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JEE Advanced 2024 Paper 2 · Chemistry · Ionic Equilibrium · medium

A weak monobasic acid HA has Ka=1.0×10^-5 at 298 K (Kw=1.0×10^-14).

One or more options correct

A weak monobasic acid HA\mathrm{HA} has Ka=1.0×10−5K_a=1.0\times10^{-5} at 298 K298\ \mathrm{K} (Kw=1.0×10−14K_w=1.0\times10^{-14}). Choose the correct statement(s).
  1. A. The pH of 0.10 M0.10\ \mathrm{M} HA\mathrm{HA} is 3.03.0 (to the first decimal place)
  2. B. When 100 mL100\ \mathrm{mL} of 0.10 M0.10\ \mathrm{M} HA is mixed with 25 mL25\ \mathrm{mL} of 0.10 M0.10\ \mathrm{M} NaOH, the pH of the resulting solution is 5.485.48
  3. C. When the 0.10 M0.10\ \mathrm{M} solution of HA is diluted four-fold, its degree of dissociation becomes half of its original value
  4. D. The base-dissociation constant KbK_b of A−\mathrm{A^-} is 1.0×10−81.0\times10^{-8}
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Answer: A

(A) Correct. x=Kac=10−5×0.10=10−3x=\sqrt{K_ac}=\sqrt{10^{-5}\times0.10}=10^{-3} M, so pH =3.0=3.0 (degree of dissociation 1%, so the approximation is valid; exact quadratic solution gives 3.002). (B) Incorrect. Moles HA =10=10 mmol, NaOH =2.5=2.5 mmol, giving 7.5 mmol HA and 2.5 mmol A−\mathrm{A^-} (buffer). pH=pKa+log⁡2.57.5=5−0.477=4.52\mathrm{pH}=\mathrm{p}K_a+\log\dfrac{2.5}{7.5}=5-0.477=4.52, not 5.48 (that value uses the inverted ratio). (C) Incorrect. Ostwald's law α≈Ka/c\alpha\approx\sqrt{K_a/c}: diluting four times (c to c/4c/4) doubles α\alpha (0.00995 to 0.0198 on exact calculation). (D) Incorrect. Kb=Kw/Ka=10−14/10−5=1.0×10−9K_b=K_w/K_a=10^{-14}/10^{-5}=1.0\times10^{-9}. Answer: A only.
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