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ChemMedAdv 2026 • Paper 2

Molar Conductivity and Solubility Calculation for Silver Chloride

At 300 K300\text{ K}, the molar conductivities of the aqueous solutions of three salts at two different concentrations are given below:

SaltConcentration(M)Molar conductivity(S cm2 mol−1)NaNO30.011110.04101NaCl0.011170.04107AgNO30.011250.04116\begin{array}{|c|c|c|} \hline \text{Salt} & \begin{array}{c} \text{Concentration} \\ \text{(M)} \end{array} & \begin{array}{c} \text{Molar conductivity} \\ (\text{S cm}^2\text{ mol}^{-1}) \end{array} \\ \hline \text{NaNO}_3 & 0.01 & 111 \\ & 0.04 & 101 \\ \hline \text{NaCl} & 0.01 & 117 \\ & 0.04 & 107 \\ \hline \text{AgNO}_3 & 0.01 & 125 \\ & 0.04 & 116 \\ \hline \end{array}

The conductivity of a saturated aqueous solution of AgCl\text{AgCl} is 1.40×10−6 S cm−11.40 \times 10^{-6}\text{ S cm}^{-1} at 300 K300\text{ K}. If the solubility of AgCl\text{AgCl} in water at 300 K300\text{ K} is X mol L−1X\text{ mol L}^{-1}, then log⁡10(X−1)\log_{10}(X^{-1}) is

(Assume that AgCl\text{AgCl} dissolved in water ionizes completely and that the molar conductivity of saturated AgCl\text{AgCl} solution is equal to its limiting molar conductivity.)

Options

A

3

B

4

C

5

Correct
D

6

Topics & Concepts

ElectrochemistryElectrochemical CellConcentration Terms

Step-by-Step Solution

To find the value of log⁡10(X−1)\log_{10}(X^{-1}), we apply Kohlrausch's law and Debye-Hückel-Onsager equation to determine the limiting molar conductivities (Λm∘\Lambda_m^\circ) of the given salts.

Using the relation Λm=Λm∘−AC\Lambda_m = \Lambda_m^\circ - A\sqrt{C}:

  1. For NaNO3\text{NaNO}_3: Λm∘(NaNO3)=121 S cm2 mol−1\Lambda_m^\circ(\text{NaNO}_3) = 121 \text{ S cm}^2\text{ mol}^{-1}

  2. For NaCl\text{NaCl}: Λm∘(NaCl)=127 S cm2 mol−1\Lambda_m^\circ(\text{NaCl}) = 127 \text{ S cm}^2\text{ mol}^{-1}

  3. For AgNO3\text{AgNO}_3: Λm∘(AgNO3)=134 S cm2 mol−1\Lambda_m^\circ(\text{AgNO}_3) = 134 \text{ S cm}^2\text{ mol}^{-1}

Using Kohlrausch's law, the limiting molar conductivity of AgCl\text{AgCl} is: Λm∘(AgCl)=Λm∘(AgNO3)+Λm∘(NaCl)−Λm∘(NaNO3)=134+127−121=140 S cm2 mol−1\Lambda_m^\circ(\text{AgCl}) = \Lambda_m^\circ(\text{AgNO}_3) + \Lambda_m^\circ(\text{NaCl}) - \Lambda_m^\circ(\text{NaNO}_3) = 134 + 127 - 121 = 140 \text{ S cm}^2\text{ mol}^{-1}

Given that for saturated AgCl\text{AgCl}, Λm=Λm∘\Lambda_m = \Lambda_m^\circ, we relate molar conductivity to solubility XX: Λm∘=κ×1000X\Lambda_m^\circ = \frac{\kappa \times 1000}{X}

Substituting the given conductivity κ=1.40×10−6 S cm−1\kappa = 1.40 \times 10^{-6} \text{ S cm}^{-1}: 140=1.40×10−6×1000X  ⟹  X=10−5 mol L−1140 = \frac{1.40 \times 10^{-6} \times 1000}{X} \implies X = 10^{-5} \text{ mol L}^{-1}

Thus, X−1=105X^{-1} = 10^5, leading to: log⁡10(X−1)=log⁡10(105)=5\log_{10}(X^{-1}) = \log_{10}(10^5) = 5

Hence, the correct option is (C).

Molar Conductivity and Solubility Calculation for Silver Chloride | Chemistry PYQ Solution - JEE Challenger