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MHT CET 202623 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
23 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

MHT CET 2026 · 23 April · Evening Shift · Chemistry · Ionic Equilibrium

Which of the following salts of weak acids and weak bases on...

Which of the following salts of weak acids and weak bases on hydrolysis turns red litmus blue?
[K_aforaceticacid=1.8×10−5for acetic acid = 1.8\times10^{-5}, KaK_a for hydrofluoric acid = 6.8×10−46.8\times10^{-4}, KaK_a for hydrogen cynide = 4.0×10−104.0\times10^{-10}, KaK_a for formic acid 1.8×10−41.8\times10^{-4} and KbK_b for ammonium hydroxide = 1.8×10−51.8\times10^{-5}]

Options

  1. AAmmonium acetate
  2. BAmmonium fluoride
  3. CAmmonium cynide
  4. DAmmonium formate

Correct answer

C. Ammonium cynide

Step-by-step solution

A salt of a weak acid and a weak base forms a basic solution if the dissociation constant of the base (K_b)isgreaterthanthedissociationconstantoftheacid(Ka) is greater than the dissociation constant of the acid (K_a). A basic solution turns red litmus blue.

For ammonium acetate: Ka=1.8×10−5K_a = 1.8 \times 10^{-5} and Kb=1.8×10−5K_b = 1.8 \times 10^{-5}. Since Ka=KbK_a = K_b, the solution is neutral.

For ammonium fluoride: Ka=6.8×10−4K_a = 6.8 \times 10^{-4} and Kb=1.8×10−5K_b = 1.8 \times 10^{-5}. Since Ka>KbK_a > K_b, the solution is acidic.

For ammonium cyanide: Ka=4.0×10−10K_a = 4.0 \times 10^{-10} and Kb=1.8×10−5K_b = 1.8 \times 10^{-5}. Since Kb>KaK_b > K_a, the solution is basic.

For ammonium formate: Ka=1.8×10−4K_a = 1.8 \times 10^{-4} and Kb=1.8×10−5K_b = 1.8 \times 10^{-5}. Since Ka>KbK_a > K_b, the solution is acidic.

Thus, ammonium cyanide forms a basic solution and turns red litmus blue.

Answer: Ammonium cynide
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