Question

Which equation is incorrect?

• A. ${\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaBr}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaCl}\right)={\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{KBr}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{KCl}\right)$
• B. ${\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaBr}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaI}\right)={\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{KBr}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaBr}\right)$
• C. ${\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{KCl}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaCl}\right)={\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{KBr}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaBr}\right)$
• D. ${\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{H}}_2\mathrm{O}\right)={\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{HCl}\right)+{\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaOH}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaBr}\right)$

Answer 1

The correct option is B

${\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaBr}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaI}\right)={\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{KBr}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaBr}\right)$

Explanation for the correct options:

Option B is the correct answer.

Kohlrauch's Law: The limiting molar conductivity (${\lambda }_m°$) for strong and weak electrolyte can be determined by using Kohlrauch's law : "the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte."

${\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaBr}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaI}\right)={\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{KBr}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaBr}\right)$

$$\begin{gathered} {\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{NaBr}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°(\mathrm{NaI})={\mathrm{\lambda }}_{\mathrm{m}}°\left(\mathrm{KBr}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°(\mathrm{NaBr}) \\ \mathrm{L}.\mathrm{H}.\mathrm{S}. \\ {\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{Na}}^{+}\right)+{\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{Br}}^{-}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{Na}}^{+}\right)-{\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{I}}^{-}\right) \\ ={\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{Br}}^{-}\right)-{\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{I}}^{-}\right) \\ \mathrm{R}.\mathrm{H}.\mathrm{S} \\ {\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{K}}^{+}\right)+{\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{Br}}^{-}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{Na}}^{+}\right)-{\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{Br}}^{-}\right) \\ ={\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{K}}^{+}\right)–{\mathrm{\lambda }}_{\mathrm{m}}°\left({\mathrm{Na}}^{+}\right) \end{gathered}$$

Therefore, option B is correct answer.