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ChemistryElectrochemistryMiscellaneous/MixedEasy2 minPYQ_2020
ChemistryEasynumerical

The molar conductivity of aqueous solution of a weak acid HA is 200  Scm 2 mol −1 and its pH is 4. What is the value of pK a of HA at 25 o C ?
(Given: λ m o NaA  = 100  Scm 2 mo1 −1 ,  λ m o HCl  = 425  Scm 2 mol −1 ,  λ m o (NaCl) = (125  Scm 2 mo1 −1 )λ m o NaA  = 100  Scm 2 mo1 −1 ,  λ m o HCl  = 425  Scm 2 mol −1 ,  λ m o (NaCl) = 125  Scm 2 mo1 −1 ))

Answer:
4.00
Solution:

∧ M o (HA)= ∧ M ∞ (HCl)+ ∧ M ∞ (NaA)− ∧ M ∞ (NaCl)
=425+100−125=400  Scm 2 mol −1
pH=4,  [H + ]= 10 −4 =αC
α= ∧m ∧ m ∞ = 200 400 =0.5
K a = (Cα)α (1−α) = 10 −4 (0.5) (1−α) = 10 −4 ,  pK a =4

Stream:NTA_ABHYASSubject:ChemistryTopic:ElectrochemistrySubtopic:Miscellaneous/Mixed
⏱ 2mℹ️ Source: PYQ_2020

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