What is the solubility of silver chloride (AgCl) when calculated to 2 decimal places in molarity?

To find the solubility of silver chloride (AgCl) in molarity, we start by recognizing that AgCl is a sparingly soluble salt. The solubility product constant (Ksp) expression reflects this:

Ksp = [Ag^+][Cl^-]

For silver chloride, Ksp is known to be approximately 1.77 x 10^-10 at room temperature. When AgCl dissolves in water, it dissociates into Ag^+ and Cl^− ions in a 1:1 ratio:

AgCl (s) ⇌ Ag^+ (aq) + Cl^- (aq)

Let the solubility of AgCl be 's' in molarity. Thus, at equilibrium, the concentrations of the ions would be:

[Ag^+] = s

[Cl^-] = s

Substituting these into the Ksp expression gives us:

Ksp = s * s = s²

This allows us to set up the equation:

s² = 1.77 x 10^-10

To find 's', we take the square root of both sides:

s = √(1.77 x 10^-10)

s ≈ 1.33 x