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Question: Equivalent conductance at infinite dilution for weak electrolyte HF: A) Can be determined by measu...

Equivalent conductance at infinite dilution for weak electrolyte HF:
A) Can be determined by measurement of equivalent conductance at infinite dilution for dilute solution of HCl, HBr and HI.
B) Can be determined by measurement of equivalent conductance at infinite dilution for very dilute HF solutions.
C) Can best be determined from measurements on dilute solutions of NaF, NaCl, HCl.
D) Cannot be calculated.

Explanation

Solution

The limiting molar conductivity of an infinite diluted solution is the sum of the limiting molar conductivity of its constituent ions. This fact will lead you to the correct answer.

Complete answer:
In the previous classes, we have studied the concepts of physical chemistry which tells about the reactions in the solutions and includes some of the basic definitions and calculations.
Let us see one of these calculations that is the limiting molar conductivity in detail.
- As we know that, the graph plotted between molar conductivity and c\sqrt c (where c is the concentration) is not a straight line for weak electrolytes.
- The molar conductivity of weak electrolyte increases steeply at lower concentrations. Therefore, limiting molar conductivity, cannot be obtained by extrapolation of molar conductivity to zero concentration.
- Hence, we use the Kohlrausch law for the determination of limiting molar conductivity for weak electrolytes. Kohlrausch law states that at infinite dilution, the limiting molar conductivity of a solution is the sum of the limiting molar conductivity of its constituent ions.
- Now, at infinite dilution, HF breaks into H+{H^ + } and F{F^ - } ions. The chemical equation for the same is given below.
HFH++FHF \to {H^ + } + {F^ - }
- Now, at infinite dilution, if we know the equivalent conductance of NaF, NaCl and HCl, we can calculate the equivalent conductance of HF by the relation:
Equivalent conductance of HF = Equivalent conductance of HCl + equivalent conductance of NaF – equivalent conductance of NaCl
i.e. λHF=λHCl+λNaFλNaCl{\lambda ^\circ }_{HF} = {\lambda ^\circ }_{HCl} + {\lambda ^\circ }_{NaF} - {\lambda ^\circ }_{NaCl}

Hence, the correct answer is (C) Can best be determined from measurements on dilute solutions of NaF, NaCl, HCl.

Note:
It is important to note that this law that is Kohlrausch’s law is applicable for both weak and strong electrolytes and is used only for the infinite dilution and is not valid for other concentrations.