JEE Advanced
Chemistry
Electrochemistry
2023
JEE Advanced 2023 (Paper 1)
JEE Advanced Chemistry Question (2023) — Solution
Question
Plotting 1 / Λ m against cΛ m for aqueous solutions of a monobasic weak acid ( HX ) resulted in a straight line with y - axis intercept of P and slope of S . The ratio P / S is Λ m =  molar conductivity  Λ m o =   limiting   molar conductivity       c =   molar concentration K a =  dissociation constant of  HX
Options
- A. K a Λ m o
- B. K a Λ m o / 2
- C. 2   K a Λ m o
- D. 1 / K a Λ m o
Step-by-step solution
The degree of dissociation is the ratio of molar conductivity to the limiting molar conductivity. α = Λ m Λ m o The equilibrium reaction is given below,                                                   HX ⇌ H + + X - initial                                     c                         At   equilibrium       c - cα         cα           cα The acid dissociation constant, K a = cα 2 1 − α ⇒ K a = c Λ m / Λ m o 2 1 − Λ m / Λ m o ⇒ K a = cΛ m 2 Λ m o Λ m o − Λ m K a Λ m o 2 − K a Λ m o Λ m = cΛ m 2 K a Λ m o 2 Λ m − K a Λ m o = CΛ m K a Λ m o 2 Λ m = cΛ m + K a Λ m o 1 Λ m = cΛ m K a Λ m o 2 + 1 Λ m o From the above line equation, slope and y-intercept can be written as follows, P = 1 Λ m o S = 1 K a Λ m o 2 P S = 1 Λ m o 1 K a Λ m o 2 = K a Λ m o
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