Calculate the molar solubility of silver chloride in 0.15 M sodium chloride.

Ksp of silver chloride = 1.6x10-10.

S AgCl = 1.066 E-9 M

Explanation:

NaCl (s) → Na + (aq) + Cl - (aq)

0.15 M 0.15 M 0.15 M

AgCl (s) → Ag + (aq) + Cl - (aq)

S S S + 0.15

∴ Ksp AgCl = 1.6 E-10 = [Ag+]*[Cl-]

⇒ Ksp = 1.6 E-10 = (S) * (S + 0.15)

⇒ S² + 0.15S - 1.6 E-10 = 0

⇒ S = 1.066 E-9 M

S = 1.1 * 10⁻⁹ M

Explanation:

NaCl is a strong electrolyte that dissociates according to the following expression.

NaCl (aq) → Na⁺ (aq) + Cl⁻ (aq)

Given the concentration of NaCl is 0.15 M, the concentration of Cl⁻ will be 0.15 M.

We can find the molar solubility (S) of AgCl using an ICE chart.

AgCl (s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

I 0 0.15

C + S + S

E S 0.15+S

The solubility product (Ksp) is:

Ksp = 1.6 * 10⁻¹⁰ = [Ag⁺].[Cl⁻] = S (0.15 + S)

If we solve the quadratic equation, the positive result is S = 1.1 * 10⁻⁹ M