Chemistry - Electrochemistry Question with Solution | TestHub

${\Lambda }_m^{\circ }({\mathrm{H}}_2\mathrm{O})={\lambda }^{\circ }({\mathrm{H}}^{+})+{\lambda }^{\circ }({\mathrm{OH}}^{-})$.

(A) $\mathrm{HBr}+\mathrm{KOH}-\mathrm{KBr}$: ${\mathrm{H}}^{+}+{\mathrm{Br}}^{-}+{\mathrm{K}}^{+}+{\mathrm{OH}}^{-}-{\mathrm{K}}^{+}-{\mathrm{Br}}^{-}={\mathrm{H}}^{+}+{\mathrm{OH}}^{-}$: true.

(B) leaves ${\mathrm{H}}^{+}+2{\mathrm{NO}}_3^{-}-{\mathrm{OH}}^{-}$: false.

(C) adds both ${\mathrm{Na}}^{+}$ and ${\mathrm{Cl}}^{-}$ twice: false.

(D) ${\mathrm{CH}}_3\mathrm{COOH}+\mathrm{NaOH}-{\mathrm{CH}}_3\mathrm{COONa}={\mathrm{H}}^{+}+{\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{Na}}^{+}+{\mathrm{OH}}^{-}-{\mathrm{CH}}_3{\mathrm{COO}}^{-}-{\mathrm{Na}}^{+}={\mathrm{H}}^{+}+{\mathrm{OH}}^{-}$: true (limiting values, so the weak acid is fully ionised).

Answer: (A), (D).