At 291k the molar conductance values at infinite dilution of nh4cl noah and nacl are 129.1,217.4,108.3scm2mol inverse respectively calculate the molar conductance of nh4oh at infinite dilution
Answers
Given:
Λ°(NH₄Cl) = 129.1 Scm²mol⁻¹
Λ°(NaOH) = 207.4 Scm²mol⁻¹
Λ°(NaCl) = 108.3 Scm²mol⁻¹
To find: Λ°(NH₄OH) = ?
Solution: At infinite dilution each ion has a definite contribution towards the molar conductance of electrolyte and it does not depend on the counter ion present.
Therefore,
∧°(NH₄Cl) = λ°(NH₄⁺) + λ°(Cl⁻)
∧ °(NaOH) = λ°(Na⁺) + λ°(OH⁻)
∧°(NaCl) = λ°(Na⁺) + λ°(Cl⁻)
Now,
∧°(NH₄OH) = λ°(NH₄⁺) + λ°(OH⁻)
⇒∧°(NH₄OH) = λ°(NH₄⁺) + λ°(Na⁺) - λ°(Na⁺) + λ°(Cl⁻) - λ°(Cl⁻) + λ°(OH⁻) [adding and substracting λ°(Na⁺) and λ°(Cl⁻)]
⇒∧°(NH₄OH) = [λ°(NH₄⁺) + λ°(Cl⁻)] + [λ°(Na⁺) + λ°(OH⁻)] - [λ°(Na⁺) + λ°(Cl⁻)]
⇒∧°(NH₄OH) = Λ°(NH₄Cl) + Λ°(NaOH) - Λ°(NaCl)
⇒∧°(NH₄OH) = (129.1 + 217.4 - 108.3) Scm²mol⁻¹
⇒∧°(NH₄OH) = 238.2 Scm²mol⁻¹
∴ Limiting molar conductance of NH₄OH at infinite dilution,
∧°(NH₄OH) = 238.2 Scm²mol⁻¹