Exams › NEET › Chemistry

The ionization constant of ammonium hydroxide is 1.77 × 10⁻⁵ at 298 K. Hydrolysis constant of ammonium chloride is:

1. 6.50 × 10⁻⁶
2. 5.65 × 10⁻¹³
3. 5.65 × 10⁻¹²
4. 5.65 × 10⁻¹⁰

Correct answer: 5.65 × 10⁻¹²

Solution

The hydrolysis constant (Kh) is related to the ionization constant (Kb) of ammonium hydroxide by the relation Kh = Kw / Kb, where Kw = 1 × 10⁻¹⁴ at 298 K. Substituting Kb = 1.77 × 10⁻⁵, we get Kh = 1 × 10⁻¹⁴ / 1.77 × 10⁻⁵ = 5.65 × 10⁻¹².

Related NEET Chemistry questions

• A beaker contains a saturated solution of copper(l) chloride. Copper(l) chloride is slightly soluble salt with a solubility product of \( 1.2 \times 10^{-6} \) Which of the following salts when added to the solution would precipitate copper(l) chloride?
• The function of 'buffer solution' is:
• When \( C a(O H)_{2} \) attains solubility equilibrium, the:
• Predict if there will be any precipitate by mixing \( 50 \mathrm{mL} \) of \( 0.01 \mathrm{M} \) NaCl and 50 mL of \( 0.01 \mathrm{M} A g N O_{3} \) solution. The solubility product of \( A g C l \) is \( 1.5 \times \) \( 10^{-10} \)
• Calculate the ratio of \( H C O O^{-} \) and \( F^{-} \) in a mixture of \( 0.2 M \) HCOOH \( \left(K_{a}=\right. \) \( \left.2 \times 10^{-4}\right) \) and \( 0.1 M H F\left(K_{a}=6.6 \times\right. \) \( \left.10^{-4}\right) \)
• What is the [OH⁻] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)₂?

⚔️ Practice NEET Chemistry free + battle 1v1 →