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Question: The specific conductivity of a saturated solution of silver oxalate is 3.8 × 10⁻⁵ Ω⁻¹ cm⁻¹ at 25 °C....

The specific conductivity of a saturated solution of silver oxalate is 3.8 × 10⁻⁵ Ω⁻¹ cm⁻¹ at 25 °C. Conductivity of water is 6.2 × 10⁻⁸ Ω⁻¹ cm⁻¹. Calculate molar conductivity of oxalate ion given molar conductivity of Ag⁺ at infinite dilution is 62 Ω⁻¹ cm² mole⁻¹ and Kₛₚ for Ag₂C₂O₄ at 25 °C is 1.2 × 10⁻¹¹.

A

139

B

263

C

379

D

178

Answer

139

Explanation

Solution

Explanation of the solution:

  1. Calculate the specific conductivity of silver oxalate (κ): The specific conductivity of the saturated solution includes the conductivity of water. To find the specific conductivity due to silver oxalate alone, subtract the conductivity of water from the total specific conductivity.

    κ=κsolution−κwater\kappa = \kappa_{\text{solution}} - \kappa_{\text{water}} κ=(3.8×10−5 Ω−1 cm−1)−(6.2×10−8 Ω−1 cm−1)\kappa = (3.8 \times 10^{-5} \ \Omega^{-1} \text{ cm}^{-1}) - (6.2 \times 10^{-8} \ \Omega^{-1} \text{ cm}^{-1}) κ=(3.8×10−5)−(0.0062×10−5) Ω−1 cm−1\kappa = (3.8 \times 10^{-5}) - (0.0062 \times 10^{-5}) \ \Omega^{-1} \text{ cm}^{-1} κ=3.7938×10−5 Ω−1 cm−1\kappa = 3.7938 \times 10^{-5} \ \Omega^{-1} \text{ cm}^{-1}

  2. Calculate the molar solubility (s) of silver oxalate: The dissociation of silver oxalate is:

    Ag2C2O4(s)⇌2Ag+(aq)+C2O42−(aq)\text{Ag}_2\text{C}_2\text{O}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{C}_2\text{O}_4^{2-}(aq)

    If 's' is the molar solubility of Ag₂C₂O₄, then at equilibrium, [Ag+]=2s[\text{Ag}^+] = 2s and [C2O42−]=s[\text{C}_2\text{O}_4^{2-}] = s. The solubility product constant (Ksp) is given by:

    Ksp=[Ag+]2[C2O42−]=(2s)2(s)=4s3K_{sp} = [\text{Ag}^+]^2[\text{C}_2\text{O}_4^{2-}] = (2s)^2(s) = 4s^3

    Given Ksp = 1.2 × 10⁻¹¹:

    4s3=1.2×10−114s^3 = 1.2 \times 10^{-11} s3=1.2×10−114=0.3×10−11=3×10−12s^3 = \frac{1.2 \times 10^{-11}}{4} = 0.3 \times 10^{-11} = 3 \times 10^{-12} s=(3×10−12)1/3=(3)1/3×10−4 mol L−1s = (3 \times 10^{-12})^{1/3} = (3)^{1/3} \times 10^{-4} \text{ mol L}^{-1}

    Using (3)1/3≈1.442(3)^{1/3} \approx 1.442:

    s=1.442×10−4 mol L−1s = 1.442 \times 10^{-4} \text{ mol L}^{-1}

  3. Calculate the molar conductivity (Λm) of the saturated Ag₂C₂O₄ solution: The molar conductivity is given by the formula:

    Λm=κ×1000s\Lambda_m = \frac{\kappa \times 1000}{s}

    (where κ is in Ω⁻¹ cm⁻¹, s is in mol L⁻¹, and Λm is in Ω⁻¹ cm² mol⁻¹)

    Λm(Ag2C2O4)=(3.7938×10−5 Ω−1 cm−1)×1000 cm3 L−1(1.442×10−4 mol L−1)\Lambda_m(\text{Ag}_2\text{C}_2\text{O}_4) = \frac{(3.7938 \times 10^{-5} \ \Omega^{-1} \text{ cm}^{-1}) \times 1000 \text{ cm}^3 \text{ L}^{-1}}{(1.442 \times 10^{-4} \text{ mol L}^{-1})} Λm(Ag2C2O4)=3.7938×10−21.442×10−4=3.79381.442×102\Lambda_m(\text{Ag}_2\text{C}_2\text{O}_4) = \frac{3.7938 \times 10^{-2}}{1.442 \times 10^{-4}} = \frac{3.7938}{1.442} \times 10^2 Λm(Ag2C2O4)≈2.6309×102=263.09 Ω−1 cm2 mol−1\Lambda_m(\text{Ag}_2\text{C}_2\text{O}_4) \approx 2.6309 \times 10^2 = 263.09 \ \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}

  4. Calculate the molar conductivity of the oxalate ion (Λm(C₂O₄²⁻)) using Kohlrausch's Law: Kohlrausch's Law states that at infinite dilution, the molar conductivity of an electrolyte is the sum of the molar conductivities of its constituent ions, each multiplied by the number of ions produced by the dissociation of one formula unit of the electrolyte.

    Λm(Ag2C2O4)=2×Λm(Ag+)+Λm(C2O42−)\Lambda_m(\text{Ag}_2\text{C}_2\text{O}_4) = 2 \times \Lambda_m(\text{Ag}^+) + \Lambda_m(\text{C}_2\text{O}_4^{2-})

    Given Λm(Ag⁺) = 62 Ω⁻¹ cm² mol⁻¹:

    263.09=(2×62)+Λm(C2O42−)263.09 = (2 \times 62) + \Lambda_m(\text{C}_2\text{O}_4^{2-}) 263.09=124+Λm(C2O42−)263.09 = 124 + \Lambda_m(\text{C}_2\text{O}_4^{2-}) Λm(C2O42−)=263.09−124\Lambda_m(\text{C}_2\text{O}_4^{2-}) = 263.09 - 124 Λm(C2O42−)=139.09 Ω−1 cm2 mol−1\Lambda_m(\text{C}_2\text{O}_4^{2-}) = 139.09 \ \Omega^{-1} \text{ cm}^2 \text{ mol}^{-1}

    Rounding to the nearest integer, the molar conductivity of the oxalate ion is 139 Ω⁻¹ cm² mol⁻¹.