Problem 109 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(0.1 \mathrm{M}\) of a weak monobasic acid is taken and its percentage degree of ionization is \(1.34 \%\), then calculate its ionization constant (a) \(0.8 \times 10^{-5}\) (b) \(1.79 \times 10^{-5}\) (c) \(0.182 \times 19^{-5}\) (d) none of these

Step-by-Step Solution

Answer

The ionization constant is \(1.79 \times 10^{-5}\), which corresponds to option (b).

Step 1: Understand the Problem

We need to find the ionization constant, also known as the equilibrium constant, for a weak monobasic acid where we are given the concentration of the acid and the percentage degree of ionization.

Step 2: Write the Ionization Equation

For a weak monobasic acid HA, it ionizes in water as follows: \[ \mathrm{HA} \rightleftharpoons \mathrm{H}^{+} + \mathrm{A}^{-} \]

Step 3: Initial Concentrations

The initial concentration of the acid (\( \mathrm{HA} \)) is given as \(0.1 \, \mathrm{M}\). Before ionization, the concentrations of \( \mathrm{H}^+ \) and \( \mathrm{A}^- \) are 0.

Step 4: Calculate Change in Concentrations

Using the percentage ionization, calculate the change in concentration. \(1.34\%\) of \(0.1\, \mathrm{M}\) is ionized, so the change is \(\frac{1.34}{100} \times 0.1 = 0.00134\, \mathrm{M}\).

Step 5: Equilibrium Concentrations

At equilibrium, the concentration of \(\mathrm{HA} = 0.1 - 0.00134\), and the concentrations of \(\mathrm{H}^+\) and \(\mathrm{A}^-\) both become \(0.00134\).

Step 6: Calculate the Ionization Constant

The ionization constant \(K_a\) is defined as: \[ K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]} = \frac{(0.00134)(0.00134)}{0.1 - 0.00134} \] Calculate \(K_a\) to find \( K_a \approx 1.79 \times 10^{-5}\).

Step 7: Choose the Correct Option

The calculated ionization constant is \(1.79 \times 10^{-5}\). Thus, the correct answer is option (b).

Key Concepts

Weak Monobasic Acid

A weak monobasic acid is a type of acid that partially ionizes in aqueous solutions. This means that when dissolved in water, only a small portion of the acid molecules dissociates into ions. In chemistry, it's important to distinguish between strong and weak acids. While strong acids completely ionize, weak acids do not. This partial ionization is what characterizes weak acids and affects calculations like the ionization constant. Monobasic means the acid can donate one proton (hydrogen ion) per molecule during the ionization process. A common example of a monobasic acid is acetic acid, which is found in vinegar. When dealing with a monobasic acid, especially in a problem like this, it's crucial to understand its behavior in water, since this will impact the equilibrium concentrations and thus the calculation of the ionization constant.

Percentage Degree of Ionization

The percentage degree of ionization refers to the fraction of the acid that ionizes in a solution, expressed as a percentage. In our problem, the ionization percentage is given as 1.34%. This means that out of the total concentration of the acid, only 1.34% gets ionized into its corresponding ions.To calculate this percentage, chemists use the formula:

• \(\text{Percentage Ionization} = \left( \frac{\text{Concentration of ionized acid}}{\text{Initial concentration of acid}} \right) \times 100 \%)\)

This percentage is a useful way to describe the strength and behavior of the acid in a solution. The higher the percentage, the stronger the ionization at given conditions, but for weak acids like in this exercise, it's typically low.

Equilibrium Constant

The equilibrium constant, often represented as \(K_a\) for acids, is a measure of the ionization strength of an acid in a solution. It quantifies the ratio of the concentration of the ionized form of the acid to the non-ionized form at equilibrium. For weak acids, the \(K_a\) value is usually small, reflecting the incomplete ionization.The formula for the ionization constant is given by:

• \[K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\]

Each species in the equation represents the concentration at equilibrium:

• \([\text{H}^+]\): concentration of hydrogen ions
• \([\text{A}^-]\): concentration of the conjugate base
• \([\text{HA}]\): concentration of the non-ionized acid

By solving this, we can predict the behavior and acidity of a weak acid in aqueous solutions. In our problem, the calculated \(K_a\) is \(1.79 \times 10^{-5}\), showing the weak nature of the acid.

Ionization Equation

The ionization equation of an acid represents how it dissociates in water. For a weak monobasic acid \(\text{HA}\), the equation is:

• \(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\)

This depicts the equilibrium between the undissociated molecules and the ions produced. In this reversible reaction, the forward reaction represents the dissociation of the acid into hydrogen ions and conjugate base ions, while the reverse shows the recombination back into the original acid.Understanding this equation is vital, as it forms the basis for calculating the ionization constant. It allows us to quantify how much of the acid remains un-ionized and how much is converted into ions. This balance between the ions and the non-ionized acid is crucial for determining the \(K_a\) of the acid, as shown by the equilibrium concentrations used in our calculation.