Question:medium

In hydrolysis of a salt of weak acid and strong base, A\(^-\) + H\(_2\)O \(\rightleftharpoons\) HA + OH\(^-\), the hydrolysis constant (\(K_h\)) is equal to

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For weak base-strong acid salt, \(K_h = K_w/K_b\).
Updated On: Jun 16, 2026
  • \(K_w/K_a\)
  • \(K_w/K_b\)
  • \(K_a/C\)
  • \(K_w \times K_a/K_b\)
Show Solution

The Correct Option is A

Solution and Explanation

To solve this problem, we need to understand the concept of hydrolysis constant (\(K_h\)) in the context of a salt derived from a weak acid and a strong base. Let’s go through the steps:

  1. When a salt of a weak acid and a strong base is dissolved in water, the anion of the weak acid (\(A^-\)) reacts with water (\(H_2O\)) to form the weak acid (\(HA\)) and hydroxide ions (\(OH^-\)). The equilibrium for this hydrolysis reaction can be written as:

\(A^- + H_2O \rightleftharpoons HA + OH^-\)

  1. The hydrolysis constant (\(K_h\)) is given by the expression:

\(K_h = \frac{[HA][OH^-]}{[A^-]}\)

  1. For this reaction, the equilibrium can be linked to the ion product of water (\(K_w\)) and the dissociation constant (\(K_a\)) of the weak acid (\(HA\)). The relationship between these constants is:

\(K_h = \frac{K_w}{K_a}\)

  1. Explanation of Options:
    • \(K_w/K_a\): This matches our derived formula for the hydrolysis constant in the given reaction. This is the correct option.
    • \(K_w/K_b\): This expression is relevant for the hydrolysis of a salt of a weak base and a strong acid, so it does not apply here.
    • \(K_a/C\): This is not related to any standard expression for hydrolysis constant. Hence, it's incorrect.
    • \(K_w \times K_a/K_b\): This is not relevant for this scenario, as it represents a different context of equilibria.
  2. Conclusion: The correct answer for the hydrolysis constant (\(K_h\)) of a salt of a weak acid and strong base is \(K_w/K_a\).

Therefore, the correct option is \(K_w/K_a\).

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