Question

\begin{aligned} &\text { The hydrolysis constant of a salt of weak acid }\\\ &\left(K_{\mathrm{a}}=2 \times 10^{-6}\right) \text { and of a weak basc }\left(K_{\mathrm{b}}=5 \times 10^{-7}\right) \text { is } \end{aligned}(1) 10 (2) \(10^{2}\) (3) \(2 \times 10^{2}\) (4) \(5 \times 10^{3}\)

Short answer

Option (1) 10

Step-by-step solution

- Identify Given Data

Given the problem, the ionization constants for a weak acid \(K_a = 2 \times 10^{-6}\) and for a weak base \(K_b = 5 \times 10^{-7}\) are provided.

- Formula to Determine Hydrolysis Constant

The hydrolysis constant \(K_{h}\) for a salt of a weak acid and weak base is determined using the formula: \[ K_{h} = \frac{K_w}{K_a \times K_b} \] where \(K_w\) is the ion-product constant for water, \(1 \times 10^{-14}\).

- Calculate the Hydrolysis Constant

Using the formula, substitute the values: \[ K_{h} = \frac{1 \times 10^{-14}}{2 \times 10^{-6} \times 5 \times 10^{-7}} \]

- Simplify the Expression

First, multiply the values in the denominator: \[ 2 \times 10^{-6} \times 5 \times 10^{-7} = 10 \times 10^{-13} = 1 \times 10^{-12} \] Then, divide by the \(K_w\): \[ K_{h} = \frac{1 \times 10^{-14}}{1 \times 10^{-12}} = 1 \times 10^{-2} \]

- Identify Closest Answer in Options

The closest option to \(1 \times 10^{-2}\) is 10. Therefore, the correct answer is option (1).

Key concepts

weak acid

A weak acid is a type of acid that only partially ionizes in solution. This means it doesn't completely dissociate into its ions in water. For example, acetic acid (CH\textsubscript{3}COOH) is a common weak acid. When dissolved in water, it partially ionizes to produce hydrogen ions (H\textsuperscript{+}) and acetate ions (CH\textsubscript{3}COO\textsuperscript{-}). The degree of ionization is represented by the acid dissociation constant, denoted as \(K_a\). The smaller the value of \(K_a\), the weaker the acid. For instance, in the given exercise, \(K_a = 2 \times 10^{-6}\), indicating that the acid does not ionize much in water. Understanding weak acids is critical since they play a significant role in the hydrolysis of salts involving weak acids and weak bases.

weak base

A weak base is similar to a weak acid in that it only partially ionizes in solution. Instead of donating protons like acids, bases accept protons. Ammonia (NH\textsubscript{3}) is a common example of a weak base. When dissolved in water, it partially ionizes to produce ammonium ions (NH\textsubscript{4}\textsuperscript{+}) and hydroxide ions (OH\textsuperscript{-}). The extent of this ionization is given by the base dissociation constant, \(K_b\). Lower \(K_b\) values indicate weaker bases. In the exercise above, \(K_b = 5 \times 10^{-7}\), showing that only a small fraction of the base ionizes in water. Weak bases play an essential role in many natural and industrial processes, including the hydrolysis of salts.

ion-product constant of water

The ion-product constant of water, represented as \(K_w\), is a vital concept in understanding acid-base chemistry. At 25°C, \(K_w\) has a value of \(1 \times 10^{-14}\). It is the product of the concentrations of hydrogen ions \( [H^+] \) and hydroxide ions \( [OH^-] \) in pure water. For example, in pure water, where \([H^+] = [OH^-]\), we can calculate \[ [H^+] \times [OH^-] = 1 \times 10^{-14} \text{ M}^2 \ [H^+] = [OH^-] = 1 \times 10^{-7} \text{ M} \] This constant is crucial in determining the pH of solutions and understanding the behavior of acids, bases, and salts in water. In the provided problem, \(K_w\) is used to find the hydrolysis constant for a salt formed from a weak acid and weak base.

acid-base equilibrium

Acid-base equilibrium refers to the balance reached in a solution when the rates of the forward and reverse reactions of acid and base ionization are equal. This equilibrium is crucial in determining the pH and reactivity of the solution. For weak acids and bases, which do not fully ionize, an equilibrium is established between the undissociated molecules and the ions produced. For instance, in the case of a weak acid HA dissociating in water: \[ HA \rightleftharpoons H^+ + A^- \] The position of this equilibrium is described by the \(K_a\) value for acids and \(K_b\) for bases. This concept is fundamental when dealing with buffer solutions, titrations, and hydrolysis reactions like the one in the exercise.

salt hydrolysis

Salt hydrolysis is the process through which an ion from a dissociated salt reacts with water to produce either an acidic or basic solution. This mainly happens with salts derived from weak acids and weak bases. For example, consider a salt formed from a weak acid (with \(K_a = 2 \times 10^{-6}\)) and a weak base (with \(K_b = 5 \times 10^{-7}\)). When such a salt dissolves in water, the resulting solution's pH depends on the relative strengths of the original acid and base. In the exercise above, we calculate the hydrolysis constant \(K_h\) using the formula \( K_h = \frac{K_w}{K_a \times K_b} \), where \(K_w\) is the ion-product constant of water, \(1 \times 10^{-14}\). This calculation helps determine whether the solution will be acidic or basic. An in-depth understanding of salt hydrolysis is crucial for predicting solution behavior and adjusting pH in various chemical applications.