Question

The solubility of ag₃po₄ in water at 25 °c is 4.3 × 10⁻⁵ m. what is ksp for ag₃po₄?

Answer 1

The solubility product constant (Ksp) for Ag₃PO₄ can be calculated using the given molar solubility and the stoichiometry of its dissolution reaction. Once the ion concentrations are found, they are substituted into the Ksp expression to obtain the value.

Step-by-step explanation:

To calculate the solubility product constant (Ksp) for Ag₃PO₄, we use the given molar solubility and the stoichiometry of the dissolution reaction.

Ag₃PO₄ (s) ↔ 3Ag⁺ (aq) + PO₄³⁻ (aq)

If the solubility of Ag₃PO₄ is 4.3 × 10⁻⁵ M, then the concentration of Ag⁺ ions will be 3 times the solubility because there are three silver ions per formula unit of Ag₃PO₄. The concentration of PO₄³⁻ ions will be equal to the solubility.

[Ag⁺] = 3 × 4.3 × 10⁻⁵ M = 1.29 × 10⁻⁴ M
[PO₄³⁻] = 4.3 × 10⁻⁵ M

Using these concentrations, we can write the expression for Ksp:

Ksp = [Ag⁺]³[PO₄³⁻] = (1.29 × 10⁻⁴)³ × (4.3 × 10⁻⁵)

After calculating the powers and multiplying, we obtain the Ksp value.

Answer 2

The solubility product constant (Ksp) for Ag3PO4 at 25 °C is calculated using the solubility and the dissociation equation, resulting in a value of 2.54 × 10−12.

Step-by-step explanation:

To find the Ksp (solubility product constant) for Ag3PO4, we need to take into account the molar ratios in which the ions dissociate in water. The dissociation of Ag3PO4 in water can be represented as:

Ag3PO4 (s) → 3Ag+ (aq) + PO43- (aq)

Given the solubility of Ag3PO4 is 4.3 × 10−5 M, we can establish the following equilibrium concentrations at solubility:

• [Ag+] = 3 × 4.3 × 10−5 M = 1.29 × 10−4 M
• [PO43-] = 4.3 × 10−5 M

Substituting these values into the Ksp expression:
Ksp = [Ag+]3 × [PO43-] = (1.29 × 10−4 M)3 × 4.3 × 10−5 M = 2.54 × 10−12

Thus, the Ksp for Ag3PO4 at 25 °C is 2.54 × 10−12.