Question

The molar ionic conductance at infinite dilution for \( \text{NaOH} \), \( \text{NaCl} \), and \( \text{BaCl}_2 \) are \( 248.1 \times 10^{-4} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \), \( 126.5 \times 10^{-4} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \), and \( 280.0 \times 10^{-4} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \), respectively. The \( \Lambda^\circ_m \) for \( \text{Ba(OH)}_2 \) is:

• A. \( 523.2 \times 10^{-4} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \)
• B. \( 252.32 \times 10^{-4} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \)
• C. \( 52.3 \times 10^{-4} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \)
• D. \( 50.5 \times 10^{-4} \, \text{S} \, \text{m}^2 \, \text{mol}^{-1} \)

Hint:

The Λ◦ m for a compound like Ba(OH)2 is calculated by summing the molar ionic conductivities of each of its constituent ions, multiplied by their respective stoichiometric coefficients in the dissociation equation.

Answer 1

The Correct Option is A

To determine the molar ionic conductance at infinite dilution (\( \Lambda_m^\circ \)) for Ba(OH)\(_2\), we analyze the dissociation of Ba(OH)\(_2\) into its constituent ions and utilize the given conductance values for related compounds.

1. Dissociation of Compounds:
   NaOH → Na\(^+\) + OH\(^-\)
   NaCl → Na\(^+\) + Cl\(^-\)
   BaCl\(_2\) → Ba\(^{2+}\) + 2Cl\(^-\)
2. Equations for Molar Ionic Conductances:
   \( \Lambda_m^\circ (\text{NaOH}) = \Lambda_m^\circ (\text{Na}^+) + \Lambda_m^\circ (\text{OH}^-) \)
   \( \Lambda_m^\circ (\text{NaCl}) = \Lambda_m^\circ (\text{Na}^+) + \Lambda_m^\circ (\text{Cl}^-) \)
   \( \Lambda_m^\circ (\text{BaCl}_2) = \Lambda_m^\circ (\text{Ba}^{2+}) + 2\Lambda_m^\circ (\text{Cl}^-) \)
3. Substituting the Given Values:
   248.1 × 10\(^{-4}\) = \( \Lambda_m^\circ (\text{Na}^+) + \Lambda_m^\circ (\text{OH}^-) \) (1)
   126.5 × 10\(^{-4}\) = \( \Lambda_m^\circ (\text{Na}^+) + \Lambda_m^\circ (\text{Cl}^-) \) (2)
   280.0 × 10\(^{-4}\) = \( \Lambda_m^\circ (\text{Ba}^{2+}) + 2\Lambda_m^\circ (\text{Cl}^-) \) (3)
4. Solving for Individual Ionic Conductivities:
   Step 1: Subtract equation (2) from equation (1) to find \( \Lambda_m^\circ (\text{OH}^-) \):
   \( 248.1 - 126.5 = \Lambda_m^\circ (\text{OH}^-) - \Lambda_m^\circ (\text{Cl}^-) \)
   \( 121.6 × 10^{-4} = \Lambda_m^\circ (\text{OH}^-) - \Lambda_m^\circ (\text{Cl}^-) \)
   Step 2: Let:
   \( \Lambda_m^\circ (\text{Na}^+) = x, \Lambda_m^\circ (\text{Cl}^-) = y, \Lambda_m^\circ (\text{OH}^-) = z, \Lambda_m^\circ (\text{Ba}^{2+}) = w \)
   From equation (2): \( x + y = 126.5 × 10^{-4} \) (4)
   From equation (1): \( x + z = 248.1 × 10^{-4} \) (5)
   From the difference: \( z - y = 121.6 × 10^{-4} \) (6)
   Step 3: Solve for \( z \) and \( y \):
   \( z = y + 121.6 × 10^{-4} \) (7)
   Substitute equation (7) into equation (5):
   \( x + y + 121.6 × 10^{-4} = 248.1 × 10^{-4} \)
   \( x + y = 126.5 × 10^{-4} \) (From equation (4))
   Step 4: From equation (3): \( w + 2y = 280.0 × 10^{-4} \) (8)
5. Calculating \( \Lambda_m^\circ \) for Ba(OH)\(_2\):
   The dissociation of Ba(OH)\(_2\) is:
   Ba(OH)\(_2\) → Ba\(^{2+}\) + 2OH\(^-\)
   Therefore:
   \( \Lambda_m^\circ (\text{Ba(OH)}_2) = \Lambda_m^\circ (\text{Ba}^{2+}) + 2\Lambda_m^\circ (\text{OH}^-) \)
   From the given correct answer and the context, we sum the conductances:
   \( \Lambda_m^\circ (\text{Ba(OH)}_2) = 280.0 × 10^{-4} + 2 × 121.6 × 10^{-4} = 280.0 + 243.2 = 523.2 × 10^{-4} \, \text{S m}^2 \text{ mol}^{-1} \)

Therefore, the molar ionic conductance at infinite dilution for Ba(OH)\(_2\) is 523.2 × 10\(^{-4}\) S m\(^2\) mol\(^{-1}\).