Question

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

Answer 1

Answer:

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

Answer 2

Answer:

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