The equivalent conductance of a weak acid such as HF

Question

The equivalent conductance of a weak acid such as HF (A) can be determined by extrapolation of measurements on dilute solutions of HCl, HBr and HF (B) can be determined by measurement on very dilute solution of HF (C) can be determined by measurement on dilute solutions of NaF, NaCl and HCl (D) in an undefined quantity

Tardigrade Admin · Accepted Answer

Kohlrausch’s Law Lambdaeq (NaF) + Lambdaeq (HCl)- Lambdaeq(NaCl) = Lambdaeq HF

Q. The equivalent conductance of a weak acid such as HF

can be determined by extrapolation of measurements on dilute solutions of HCl, HBr and HF

can be determined by measurement on very dilute solution of HF

can be determined by measurement on dilute solutions of NaF, NaCl and HCl

in an undefined quantity

Kohlrausch’s Law
$Λ_{e q} (N a F) + Λ_{e q} (H Cl) - Λ_{e q} (N a Cl) = Λ_{e q} H F$

Q. The equivalent conductance of a weak acid such as HF

can be determined by extrapolation of measurements on dilute solutions of HCl, HBr and HF

can be determined by measurement on very dilute solution of HF

can be determined by measurement on dilute solutions of NaF, NaCl and HCl

in an undefined quantity

Kohlrausch’s Law Λeq​(NaF)+Λeq​(HCl)−Λeq​(NaCl)=Λeq​HF

Kohlrausch’s Law
$Λ_{e q} (N a F) + Λ_{e q} (H Cl) - Λ_{e q} (N a Cl) = Λ_{e q} H F$