Question

The molar conductivities $\Lambda_m^\circ$ (NaOAc) and $\Lambda_m^\circ$ (HCl) at infinite dilution in water are 91.0 and 426.2 S cm$^2$ mol$^{-1}$ respectively. To calculate $\Lambda_m^\circ$ (HOAc), the additional value required is

• A. $\Lambda_m^\circ$ (H$_2$O)
• B. $\Lambda_m^\circ$ (NaCl)
• C. $\Lambda_m^\circ$ (NaOH)
• D. $\Lambda_m^\circ$ (NaNO$_3$)

Answer 1

Answer: (B)

$\Lambda_m^\circ$ (NaCl)

Answer 2

According to Kohlrausch's Law, the molar conductivity at infinite dilution of an electrolyte is the sum of the limiting molar ionic conductivities of its constituent ions.

We are given: $\Lambda_m^\circ$(NaOAc) = $\lambda_m^\circ$(Na$^+$) + $\lambda_m^\circ$(OAc$^-$) $\Lambda_m^\circ$(HCl) = $\lambda_m^\circ$(H$^+$) + $\lambda_m^\circ$(Cl$^-$)

We want to find $\Lambda_m^\circ$(HOAc) = $\lambda_m^\circ$(H$^+$) + $\lambda_m^\circ$(OAc$^-$).

Adding the given equations: $\Lambda_m^\circ$(NaOAc) + $\Lambda_m^\circ$(HCl) = $\lambda_m^\circ$(Na$^+$) + $\lambda_m^\circ$(OAc$^-$) + $\lambda_m^\circ$(H$^+$) + $\lambda_m^\circ$(Cl$^-$)

To obtain $\Lambda_m^\circ$(HOAc), we need to subtract $\lambda_m^\circ$(Na$^+$) + $\lambda_m^\circ$(Cl$^-$), which is $\Lambda_m^\circ$(NaCl).

Therefore: $\Lambda_m^\circ$(HOAc) = $\Lambda_m^\circ$(NaOAc) + $\Lambda_m^\circ$(HCl) - $\Lambda_m^\circ$(NaCl)

The additional value required is $\Lambda_m^\circ$(NaCl).