Question

Calculate the solubility of silver phosphate, Ag3PO4, in pure water. Ksp = 2.6 x 10-18 O 1.5 x 10-5 M O 4.0 x 10-5 M O 4.0 x 10-6 M O 1.8 x 10-5 M O < 1.0 x 10-5M

Answer 1

The solubility of silver phosphate, Ag3PO4, in pure water is approximately 1.6 x 10^-6 M.

Step-by-step explanation:

The solubility of silver phosphate, Ag3PO4, in pure water can be calculated using the solubility product constant (Ksp). The Ksp for Ag3PO4 is given as 2.6 x 10^-18. The solubility of Ag3PO4 can be determined by finding the equilibrium concentration of the silver ion (Ag+) and the phosphate ion (PO4^3-).

To calculate the solubility, we can assume x mol/L to represent the concentration of Ag3PO4. Since the stoichiometry of Ag3PO4 is 1:1, both Ag+ and PO4^3- will have a concentration of x mol/L.

Thus, the solubility of Ag3PO4 in pure water is approximately 1.6 x 10^-6 M.

Answer 2

The solubility of silver phosphate, Ag3PO4, in pure water is approximately 1.6 × 10⁻¹⁹M.

Step-by-step explanation:

The solubility of silver phosphate, Ag3PO4, can be calculated using the solubility product constant (Ksp). The Ksp for Ag₃PO₄is given as 2.6 × 10⁻¹⁸. To find the solubility, let's assume x moles of Ag3PO4 dissolve in pure water. Since the stoichiometry of the reaction is 1:1, the concentrations of Ag+ and PO³₄ ions in solution will also be x M. Therefore, the Ksp expression for Ag3PO4 is [Ag+][PO4⁻³] = (x)(x) = x² Equating this expression to the given Ksp value and solving for x gives us the solubility of Ag3PO4 as approximately 1.6 × 10⁻⁹ M.