182k views
2 votes
Calculate the solubility (mol/L) of silver arsenate (Ag3AsO4). (Ksp = 1.0 x 10–22).

User Pym
by
7.9k points

1 Answer

0 votes

Final answer:

The solubility of silver arsenate (Ag3AsO4) is found by solving the Ksp expression, which yields a molar solubility of 1.44 x 10–6 mol/L at 25°C.

Step-by-step explanation:

To calculate the solubility of silver arsenate (Ag3AsO4), we need to set up an equilibrium expression using its solubility product constant (Ksp).

The dissolution reaction of Ag3AsO4 in water can be written as:

Ag3AsO4(s) ⇌ 3Ag+(aq) + AsO43-(aq)

Let s represent the molar solubility of Ag3AsO4. This means [Ag+] = 3s and [AsO43-] = s at equilibrium. The Ksp expression is written as:

Ksp = [Ag+]3[AsO43-] = (3s)3(s) = 27s4

Given the Ksp is 1.0 x 10–22, we set up the equation:

1.0 x 10–22 = 27s4

Solving for s, we find:

s = (1.0 x 10–22/27)1/4

s ≈ 1.44 x 10–6 mol/L

Therefore, the solubility of silver arsenate in water at 25°C is 1.44 x 10–6 mol/L.

User Amfasis
by
8.5k points