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Calculate the solubility product constant (K_sp) for silver chloride (AgCl) given its molar solubility in water is 1.30 imes 10⁻⁵ mol dm⁻³.

Chemistry

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C. 1.69×10−10 mol2 dm−61.69 \times 10^{-10} \text{ mol}_2 \text{ dm}^{-6}

  1. Write the dissociation equation for silver chloride (AgCl).

    Silver chloride is a sparingly soluble ionic compound that dissociates in water as follows:

    AgCl(s)⇌Ag+(aq)+Cl−(aq)\text{AgCl(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{Cl}^-\text{(aq)}

  2. Relate the concentration of the saturated solution to the solubility (ss).

    The concentration of a saturated solution of AgCl is given as 1.30×10−5 mol dm−31.30 \times 10^{-5} \text{ mol dm}^{-3}. This value represents the molar solubility (ss) of AgCl.
    From the dissociation equation, for every mole of AgCl that dissolves, one mole of Ag+\text{Ag}^+ ions and one mole of Cl−\text{Cl}^- ions are produced.
    Therefore, the concentration of Ag+\text{Ag}^+ ions is [Ag+]=s=1.30×10−5 mol dm−3[\text{Ag}^+] = s = 1.30 \times 10^{-5} \text{ mol dm}^{-3}.
    And the concentration of Cl−\text{Cl}^- ions is [Cl−]=s=1.30×10−5 mol dm−3[\text{Cl}^-] = s = 1.30 \times 10^{-5} \text{ mol dm}^{-3}.

  3. Write the expression for the solubility product constant (KspK_{sp}).

    For the dissociation of AgCl, the solubility product expression is:

    Ksp=[Ag+][Cl−]K_{sp} = [\text{Ag}^+][\text{Cl}^-]

  4. Substitute the solubility (ss) into the KspK_{sp} expression and calculate the value.

    Ksp=(s)(s)=s2K_{sp} = (s)(s) = s^2 Ksp=(1.30×10−5 mol dm−3)2K_{sp} = (1.30 \times 10^{-5} \text{ mol dm}^{-3})^2 Ksp=(1.30)2×(10−5)2 mol2 dm−6K_{sp} = (1.30)^2 \times (10^{-5})^2 \text{ mol}^2 \text{ dm}^{-6} Ksp=1.69×10−10 mol2 dm−6K_{sp} = 1.69 \times 10^{-10} \text{ mol}^2 \text{ dm}^{-6}

    The calculated solubility product matches option C.

    The final answer is \boxed{\text{1.69 \times 10^{-10} mol}_2 \text{ dm}^{-6}}.

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