NTA Abhyas JEE Main2020ChemistryElectrochemistryPractice
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 ) )
Correct answer
4
Step-by-step solution
∧ M o (HA) = ∧ M ∞ (HCl) + ∧ M ∞ (NaA) − ∧ M ∞ (NaCl) = 425 + 100 − 125 = 400 Scm 2 mol − 1 pH = 4 , [H + ] = 10 − 4 = α C α = ∧ m ∧ m ∞ = 200 400 = 0.5 K a = ( C α ) α ( 1 − α ) = 10 − 4 ( 0.5 ) ( 1 − α ) = 10 − 4 ,   pK a = 4