Answer:
1.34x10⁻⁵ mol / L is the solubility of AgCl
Step-by-step explanation:
Ksp of AgCl is defined as:
AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)
If 0.010M of AgCl is added, some Ag⁺ and Cl⁻ will be produced until:
Ksp = [Ag⁺] [Cl⁻]
Ksp for AgCl = 1.8x10⁻¹⁰ (Taken from ALEKS Data tab):
Some Ag⁺ and Cl⁻ are produced, you can take this "some" as X:
[Ag⁺] = X
[Cl⁻] = X
Where X is the amount of AgCl that dissolvesin water. X = solubility:
Ksp = 1.8x10⁻¹⁰ = [Ag⁺] [Cl⁻]
1.8x10⁻¹⁰ = [X] [X]
1.8x10⁻¹⁰ = X²
X =
1.34x10⁻⁵ mol / L is the solubility of AgCl