Problem 146 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant for the reaction given is \(3.6 \times 10^{-7}\) \(\mathrm{OCl}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{HOCl}(\mathrm{aq})+\mathrm{OH}^{-}\)(aq) What is \(\mathrm{Ka}\) for HOCl? (a) \(2.8 \times 10^{-8}\) (b) \(3.6 \times 10^{-7}\) (c) \(6 \times 10^{-4}\) (d) \(2.8 \times 10^{-6}\)

Step-by-Step Solution

Answer

The answer is (a) \(2.8 \times 10^{-8}\).

Step 1: Understanding the Problem

We are given the equilibrium constant \(K_c\) for the reaction \(\mathrm{OCl}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{HOCl} + \mathrm{OH}^{-}\), and we need to find the acid dissociation constant \(K_a\) for \(\mathrm{HOCl}\).

Step 2: Identify Reaction Relationships

Recognize that the reverse reaction of the given equilibrium is the acid dissociation for \(\mathrm{HOCl}\), where \(\mathrm{HOCl}\) dissociates into \(\mathrm{H}^+\) and \(\mathrm{OCl}^-\). The relationship between \(K_a\) and \(K_c\) is given by the ion product of water \(K_w = 10^{-14}\).

Step 3: Use the Relationship between Equilibrium Constants

For two equilibrium reactions related by reverse processes, the relationship is \(K_w = K_c \cdot K_a\). Use this formula to solve for \(K_a\).

Step 4: Calculate \(K_a\)

Calculate \(K_a\) using \(K_w = K_c \cdot K_a\), therefore, \(K_a = \frac{K_w}{K_c}\). Substitute the known values: \(K_a = \frac{10^{-14}}{3.6 \times 10^{-7}}\).

Step 5: Simplify the Calculation

Divide \(10^{-14}\) by \(3.6 \times 10^{-7}\):\[K_a = \frac{10^{-14}}{3.6 \times 10^{-7}} \approx 2.8 \times 10^{-8}\]

Step 6: Select the Correct Answer

Compare the calculated \(K_a\) to the given options and find that the answer closest to \(2.8 \times 10^{-8}\) is option (a).

Key Concepts

Acid Dissociation Constant

The acid dissociation constant, abbreviated as \(K_a\), is a crucial concept when discussing acids and their behavior in water. It quantifies the strength of a weak acid in a solution by showing the degree to which the acid dissociates into its ions. Specifically, for an acid \(HA\), the dissociation can be described as:

\[HA (aq) \rightleftharpoons H^+ (aq) + A^- (aq)\]

Here, \(HA\) represents the acid, \(H^+\) the hydrogen ion, and \(A^-\) the conjugate base. The \(K_a\) is given by the expression:

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

A larger \(K_a\) value indicates a stronger acid, which means it dissociates more completely in water.
Conversely, a smaller \(K_a\) signifies a weaker acid, with less tendency to release its hydrogen ions. This concept is crucial when determining the acidic strength of a substance used in a reaction.

Equilibrium Constant

When a chemical reaction reaches a state where the reactants and products no longer change in concentration, it is said to be in equilibrium. The equilibrium constant, \(K_c\), numerically describes the state of equilibrium for a reversible reaction. It's given by the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients. Considering the reaction:

\[aA + bB \rightleftharpoons cC + dD\]

The expression for \(K_c\) is:

\[K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]

Where [C], [D], [A], and [B] are the molar concentrations of the substances at equilibrium. The equilibrium constant gives insight into the position of equilibrium:

• If \(K_c >> 1\), the reaction favors products.
• If \(K_c << 1\), the reaction predominantly contains reactants at equilibrium.

Knowing the \(K_c\) of a reaction helps predict how changes in conditions (like concentration or temperature) could affect the system's equilibrium position.

Ion Product of Water

Water is considered neutral, but it auto-ionizes slightly, meaning a small portion of water molecules dissociate into ions. The ion product of water, \(K_w\), represents this auto-ionization. The reaction can be expressed as:

\[2H_2O (l) \rightleftharpoons H_3O^+ (aq) + OH^- (aq)\]

Also simplified to:

\[H_2O (l) \rightleftharpoons H^+ (aq) + OH^- (aq)\]

The \(K_w\) at 25°C (298 K) is constantly \(1.0 \times 10^{-14}\). This is given by:

\[K_w = [H^+][OH^-]\]

This ion product is a fundamental feature of aqueous chemistry and plays a vital role in understanding concepts like pH and acid-base reactions. Since \(K_w\) is constant at a given temperature, any shift in \([H^+]\) or \([OH^-]\) can be used to calculate the pH or pOH of a solution. Furthermore, it links \(K_a\) and \(K_b\), the equilibrium constant for the base reactions, illustrating their interdependence in aqueous solutions.