Problem 64 - A Complete Resource Book in Chemistry for JEE Main

Question

If pKb for fluoride ion at \(25^{\circ} \mathrm{C}\) is \(10.83\), the ionization constant of hydrofluoric acid at this temperature is (a) \(1.74 \times 10^{-5}\) (b) \(3.52 \times 10^{-3}\) (c) \(6.75 \times 10^{-4}\) (d) \(5.38 \times 10^{-2}\)

Step-by-Step Solution

Answer

The ionization constant of hydrofluoric acid is \(6.75 \times 10^{-4}\), so the answer is (c).

Step 1: Understanding the Relationship Between Constants

The dissociation constant of an acid, known as the acid dissociation constant (Ka), is related to the base dissociation constant (Kb) of its conjugate base by the water ion-product constant (Kw). The relationship between these constants is given by the equation: \[ K_a \times K_b = K_w \] where \( K_w \) is the ion-product constant of water at 25°C, which is \( 1 \times 10^{-14} \).

Step 2: Calculation of Kb for Hydrofluoric Acid

First, convert the provided pKb value for the fluoride ion to its Kb value using the formula: \[ K_b = 10^{-pK_b} \] Substitute \(pK_b = 10.83\) into the formula to get \( K_b = 10^{-10.83} \).

Step 3: Determine Ka for Hydrofluoric Acid

Using the relationship \( K_a \times K_b = K_w \), solve for \( K_a \) by rearranging the equation to \( K_a = \frac{K_w}{K_b} \). Substitute \(K_w = 1 \times 10^{-14}\) and the calculated \(K_b\) from the previous step into the equation to find \(K_a\).

Step 4: Calculate the Numeric Value of Ka

Convert \(pK_b = 10.83\) to \(K_b\): \[ K_b \approx 1.479 \times 10^{-11} \]. Now substitute into \( K_a = \frac{1 \times 10^{-14}}{1.479 \times 10^{-11}} \) to find \( K_a \).

Step 5: Choose the Correct Answer from the Options

From the calculation, \( K_a \approx 6.75 \times 10^{-4} \). This value matches option (c) from the list provided.

Key Concepts

Acid Dissociation Constant (Ka)

When an acid dissolves in water, it partially dissociates into ions. This process is characterized by the acid dissociation constant, abbreviated as \( K_a \). The value of \( K_a \) gives us an idea of the strength of the acid. Strong acids have high \( K_a \) values because they dissociate more completely in water, while weak acids have lower \( K_a \) values due to only partial dissociation.

To calculate \( K_a \), we examine the equilibrium concentration of the products and reactants. For a general acid \( HA \) dissociating into \( H^+ \) and \( A^- \), the expression is:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Understanding \( K_a \) is crucial for predicting how an acid behaves in solution, affecting pH and reactivity. In the context of hydrofluoric acid, calculating \( K_a \) involves using the related \( K_b \) and the water ion-product constant \( K_w \). These relationships are key for chemical equilibrium calculations.

Base Dissociation Constant (Kb)

The base dissociation constant, or \( K_b \), provides a measure of the base's strength. It quantifies how effectively a base dissociates in solution to form its conjugate acid and hydroxide ions. Bases with higher \( K_b \) values are stronger, indicating greater dissociation in water.

For a base \( B \) dissociating into its conjugate acid \( BH^+ \) and \( OH^- \), \( K_b \) is calculated as:
\[ K_b = \frac{[BH^+][OH^-]}{[B]} \]

When determining \( K_a \) for an acid like hydrofluoric acid, we first consider the \( K_b \) of its conjugate base, the fluoride ion \( F^- \). The exercise provides a \( pK_b \) value, which is converted to \( K_b \). This converted \( K_b \) is then used along with \( K_w \) to find the acid's \( K_a \) via \( K_a \times K_b = K_w \). Understanding \( K_b \) helps in assessing the overall equilibrium in acid-base reactions.

Ion-Product Constant of Water (Kw)

Water naturally dissociates into hydrogen ions \( H^+ \) and hydroxide ions \( OH^- \). The ion-product constant of water, represented as \( K_w \), is the equilibrium constant for this self-ionization of water. At \( 25^\circ C \), \( K_w \) equals \( 1 \times 10^{-14} \).

This constant is fundamental in connecting \( K_a \) and \( K_b \) for acid-base pairs. The equation \( K_a \times K_b = K_w \) highlights how these constants interplay. If one value is known (either \( K_a \) or \( K_b \)), the other can be easily determined using \( K_w \).

\( K_w \) is pivotal in understanding the neutral point of water in terms of pH, which is 7 at \( 25^\circ C \). It also underpins our ability to calculate pH, pOH, and concentrations of ions in any aqueous solution. Thus, in calculating the \( K_a \) of hydrofluoric acid, \( K_w \) offers the crucial link in the relationship between its \( K_b \) and \( K_a \).