The molar conductivities at infinite dilution of BaCl2, NaCl and NaOH are 280 × 10−4, 126.5 × 10−4 and 248 × 10−4 S m2 mol−1 respectively. What is the molar conductivity of Ba(OH)2 at infinite dilution?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 523 × 10−4 S m2 mol−1
By Kohlrausch’s law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the contributions of its ions, each counted as many times as it appears in the formula. For \( \ce{Ba(OH)2} \), which gives one \( \ce{Ba^2+} \) and two \( \ce{OH^-} \):
\( \displaystyle \Lambda^{\circ}_{m}(\ce{Ba(OH)2})=\lambda^{\circ}_{\ce{Ba^2+}}+2\lambda^{\circ}_{\ce{OH^-}} \)
The given electrolytes are combined so that Na+ and Cl− cancel, leaving one \( \ce{Ba^2+} \) and two \( \ce{OH^-} \):
\( \displaystyle \Lambda^{\circ}_{m}(\ce{BaCl2})+2\Lambda^{\circ}_{m}(\ce{NaOH})-2\Lambda^{\circ}_{m}(\ce{NaCl}) \)
\( \displaystyle =(\lambda^{\circ}_{\ce{Ba^2+}}+2\lambda^{\circ}_{\ce{Cl^-}})+(2\lambda^{\circ}_{\ce{Na^+}}+2\lambda^{\circ}_{\ce{OH^-}})-(2\lambda^{\circ}_{\ce{Na^+}}+2\lambda^{\circ}_{\ce{Cl^-}}) \)
\( \displaystyle =\lambda^{\circ}_{\ce{Ba^2+}}+2\lambda^{\circ}_{\ce{OH^-}}=\Lambda^{\circ}_{m}(\ce{Ba(OH)2}) \)
Substituting the values (all in \( \mathrm{S\,m^{2}\,mol^{-1}} \)):
\( \displaystyle \Lambda^{\circ}_{m}(\ce{Ba(OH)2})=(280+2\times248-2\times126.5)\times10^{-4} \)
\( \displaystyle =(280+496-253)\times10^{-4} \)
\( \displaystyle =523\times10^{-4}\,\mathrm{S\,m^{2}\,mol^{-1}} \)
NCERT (2026-27 edition):
2.4.2 Variation of Conductivity and Molar Conductivity with Concentration – “Kohlrausch examined Λ°m values for a number of strong electrolytes and observed certain regularities. … On the basis of the above observations he enunciated Kohlrausch law of independent migration of ions. The law states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. … In general, if an electrolyte on dissociation gives ν+ cations and ν− anions then its limiting molar conductivity is given by: Λ°m = ν+λ°+ + ν−λ°− Here, λ°+ and λ°− are the limiting molar conductivities of the cation and anion respectively.”