Question

The limiting equivalent conductivity of NaCl, KCl, and KBr are \( 126.5 \), \( 150.0 \), and \( 151.5 \) S cm\(^2\) eq\(^{-1}\), respectively. The limiting equivalent ionic conductivity for Br\(^{-}\) is \( 78 \) S cm\(^2\) eq\(^{-1}\). The limiting equivalent ionic conductivity for Na\(^+\) ions would be:

• A. 128
• B. 125
• C. 49
• D. 50

Hint:

The limiting ionic conductivities of individual ions can be calculated using the difference in the limiting conductivities of different salts and ions.

Answer 1

The Correct Option is D

Step 1: The limiting equivalent conductivity of the electrolyte is the sum of the limiting ionic conductivities of the individual ions. \[ \lambda_{{m}} ({NaCl}) = \lambda_{{m}} ({Na}^+) + \lambda_{{m}} ({Cl}^-) \] Similarly for \( {KCl} \) and \( {KBr} \), we have: \[ \lambda_{{m}} ({KCl}) = \lambda_{{m}} ({K}^+) + \lambda_{{m}} ({Cl}^-) \] \[ \lambda_{{m}} ({KBr}) = \lambda_{{m}} ({K}^+) + \lambda_{{m}} ({Br}^-) \] Step 2: Given values: \[ \lambda_{{m}} ({NaCl}) = 126.5 \, {S cm}^2 {eq}^{-1}, \quad \lambda_{{m}} ({KCl}) = 150.0 \, {S cm}^2 {eq}^{-1}, \quad \lambda_{{m}} ({KBr}) = 151.5 \, {S cm}^2 {eq}^{-1} \] \[ \lambda_{{m}} ({Br}^-) = 78 \, {S cm}^2 {eq}^{-1} \] Step 3: Substituting the known values into the equation for \( \lambda_{{m}} ({NaCl}) \), we get: \[ 126.5 = \lambda_{{m}} ({Na}^+) + \lambda_{{m}} ({Cl}^-) \] Substituting \( \lambda_{{m}} ({Cl}^-) \) from \( \lambda_{{m}} ({KCl}) \): \[ 150.0 = \lambda_{{m}} ({K}^+) + \lambda_{{m}} ({Cl}^-) \] Now, subtract the two equations: \[ 150.0 - 126.5 = \lambda_{{m}} ({K}^+) - \lambda_{{m}} ({Na}^+) \] \[ \Rightarrow 23.5 = \lambda_{{m}} ({K}^+) - \lambda_{{m}} ({Na}^+) \] \[ \Rightarrow \lambda_{{m}} ({Na}^+) = \lambda_{{m}} ({K}^+) - 23.5 \] Substitute this into the equation for \( \lambda_{{m}} ({KBr}) \): \[ 151.5 = \lambda_{{m}} ({K}^+) + 78 \] \[ \Rightarrow \lambda_{{m}} ({K}^+) = 151.5 - 78 = 73.5 \] Now, substitute this into the equation for \( \lambda_{{m}} ({Na}^+) \): \[ \lambda_{{m}} ({Na}^+) = 73.5 - 23.5 = 50.0 \]