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Question: At a certain temperature the saturated solution of AgCl has conductivity 1.8 × 10⁻⁶ S cm⁻¹. The ioni...

At a certain temperature the saturated solution of AgCl has conductivity 1.8 × 10⁻⁶ S cm⁻¹. The ionic conductance of Ag⁺ and Cl⁻ at infinite dilution are 54.5 and 65.5 S cm² mol⁻¹. If the solubility product of AgCl at this temperature is 2.25 × 10⁻ʸ, then value of y is ________.

Answer

10

Explanation

Solution

To determine the value of 'y' in the solubility product expression for AgCl, we follow these steps:

  1. Calculate the molar conductivity at infinite dilution for AgCl (Λm,AgCl0\Lambda^0_{m, AgCl}): Using Kohlrausch's Law of independent migration of ions: Λm,AgCl0=λAg+0+λCl−0\Lambda^0_{m, AgCl} = \lambda^0_{Ag+} + \lambda^0_{Cl-} Given: λAg+0=54.5 S cm² mol⁻¹\lambda^0_{Ag+} = 54.5 \text{ S cm² mol⁻¹} λCl−0=65.5 S cm² mol⁻¹\lambda^0_{Cl-} = 65.5 \text{ S cm² mol⁻¹} Λm,AgCl0=54.5+65.5=120.0 S cm² mol⁻¹\Lambda^0_{m, AgCl} = 54.5 + 65.5 = 120.0 \text{ S cm² mol⁻¹}

  2. Calculate the solubility (S) of AgCl: For a sparingly soluble salt like AgCl, the molar conductivity of its saturated solution (Λm\Lambda_m) is approximately equal to its molar conductivity at infinite dilution (Λm0\Lambda^0_m). The relationship between molar conductivity, specific conductivity (κ\kappa), and solubility (S, in mol L⁻¹) is: Λm=κ×1000S\Lambda_m = \frac{\kappa \times 1000}{S} Rearranging for S: S=κ×1000Λm0S = \frac{\kappa \times 1000}{\Lambda^0_m} Given: κ=1.8×10−6 S cm⁻¹\kappa = 1.8 \times 10^{-6} \text{ S cm⁻¹} S=1.8×10−6 S cm⁻¹×1000 cm³ L⁻¹120.0 S cm² mol⁻¹S = \frac{1.8 \times 10^{-6} \text{ S cm⁻¹} \times 1000 \text{ cm³ L⁻¹}}{120.0 \text{ S cm² mol⁻¹}} S=1.8×10−3120.0 mol L⁻¹S = \frac{1.8 \times 10^{-3}}{120.0} \text{ mol L⁻¹} S=0.015×10−3 mol L⁻¹S = 0.015 \times 10^{-3} \text{ mol L⁻¹} S=1.5×10−5 mol L⁻¹S = 1.5 \times 10^{-5} \text{ mol L⁻¹}

  3. Calculate the solubility product (KspK_{sp}) of AgCl: For the dissolution of AgCl: AgCl(s) ⇌\rightleftharpoons Ag⁺(aq) + Cl⁻(aq) If S is the solubility, then [Ag+]=S[Ag⁺] = S and [Cl−]=S[Cl⁻] = S. Ksp=[Ag+][Cl−]=S×S=S2K_{sp} = [Ag⁺][Cl⁻] = S \times S = S^2 Ksp=(1.5×10−5)2K_{sp} = (1.5 \times 10^{-5})^2 Ksp=(1.5)2×(10−5)2K_{sp} = (1.5)^2 \times (10^{-5})^2 Ksp=2.25×10−10K_{sp} = 2.25 \times 10^{-10}

  4. Determine the value of y: The problem states that the solubility product of AgCl is 2.25×10−y2.25 \times 10^{-y}. Comparing our calculated value (Ksp=2.25×10−10K_{sp} = 2.25 \times 10^{-10}) with the given expression, we find: y=10y = 10