Problem 224 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{K}_{\mathrm{a}_{1}} \times \mathrm{K}_{\mathrm{a}_{2}}\) for \(\mathrm{H}_{2} \mathrm{~S}=1.0 \times 10^{-21} \mathrm{M}^{2} .\) The concentration of \(\left[\mathrm{S}^{2-}\right]\) ion present in \(1 \mathrm{~L}\) of \(0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\) having \(\left[\mathrm{H}^{+}\right]\) equal to \(0.1 \mathrm{M}\) is \(\mathrm{x} \times 10^{-20}\). The value of \(\mathrm{x}\) is

Step-by-Step Solution

Answer

The value of \( x \) is 10.

Step 1: Writing the Given Information

We have the dissociation constant, \( K_{a_1} \times K_{a_2} = 1.0 \times 10^{-21} \text{ M}^2 \), and the initial concentration of \( H_2S \) is 0.1 M. The concentration of \( [H^+] \) is given as 0.1 M. We need to find the concentration of \([S^{2-}]\) in a 1 L solution and determine the value of \( x \) in \( x \times 10^{-20} \).

Step 2: Understanding Chemical Equilibrium

The dissociation of \( H_2S \) in water can be represented as: \( H_2S \rightleftharpoons H^+ + HS^- \) and then \( HS^- \rightleftharpoons H^+ + S^{2-} \). The \( K_{a_1} \) and \( K_{a_2} \) refer to these two dissociation reactions.

Step 3: Writing the Expression for \([S^{2-}]\)

Using the given \( K_{a_1} \times K_{a_2} \) value,\[ [S^{2-}] = \frac{K_{a_1} \times K_{a_2}}{[H^+]^2} \].

Step 4: Calculating \([S^{2-}]\) Concentration

Substitute \( [H^+] = 0.1 \text{ M} \) into the equation:\[ [S^{2-}] = \frac{1.0 \times 10^{-21}}{(0.1)^2} = \frac{1.0 \times 10^{-21}}{0.01} = 1.0 \times 10^{-19} \].

Step 5: Finding the Value of \( x \)

Given that \([S^{2-}] = x \times 10^{-20}\), from our calculation, \([S^{2-}] = 1.0 \times 10^{-19}\). Therefore, \( x = 10 \).

Key Concepts

Dissociation Constant

The dissociation constant, often represented as \( K_a \), is a crucial concept in understanding chemical equilibrium, particularly in acid-base reactions. This constant provides insight into the extent of dissociation for a weak acid or base in an aqueous solution. When we discuss the dissociation constant in the context of a compound like hydrogen sulfide \( (H_2S) \), it allows us to evaluate how much of the \( H_2S \) dissociates into its ions. Dissociation occurs when a compound splits into smaller chemical species. In the case of \( H_2S \), it dissociates in two steps:

• First, \( H_2S \) dissociates into \( H^+ \) and \( HS^- \).
• Second, \( HS^- \) further dissociates into \( H^+ \) and \( S^{2-} \).

Both phases of this process are characterized by their respective dissociation constants, \( K_{a_1} \) and \( K_{a_2} \). The product of these constants, \( K_{a_1} \times K_{a_2} \), relates to the overall level of dissociation from \( H_2S \) to \( S^{2-} \). Understanding these constants is vital for predicting the behavior of \( H_2S \) in solution and is crucial for performing accurate chemical calculations, especially when dealing with equilibrium dynamics.

Hydrogen Sulfide

Hydrogen sulfide \((H_2S)\) is a colorless gas known for its pungent smell, reminiscent of rotten eggs. It is a weak dibasic acid and can dissolve in water to form a slightly acidic solution. Understanding the properties of hydrogen sulfide helps in grasping how it behaves in solution.When \( H_2S \) is added to water, it partially dissociates, meaning it does not completely split into ions. Instead, it reaches a point of equilibrium where the rate of dissociation of \( H_2S \) into \( H^+ \) and \( HS^- \) ions equals the rate at which these ions combine to reform \( H_2S \). The dissociation is a two-step process, reflecting its nature as a dibasic acid:

• Initially, \( H_2S \rightarrow H^+ + HS^- \)
• This is followed by \( HS^- \rightarrow H^+ + S^{2-} \)

The two dissociation steps emphasize its weak acid behavior, indicating that even in solutions, not all \( H_2S \) will dissociate, affecting the overall ionic concentration. Comprehending these dissociation pathways is key to solving chemical equilibrium problems involving \( H_2S \).

Ionic Concentration

Ionic concentration refers to the amount of individual ion species in a solution, and it's a fundamental concept in chemistry when dealing with dissolution and dissociation processes. When a compound like hydrogen sulfide \((H_2S)\) dissolves in water, it partially disassociates into ions. The concentration of these ions determines many properties of the solution, such as pH and electrical conductivity.In the dissociation of \( H_2S \), ionic concentration becomes significant because:

• The solution starts with a given concentration of \([H_2S]\simeq 0.1\,\text{M}\).
• When the equilibrium is reached, there will be concentrations of \([H^+]\), \([HS^-]\), and \([S^{2-}]\).

Calculating the concentration of \([S^{2-}]\) involves using the equilibrium constants \( K_{a_1} \) and \( K_{a_2} \). Through these calculations, one can find the concentration of each ion, allowing us to better understand and predict the behavior of the solution. These values are crucial for reactions where ionic strength impacts the outcome, such as in electrochemistry and biological systems.

Chemical Calculation

Chemical calculations are essential in predicting the outcome and progression of chemical reactions. They involve using known values and constants to find unknown quantities. In our example with hydrogen sulfide \((H_2S)\), we apply chemical calculations to determine the concentration of sulfate ions \([S^{2-}]\) in the solution.To perform such calculations:

• First, write down what is known, such as concentrations and equilibrium constants.
• Use the given equilibrium constant product \( K_{a_1} \times K_{a_2} \) to set up an equation representing the dissociation.

For \([S^{2-}]\), the equation is:\[[S^{2-}] = \frac{K_{a_1} \times K_{a_2}}{[H^+]^2}\]Plug in the known values, which in our exercise are \( K_{a_1} \times K_{a_2} = 1.0 \times 10^{-21} \text{ M}^2 \) and \( [H^+] = 0.1 \text{ M} \), to find \([S^{2-}]\). Substituting these into the equation yields \( 1.0 \times 10^{-19} \text{ M} \). From this, students deduce the value of \( x \) in \( x \times 10^{-20} \), an essential skill in approaching quantitative chemistry problems.