Problem 125 - A Complete Resource Book in Chemistry for JEE Main

Question

Ionization constant of \(\mathrm{CH}_{3} \mathrm{COOH}\) is \(1.7 \times 10^{-5}\) and concentration of \(\mathrm{H}^{+}\)ion is \(3.4 \times 10^{-4}\). Then initial concentration of \(\mathrm{CH}_{3} \mathrm{COOH}\) is (a) \(3.4 \times 10^{-4}\) (b) \(3.4 \times 10^{-3}\) (c) \(6.8 \times 10^{-4}\) (d) \(6.8 \times 10^{-3}\)

Step-by-Step Solution

Answer

The initial concentration of \\(\mathrm{CH}_3\mathrm{COOH}\\) is \\(6.8 \times 10^{-3}\\).

Step 1: Write the Ionization Equation

The ionization of acetic acid, \(\mathrm{CH}_3\mathrm{COOH}\), is represented by: \(\mathrm{CH}_3\mathrm{COOH} \rightleftharpoons \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+\). This shows the dissociation into acetate ion and proton.

Step 2: Write the Expression for the Ionization Constant

The ionization constant expression \(K_a\) for acetic acid is given by: \[K_a = \frac{[\mathrm{CH}_3\mathrm{COO}^-][\mathrm{H}^+]}{[\mathrm{CH}_3\mathrm{COOH}]} \].

Step 3: Substitute the Given Values

We have \(K_a = 1.7 \times 10^{-5}\) and \([\mathrm{H}^+] = 3.4 \times 10^{-4}\) which implies \([\mathrm{CH}_3\mathrm{COO}^-] = [\mathrm{H}^+]\) in the expression. Substitute these values into the ionization constant equation: \[1.7 \times 10^{-5} = \frac{(3.4 \times 10^{-4})(3.4 \times 10^{-4})}{[\mathrm{CH}_3\mathrm{COOH}]_0} \].

Step 4: Solve for Initial Concentration

Rearrange the equation to find the initial concentration of acetic acid: \[ [\mathrm{CH}_3\mathrm{COOH}]_0 = \frac{(3.4 \times 10^{-4})^2}{1.7 \times 10^{-5}} \]. Calculating the numerator: \(3.4 \times 10^{-4} \times 3.4 \times 10^{-4} = 1.156 \times 10^{-7}\). Divide this by the ionization constant: \[ [\mathrm{CH}_3\mathrm{COOH}]_0 \approx \frac{1.156 \times 10^{-7}}{1.7 \times 10^{-5}} = 6.8 \times 10^{-3} \].

Key Concepts

Acetic Acid

Acetic Acid, scientifically known as \(\mathrm{CH}_3\mathrm{COOH}\), is a common weak acid found in vinegar.It is a simple carboxylic acid with a distinct sour taste and pungent smell.As it is classified as a weak acid, acetic acid does not completely dissociate into ions in water.This partial dissociation is an important concept, as it allows us to use equilibrium expressions to understand its behavior in solutions.
When discussing acetic acid, it's crucial to be familiar with the idea of acid dissociation, which refers to how many of the acid molecules release hydrogen ions into the solution. This is fundamental in calculating its ionization constant and initial concentration during chemical reactions.
Some key points about acetic acid include:

• It participates in the process of equilibrium, meaning the reaction doesn't go to completion.
• It is used widely in the food industry, as well as in chemical synthesis and even as a reagent in laboratories.
• The ionization of acetic acid can be measured using its ratio of dissociated ions to the undissociated molecule in a solution.

Ionization Equation

A crucial aspect to understanding the behavior of acetic acid in a solution is through the Ionization Equation.The Ionization Equation provides a mathematical representation of how acetic acid dissociates in water.
For acetic acid, the equation is:\[ \mathrm{CH}_3\mathrm{COOH} \rightleftharpoons \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+ \]This equation shows that acetic acid (78COOH) breaks into an acetate ion (78COO9) and a hydrogen ion (9).The double arrows indicate that this process is reversible and does not go to completion. Understanding this equilibrium is key to discerning the degree to which acetic acid ionizes in water. This equilibrium dynamic indicates the proportion of the undissociated acetic acid remaining compared to the ions that have formed.This balance is critical in calculating both the ionization constant and the initial concentration.

Ionization Constant Expression

The Ionization Constant Expression, denoted as \(K_a\), is a vital metric in gauging the strength of a weak acid such as acetic acid.This constant quantifies the extent to which acetic acid dissociates in the solution.For acetic acid, the expression is given by:\[ K_a = \frac{[\mathrm{CH}_3\mathrm{COO}^-][\mathrm{H}^+]}{[\mathrm{CH}_3\mathrm{COOH}]} \]
Here, \([\mathrm{CH}_3\mathrm{COO}^-]\) and \([\mathrm{H}^+]\) represent the concentrations of the acetate and hydrogen ions, respectively.Whereas \([\mathrm{CH}_3\mathrm{COOH}]\) is the concentration of undissociated acetic acid.
Evaluating \(K_a\) helps us understand the extent of ionization of acetic acid under different conditions.A small \(K_a\) value indicates that only a minor fraction of acetic acid is ionized in water, confirming its status as a weak acid.By substituting specific concentration values into this expression, we can solve for unknown quantities such as the initial concentration of acetic acid.

Initial Concentration Calculation

Calculating the Initial Concentration of acetic acid involves leveraging the ionization constant and known concentrations of ions in solution.Given that the ionization constant \(K_a\) for acetic acid is \(1.7 \times 10^{-5}\), and the concentration of \([\mathrm{H}^+]\) is \(3.4 \times 10^{-4}\), we assume \([\mathrm{CH}_3\mathrm{COO}^-] = [\mathrm{H}^+]\).
From this setup, the expression for the ionization constant can be rearranged to find the initial concentration \([\mathrm{CH}_3\mathrm{COOH}]_0\):\[1.7 \times 10^{-5} = \frac{(3.4 \times 10^{-4})(3.4 \times 10^{-4})}{[\mathrm{CH}_3\mathrm{COOH}]_0} \]
Now, solving for \([\mathrm{CH}_3\mathrm{COOH}]_0\) involves determining:

• The product of \(3.4 \times 10^{-4} \) by itself, yielding \(1.156 \times 10^{-7}\).
• Dividing this product by the ionization constant, \(1.7 \times 10^{-5}\), yielding \(6.8 \times 10^{-3}\).

The initial concentration of acetic acid can thus be determined, enhancing our understanding of the solution's behavior.