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NEETChemistryIonic Equilibrium

Match List-I with List-II for buffer solutions at 298 K containing equal concentrations of their respective conjugate acid-base pairs. List-I (Buffer system and given constant) List-II (pH of the buffer) (A) HA / A⁻ with K_b(A⁻) = 10⁻⁹ (I) 5 (B) BOH / B⁺ with K_a(B⁺) = 10⁻⁸ (II) 6 (C) HX / X⁻ with K_a(HX) = 10⁻⁶ (III) 8 (D) YOH / Y⁺ with K_b(YOH) = 10⁻⁵ (IV) 9 Choose the correct answer from the options given below:

Options

  1. A(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
  2. B(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
  3. C(A)-(I), (B)-(II), (C)-(II), (D)-(I)
  4. D(A)-(IV), (B)-(III), (C)-(II), (D)-(I)

Correct answer

B. (A)-(I), (B)-(III), (C)-(II), (D)-(IV)

Step-by-step solution

For buffer solutions with equal concentrations of the conjugate acid-base pair, the logarithmic term in the Henderson-Hasselbalch equation becomes zero ( 1 = 0 ). For acidic buffers: pH = pK_a For basic buffers: pOH = pK_b and pH = 14 - pOH (A) Acidic buffer HA / A⁻ : Given K_b(A⁻) = 10⁻⁹ . K_a(HA) = 10⁻¹⁴ 10⁻⁹ = 10⁻⁵ pK_a = 5 . pH = 5 . Matches (I). (B) Basic buffer BOH / B⁺ : Given K_a(B⁺) = 10⁻⁸ . K_b(BOH) = 10⁻¹⁴ 10⁻⁸ = 10⁻⁶ pK_b = 6 . pOH = 6 pH = 14 - 6 = 8 . Matches (III). (C) Acidic buffer HX / X⁻ : Given K

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