Problem 123 - Chemistry: An Atoms-Focused Approach

Question

Flements of group 16 form hydrides with the generic formula \(\mathrm{H}_{2} \mathrm{X}\). When gaseous \(\mathrm{H}_{2} \mathrm{X}\) is bubbled through a solution containing \(0.3 M\) hydrochloric acid, the solution becomes saturated and \(\left[\mathrm{H}_{2} \mathrm{X}\right]=0.1 \mathrm{M} .\) The following equilibria exist in this solution: \(\mathrm{H}_{2} \mathrm{X}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{HX}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \quad K_{1}=8.3 \times 10^{-8}\) \(\mathrm{HX}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{X}^{2-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \quad K_{2}=1 \times 10^{-14}\) Calculate the concentration of \(\mathrm{X}^{2-}\) in the solution.

Step-by-Step Solution

Answer

Question: Calculate the concentration of \(\mathrm{X}^{2-}\) ions when group 16 hydrides (\(\mathrm{H}_{2}\mathrm{X}\)) dissolve in a \(0.3\,\mathrm{M}\) hydrochloric acid solution, given the equilibrium concentrations, constants, and reactions mentioned above. Answer: The concentration of \(\mathrm{X}^{2-}\) ions in the solution is \(2.77 \times 10^{-23}\,\mathrm{M}\).

Step 1: Write the equilibrium expressions for the two reactions

Based on the given reactions, we can write the equilibrium constant expressions for each reaction: For the first reaction: \(K_{1} = \frac{\left[\mathrm{HX}^{-}\right]\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]}{\left[\mathrm{H}_{2}\mathrm{X}\right]}\) For the second reaction: \(K_{2} = \frac{\left[\mathrm{X}^{2-}\right]\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]}{\left[\mathrm{HX}^{-}\right]}\)

Step 2: Find the concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\) from the given information

The concentration of hydrochloric acid (\(\mathrm{HCl}\)) is given as \(0.3\,\mathrm{M}\). When \(\mathrm{HCl}\) dissociates in water, it forms equal concentrations of \(\mathrm{H}_{3}\mathrm{O}^{+}\) and \(\mathrm{Cl}^{-}\) ions. Therefore, the initial concentration of \(\mathrm{H}_{3}\mathrm{O}^{+}\) ions can be taken as equal to the concentration of \(\mathrm{HCl}\), which is \(0.3\,\mathrm{M}\).

Step 3: Solve the first equilibrium expression for \(\left[\mathrm{HX}^{-}\right]\) in terms of \(\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]\) and \(K_{1}\)

Rearrange the first equilibrium expression to solve for \(\left[\mathrm{HX}^{-}\right]\): \(\left[\mathrm{HX}^{-}\right] = \frac{\left[\mathrm{H}_{2}\mathrm{X}\right]K_{1}}{\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]}\) We are given that at equilibrium \(\left[\mathrm{H}_{2}\mathrm{X}\right]=0.1\,\mathrm{M}\) and \(K_{1}=8.3 \times 10^{-8}\). Thus, \(\left[\mathrm{HX}^{-}\right] = \frac{(0.1)(8.3 \times 10^{-8})}{0.3} = \frac{8.3 \times 10^{-9}}{3}\)

Step 4: Solve the second equilibrium expression for \(\left[\mathrm{X}^{2-}\right]\) in terms of \(\left[\mathrm{HX}^{-}\right]\), \(\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]\) and \(K_{2}\)

Rearrange the second equilibrium expression to solve for \(\left[\mathrm{X}^{2-}\right]\): \(\left[\mathrm{X}^{2-}\right]= \frac{\left[\mathrm{HX}^{-}\right]K_{2}}{\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]}\) We are given that \(K_{2}=1 \times 10^{-14}\). Thus, \(\left[\mathrm{X}^{2-}\right]= \frac{(8.3 \times 10^{-9}/3)(1 \times 10^{-14})}{0.3} =\frac{(8.3 \times 10^{-23})}{3}\)

Step 5: Calculate the concentration of \(\mathrm{X}^{2-}\)

Now, calculate the concentration of \(\mathrm{X}^{2-}\): \(\left[\mathrm{X}^{2-}\right] = \frac{8.3 \times 10^{-23}}{3} = 2.77 \times 10^{-23}\,\mathrm{M}\) The concentration of \(\mathrm{X}^{2-}\) in the solution is \(2.77 \times 10^{-23}\,\mathrm{M}\).

