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Ksp of a sparingly soluble salt from the conductivity of its saturated solution

This type of question, asked 3 times: 28 Jun 2022, 1 Feb 2023, 6 Apr 2026 S1. All 3 questions below, oldest first.

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  1. Question 1

    28 Jun 2022

    56
    The solubility product of a sparingly soluble salt A2X3A_{2}X_{3} is 1.1×10−231.1 \times10^{- 23}. If specific conductance of the solution is 3×10−5 S m−13 \times10^{- 5}\text{ }S{\text{ }m}^{- 1}, the limiting molar conductivity of the solution is x×10−3 S m2 mol−1x \times10^{- 3}\text{ }S{\text{ }m}^{2}{\text{ }mol}^{- 1}. The value of xx is

  2. Question 2

    1 Feb 2023

    146
    1×10−5MAgNO31 \times10^{- 5}MAgNO_{3} is added to 1 L1\text{ }L of saturated solution of AgBrAgBr. The conductivity of this solution at 298 K298\text{ }K is______ ×10−8 S m−1\times10^{- 8}\text{ }S{\text{ }m}^{- 1} [Given: KSP(AgBr)=4.9×10−13K_{SP}(AgBr) = 4.9 \times10^{- 13} at 298 K298\text{ }K
    λAg+0=6×10−3 S m2 mol−1λBr−0=8×10−3 S m2 mol−1 λNO3−0=7×10−3 S m2 mol−1]\begin{matrix} & \lambda_{Ag^{+}}^{0}= 6 \times10^{- 3}\text{ }S{\text{ }m}^{2}{\text{ }mol}^{- 1} \\ & \lambda_{Br^{-}}^{0}= 8 \times10^{- 3}\text{ }S{\text{ }m}^{2}{\text{ }mol}^{- 1} \\ & \left. \ \lambda_{NO_{3}^{-}}^{0}= 7 \times10^{- 3}\text{ }S{\text{ }m}^{2}{\text{ }mol}^{- 1} \right\rbrack \end{matrix}

  3. Question 3

    6 Apr 2026 S1

    33
    Consider the following data :
    ElectrolyteΛm∘\Lambda_{m}^{\circ} (S cm2 mol−1S\ cm^{2}\ mol^{-1})
    BaCl2BaCl_{2}x1x_{1}
    H2SO4H_{2}SO_{4}x2x_{2}
    HClHClx3x_{3}
    BaSO4BaSO_{4} is sparingly soluble in water. If the conductivity of the saturated BaSO4BaSO_{4} solution is x S cm−1x\ S\ cm^{-1}, then the solubility product of BaSO4BaSO_{4} can be given as (here Λm=Λm∘\Lambda_{m} = \Lambda_{m}^{\circ}) :

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