08 mol / 0.900 L = 0.089 M

Q = (0.167)³ x 0.089 = 4.1 x 10⁻⁴

Q > Ksp ( 4.1 x 10⁻⁴ > 2.3 x 10⁻⁴ )  will precipitate some Li₃PO₄

b. mol Li⁺ = 0.400 L x 0.20mol Li⁺/L = 0.08 mol

[Li⁺] = 0.08 mol / 0.9 L = 8.9 x 10⁻² M

mol PO₄³⁻ = 0.500 L x 0.30 mol/L = 0.15 mol

[PO₄³⁻] = 0.15 L / 0.9 L = 0.17 M

Q = (8.9 x 10⁻²)³ x 0.15 = 1.1 x 10⁻⁴

Q < Ksp it will not precipitate

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we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).