Question

The equivalent conductance at infinite dilution of a weak acid such as HF

• A. can be determined by extrapolation of measurements on dilute solutions of HCl, HBr and HI
• B. can be determined by measurement on very dilute HF solutions
• C. can best be determined from measurements on dilute solutions of NaF, NaCl and HCl
• D. is an undefined quantity

Hint:

For weak electrolytes, \(\Lambda^0\) cannot be measured directly. Use Kohlrausch’s law: \(\Lambda^0(\text{HF})=\Lambda^0(\text{NaF})+\Lambda^0(\text{HCl})-\Lambda^0(\text{NaCl})\).

Answer 1

The Correct Option is C

Step 1: Problem with weak electrolytes.
HF is a weak acid and therefore a weak electrolyte.
Weak electrolytes do not ionize completely, hence their equivalent conductance does not show a linear relation with \(\sqrt{c}\).
So direct extrapolation to infinite dilution is not accurate.
Step 2: Use Kohlrausch’s law of independent migration of ions.
At infinite dilution:
\[ \Lambda^0(\text{HF}) = \lambda^0(H^+) + \lambda^0(F^-) \] Step 3: Obtain ionic conductances indirectly.
We can obtain \(\lambda^0(F^-)\) and \(\lambda^0(H^+)\) by using strong electrolytes:
\[ \Lambda^0(\text{NaF}) = \lambda^0(Na^+) + \lambda^0(F^-) \] \[ \Lambda^0(\text{HCl}) = \lambda^0(H^+) + \lambda^0(Cl^-) \] \[ \Lambda^0(\text{NaCl}) = \lambda^0(Na^+) + \lambda^0(Cl^-) \] Step 4: Combine these equations.
\[ \Lambda^0(\text{HF}) = \Lambda^0(\text{NaF}) + \Lambda^0(\text{HCl}) - \Lambda^0(\text{NaCl}) \] Thus, measurements on NaF, NaCl and HCl best determine \(\Lambda^0\) of HF.
Final Answer: \[ \boxed{\text{can best be determined from measurements on dilute solutions of NaF, NaCl and HCl}} \]