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NCERT Exemplar · Q47

Q.Λm(H2O)0\Lambda^0_{m(H_2O)} is equal to _______________. (Two or more than two options may be correct.)

(i) Λm(HCl)0+Λm(NaOH)0−Λm(NaCl)0\Lambda^0_{m(HCl)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaCl)}
(ii) Λm(HNO3)0+Λm(NaNO3)0−Λm(NaOH)0\Lambda^0_{m(HNO_3)} + \Lambda^0_{m(NaNO_3)} - \Lambda^0_{m(NaOH)}
(iii) Λm(HNO3)0+Λm(NaOH)0−Λm(NaNO3)0\Lambda^0_{m(HNO_3)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaNO_3)}
(iv) Λm(NH4OH)0+Λm(HCl)0−Λm(NH4Cl)0\Lambda^0_{m(NH_4OH)} + \Lambda^0_{m(HCl)} - \Lambda^0_{m(NH_4Cl)}
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The limiting molar conductivity of water, Λm(H2O)0\Lambda^0_{m(H_2O)}, is found by applying Kohlrausch’s law of independent migration of ions. It equals the sum of the limiting conductivities of its constituent ions, H+H^+ and OH−OH^-, which can be obtained by combining the conductivities of strong electrolytes. The correct expressions are (i) and (iii).

The key idea here is Kohlrausch’s law: at infinite dilution, each ion contributes a fixed amount to the molar conductivity of an electrolyte, independent of the other ion it travels with. So Λm0\Lambda^0_m for any electrolyte is simply the sum of the limiting conductivities of its cation and anion.

For water, which dissociates as H2O⇌H++OH−H_2O \rightleftharpoons H^+ + OH^-, its limiting molar conductivity is:

Λm(H2O)0=λH+0+λOH−0\Lambda^0_{m(H_2O)} = \lambda^0_{H^+} + \lambda^0_{OH^-}

We don’t know these individual ionic conductivities directly, but we can get them by combining data from strong electrolytes that contain these ions — strong electrolytes are the ones whose Λm0\Lambda_m^0 can actually be measured directly, by extrapolating Λm\Lambda_m vs c\sqrt{c} to zero concentration.

Let’s check each option step by step.

  1. Option (i): Λm(HCl)0+Λm(NaOH)0−Λm(NaCl)0\Lambda^0_{m(HCl)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaCl)}

    Write each in terms of ionic conductivities:

    • Λm(HCl)0=λH+0+λCl−0\Lambda^0_{m(HCl)} = \lambda^0_{H^+} + \lambda^0_{Cl^-}
    • Λm(NaOH)0=λNa+0+λOH−0\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}
    • Λm(NaCl)0=λNa+0+λCl−0\Lambda^0_{m(NaCl)} = \lambda^0_{Na^+} + \lambda^0_{Cl^-}

    Adding the first two and subtracting the third:

(λH+0+λCl−0)+(λNa+0+λOH−0)−(λNa+0+λCl−0)(\lambda^0_{H^+} + \lambda^0_{Cl^-}) + (\lambda^0_{Na^+} + \lambda^0_{OH^-}) - (\lambda^0_{Na^+} + \lambda^0_{Cl^-})

The λNa+0\lambda^0_{Na^+} and λCl−0\lambda^0_{Cl^-} cancel, leaving λH+0+λOH−0\lambda^0_{H^+} + \lambda^0_{OH^-}, which is exactly Λm(H2O)0\Lambda^0_{m(H_2O)}. HCl, NaOH and NaCl are all strong electrolytes, so this is a legitimate calculation. (i) is correct.

  1. Option (ii): Λm(HNO3)0+Λm(NaNO3)0−Λm(NaOH)0\Lambda^0_{m(HNO_3)} + \Lambda^0_{m(NaNO_3)} - \Lambda^0_{m(NaOH)}

    Write them out:

    • Λm(HNO3)0=λH+0+λNO3−0\Lambda^0_{m(HNO_3)} = \lambda^0_{H^+} + \lambda^0_{NO_3^-}
    • Λm(NaNO3)0=λNa+0+λNO3−0\Lambda^0_{m(NaNO_3)} = \lambda^0_{Na^+} + \lambda^0_{NO_3^-}
    • Λm(NaOH)0=λNa+0+λOH−0\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}

    Sum the first two and subtract the third:

(λH+0+λNO3−0)+(λNa+0+λNO3−0)−(λNa+0+λOH−0)(\lambda^0_{H^+} + \lambda^0_{NO_3^-}) + (\lambda^0_{Na^+} + \lambda^0_{NO_3^-}) - (\lambda^0_{Na^+} + \lambda^0_{OH^-})

The λNa+0\lambda^0_{Na^+} cancels, but we get λH+0+2λNO3−0−λOH−0\lambda^0_{H^+} + 2\lambda^0_{NO_3^-} - \lambda^0_{OH^-}, which is not λH+0+λOH−0\lambda^0_{H^+} + \lambda^0_{OH^-}. So (ii) is incorrect.

  1. Option (iii): Λm(HNO3)0+Λm(NaOH)0−Λm(NaNO3)0\Lambda^0_{m(HNO_3)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaNO_3)}

    Write:

    • Λm(HNO3)0=λH+0+λNO3−0\Lambda^0_{m(HNO_3)} = \lambda^0_{H^+} + \lambda^0_{NO_3^-}
    • Λm(NaOH)0=λNa+0+λOH−0\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}
    • Λm(NaNO3)0=λNa+0+λNO3−0\Lambda^0_{m(NaNO_3)} = \lambda^0_{Na^+} + \lambda^0_{NO_3^-}

    Adding the first two and subtracting the third:

(λH+0+λNO3−0)+(λNa+0+λOH−0)−(λNa+0+λNO3−0)(\lambda^0_{H^+} + \lambda^0_{NO_3^-}) + (\lambda^0_{Na^+} + \lambda^0_{OH^-}) - (\lambda^0_{Na^+} + \lambda^0_{NO_3^-})

The λNa+0\lambda^0_{Na^+} and λNO3−0\lambda^0_{NO_3^-} cancel, leaving λH+0+λOH−0\lambda^0_{H^+} + \lambda^0_{OH^-}. HNO₃, NaOH and NaNO₃ are all strong electrolytes, so this is also legitimate. (iii) is correct.

  1. Option (iv): Λm(NH4OH)0+Λm(HCl)0−Λm(NH4Cl)0\Lambda^0_{m(NH_4OH)} + \Lambda^0_{m(HCl)} - \Lambda^0_{m(NH_4Cl)} Write:
    • Λm(NH4OH)0=λNH4+0+λOH−0\Lambda^0_{m(NH_4OH)} = \lambda^0_{NH_4^+} + \lambda^0_{OH^-}
    • Λm(HCl)0=λH+0+λCl−0\Lambda^0_{m(HCl)} = \lambda^0_{H^+} + \lambda^0_{Cl^-}
    • Λm(NH4Cl)0=λNH4+0+λCl−0\Lambda^0_{m(NH_4Cl)} = \lambda^0_{NH_4^+} + \lambda^0_{Cl^-} …
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