Problem 246 - A Complete Resource Book in Chemistry for JEE Main

Question

A monoprotic acid in a \(0.1 \mathrm{M}\) solution ionizes to \(0.001 \%\). If its ionization constant is \(\frac{10^{-\mathrm{x}}}{100}\), the value of \(\mathrm{x}\) is

Step-by-Step Solution

Answer

The value of \(\mathrm{x}\) is 9.

Step 1: Understanding the Problem

We have a monoprotic acid in a solution with a concentration of \(0.1 \text{ M}\). The acid ionizes to \(0.001\%\). We need to find the value of \(\mathrm{x}\) such that the ionization constant \(K_a\) is given by \(\frac{10^{- ext{x}}}{100}\).

Step 2: Calculate the Ionization Concentration

First, calculate the concentration of ions produced by the acid. Since it ionizes to \(0.001\%\), the ion concentration is \(0.001\%\) of \(0.1 \text{ M}\). This is calculated as follows: \[ [\text{H}^+] = 0.1 \times \frac{0.001}{100} = 10^{-6} \text{ M} \]

Step 3: Write the Ionization Constant Expression

The ionization of a monoprotic acid \(HA\) can be expressed as \(HA \rightleftharpoons H^+ + A^-\). The original concentration is approximately \(0.1 \text{ M}\), and the concentration of \(H^+\) ions is \(10^{-6} \text{ M}\). Therefore, \(K_a\) is given by:\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(10^{-6})(10^{-6})}{0.1} = 10^{-11} \]

Step 4: Determine the Value of x

We are given that \(K_a = \frac{10^{- ext{x}}}{100} = 10^{-11}\). Simplifying, we multiply both sides by 100 to get:\[ 10^{- ext{x}} = 10^{-9} \]This implies that \(-\text{x} = -9\), hence \(\text{x} = 9\).

Key Concepts

Monoprotic acid

A monoprotic acid is a type of acid that can donate only one proton (hydrogen ion, \(H^+\)) per molecule in a chemical reaction. This characteristic is important when considering the acid's ionization in water. Upon dissolution, a monoprotic acid will achieve equilibrium by partially dissociating into \(H^+\) ions and its conjugate base.

• An example of a monoprotic acid is hydrochloric acid (\(HCl\)).
• In general, the reaction of a monoprotic acid (\(HA\)) in water can be represented as: \(HA ightleftharpoons H^+ + A^-\).

Understanding monoprotic acids allows us to calculate the ionization percentage which shows how much of the acid dissociates, leading us to insights on the concentration of ions and the strength of the acid.

Ionization percentage

The ionization percentage refers to the extent to which an acid dissociates into ions in a solution. It is the ratio of the concentration of ions produced to the initial concentration of the acid, multiplied by 100. It tells us how much of the acid becomes ionized in the solution.

• For weak acids, the ionization percentage is typically low because they do not fully dissociate in solution.
• In the given exercise, the monoprotic acid ionizes to 0.001% of its initial concentration, which is very low, indicating it is a weak acid.

Understanding the ionization percentage helps us determine the concentration of \(H^+\) ions, which is crucial for calculating the ionization constant \(K_a\).

Concentration of ions

The concentration of ions, specifically \([H^+]\), in a solution is pivotal when evaluating the strength of an acid and understanding its chemical behavior. It is directly affected by the degree of ionization.

• For the monoprotic acid in the problem, with an initial concentration of \(0.1 \text{ M}\) and ionization of 0.001%, the concentration of \(H^+\) ions is calculated as follows:

\[ [H^+] = 0.1 \times \frac{0.001}{100} = 10^{-6} \text{ M} \]
This means that very few \(H^+\) ions are produced in the solution, once again highlighting the weak nature of the acid. Determining the concentration of ions is an essential step in the path to finding the ionization constant and understanding chemical equilibrium.

Chemical equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time, given that the forward and reverse reactions occur at the same rate. For a monoprotic acid in solution, equilibrium is reached when the rate of dissociation of \(HA\) into \(H^+\) and \(A^-\) balances the rate of recombination back into \(HA\).

• The equation \(K_a = \frac{[H^+][A^-]}{[HA]}\) helps to express this balance mathematically.
• In the exercise, the equilibrium concentration of \(H^+\) and \(A^-\) is \(10^{-6}\text{ M}\), while the remaining \(HA\) is approximately \(0.1 \text{ M}\), leading us to an ionization constant \(K_a = 10^{-11}\).

Understanding chemical equilibrium is crucial to calculating the ionization constant \(K_a\) and further understanding the behavior of acids in aqueous solutions.