Question

You add 10.0 grams of solid copper(II) phosphate, Cu₃(PO₄)₂, to a beaker and then add 100.0 mL of water to the beaker at T = 298 K. The solid does not appear to dissolve. You wait a long time, with occasional stirring and eventually, with some solid Cu₃(PO₄)₂ remaining at the bottom of the beaker, you measured the equilibrium concentration of Cu²⁺ (aq) in the water to be 7.01 x 10⁻⁸ M. What is the Ksp of copper(II) phosphate?

a) 1.40 x 10⁻³⁷
b) 1.61 x 10⁻²⁹
c) 1.69 x 10⁻³⁶
d) 7.52 x 10⁻³⁷
e) 2.41 x 10⁻²⁹

Answer 1

The Ksp of copper(II) phosphate, given the equilibrium concentration of Cu²⁺ ions, is calculated using the solubility product constant expression and is found to be 1.40 × 10⁻37.

Step-by-step explanation:

The question asks for the calculation of the solubility product constant (Ksp) of copper(II) phosphate, Cu₃(PO₄)₂, given the equilibrium concentration of Cu²⁺ ions in a saturated solution. To calculate the Ksp for copper(II) phosphate, we will use the equilibrium reaction:

Cu₃(PO₄)₂(s) ⇌ 3Cu²⁺(aq) + 2PO₄³⁺(aq)

The equilibrium expression for the solubility product is:

Ksp = [Cu²⁺]³[PO₄³⁺]²

Since the measured equilibrium concentration of Cu²⁺ is 7.01 × 10⁻⁸ M, and the stoichiometry of the dissolution reaction indicates that there are three moles of Cu²⁺ for every one mole of Cu₃(PO₄)₂ that dissolves, and two moles of PO₄³⁺ for every mole of Cu₃(PO₄)₂, we can express the concentration of PO₄³⁺ in terms of the concentration of Cu²⁺:

[PO₄³⁺] = (2/3) x [Cu²⁺]

Plugging the values into the Ksp expression gives us:

Ksp = (7.01 × 10⁻⁸)³ x ((2/3) x (7.01 × 10⁻⁸))²

After calculating this expression, it results in:

Ksp = 1.40 × 10⁻37

Therefore, the Ksp of copper(II) phosphate is 1.40 × 10⁻37, which corresponds to option a).