Key Concepts

Group 16 Elements

Group 16 elements, also known as chalcogens, include oxygen, sulfur, selenium, tellurium, and polonium. These elements are found in the periodic table's 16th column and are known for their ability to form compounds with hydrogen, resulting in hydrides. Examples include water (\(\mathrm{H}_2\mathrm{O}\) ) from oxygen and hydrogen, and hydrogen sulfide (\(\mathrm{H}_2\mathrm{S}\) ) from sulfur and hydrogen.

• The most chemically significant property of Group 16 elements is their tendency to form stable hydrides.
• The reactivity of these hydrides typically increases as you move down the group due to increasing atomic size and decreasing electronegativity.
• For example, while water is a stable compound under normal conditions, hydrogen sulfide is more reactive and exhibits acidic properties.

Understanding these elements and their behavior in forming hydrides is crucial to mastering concepts of chemical equilibrium and acid-base reactions involving these compounds.

Equilibrium Constant

The equilibrium constant, represented as \( K \) , is an essential concept in chemical equilibrium. It quantifies the ratio of concentrations of products to reactants at equilibrium for a given reversible reaction. This constant helps chemists predict the direction of the chemical reaction and the extent to which reactants are converted to products.

• An equilibrium constant with a value greater than one (\( K > 1 \) ) indicates that the products are favored at equilibrium.
• Conversely, an equilibrium constant less than one (\( K < 1 \) ) shows that the reactants are favored.
• For example, in the exercise above, we deal with two reactions, each having its own equilibrium constant (\( K_1 \) and \( K_2 \) ).

The concept of equilibrium constants is crucial for determining concentrations of compounds in a chemical mixture, as in the given exercise, where it aids in calculating the concentration of \( \mathrm{X}^{2-} \) .

Acid-Base Equilibrium

Acid-base equilibrium involves the balance between acids and bases in a solution, which often revolves around the transfer of protons (\( \mathrm{H}^+ \) ions) between molecules. In relation to the exercise, when \( \mathrm{H}_{2} \mathrm{X} \) (a hydride) dissolves in water, it establishes an acid-base equilibrium involving the formation of hydronium ions (\( \mathrm{H}_{3} \mathrm{O}^{+} \) ).

• The first equilibrium reaction shown (\( \mathrm{H}_{2} \mathrm{X} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{HX}^{-} + \mathrm{H}_{3} \mathrm{O}^{+} \) ) represents the acid (\( \mathrm{H}_{2} \mathrm{X} \) ) donating a proton to water, acting as a base.
• The second reaction (\( \mathrm{HX}^{-} + \mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{X}^{2-} + \mathrm{H}_{3} \mathrm{O}^{+} \) ) continues this trend where \( \mathrm{HX}^{-} \) acts as an acid.

By analyzing these equilibria, chemists can predict the behavior of acids and bases in various solutions, a foundational aspect of chemistry that is crucial for both academic study and practical applications.

Hydride Formation

Hydrides are compounds formed by the combination of hydrogen with other elements. For Group 16 elements, the formation of hydrides is an important chemical process. These hydrides generally have the formula \( \mathrm{H}_2\mathrm{X} \) , where \( \mathrm{X} \) is a chalcogen.

• Hydrides can be classified into different types depending on their bonding and properties such as ionic, covalent, and metallic hydrides.
• In the case of the exercise, the hydride \( \mathrm{H}_{2}\mathrm{X} \) represents a typical behavior of Group 16 hydrides, usually covalent and gaseous in nature like \( \mathrm{H}_{2}\mathrm{S} \) .
• The intriguing chemical behavior of these hydrides when they interact with water, leading to the formation of different ionic species, underlines their reactive nature and role in chemical equilibria.

The ability of hydrides to participate in acid-base reactions and form varying products in solution makes them a critical component for understanding complex chemical systems.