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9 March, 15:52

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

Ksp of silver chloride = 1.6x10-10.

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Answers (2)
  1. 9 March, 16:06
    0
    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
  2. 9 March, 16:21
    0
    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
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