Problem 95 - Chemistry The Science in Context

Question

Elements of group 16 form hydrides with the generic formula \(\mathrm{H}_{2} \mathrm{X}\). At a certain temperature, when gaseous \(\mathrm{H}_{2} \mathrm{X}\) is bubbled through a solution containing \(0.3 \mathrm{M}\) 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

The concentration of the X²⁻ ion in the solution is approximately 8.6 × 10⁻¹⁵ M.

Step 1: Write the expressions for the equilibrium constants K1 and K2

For each equilibrium equation, we can write expressions for the constants K1 and K2: For the first equilibrium: $$K_1 = \frac{[\mathrm{HX}^{-}][\mathrm{H}_{3}\mathrm{O}^{+}]}{[\mathrm{H}_{2}\mathrm{X}]}$$ For the second equilibrium: $$K_2 = \frac{[\mathrm{X}^{2-}][\mathrm{H}_{3}\mathrm{O}^{+}]}{[\mathrm{HX}^{-}]}$$ We know that at saturation, the concentration of \(\mathrm{H}_{2}\mathrm{X}\) is 0.1 M. Our goal is to find \([\mathrm{X}^{2-}]\) in the solution.

Step 2: Relate the concentrations of the species in solution

Since the hydrogen ion, \(\mathrm{H}_{3}\mathrm{O}^{+}\), comes from both the equilibrium reactions and the hydrochloric acid, we can write its concentration as follows: $$[\mathrm{H}_{3}\mathrm{O}^{+}] = [\mathrm{HX}^{-}] + 0.3$$ Now, we can substitute this expression into the expression for \(K_1\): $$K_1 = \frac{[\mathrm{HX}^{-}][\mathrm{HX}^{-} + 0.3]}{0.1}$$ And into the expression for \(K_2\): $$K_2 = \frac{[\mathrm{X}^{2-}][\mathrm{HX}^{-} + 0.3]}{[\mathrm{HX}^{-}]}$$

Step 3: Solve for the concentrations of HX- and X2-

First, we can solve the equation for \(K_1\) to find the concentration of \(\mathrm{HX}^{-}\). $$[\mathrm{HX}^{-}]^2 + 0.3[\mathrm{HX}^{-}] - 8.3\times 10^{-8} \cdot 0.1 = 0$$ This is a quadratic equation in \([\mathrm{HX}^{-}]\). Solving for \([\mathrm{HX}^{-}]\) (selecting the positive root as the concentration cannot be negative), we get: $$[\mathrm{HX}^{-}] \approx 2.58 \times 10^{-8} \mathrm{M}$$ Now, we can substitute this value back into the equation for \(K_2\): $$K_2 = \frac{[\mathrm{X}^{2-}](2.58 \times 10^{-8} + 0.3)}{2.58 \times 10^{-8}}$$

Step 4: Calculate the concentration of X2-

Finally, we can solve for \([\mathrm{X}^{2-}]\): $$[\mathrm{X}^{2-}] = K_2 \cdot \frac{2.58 \times 10^{-8}}{2.58 \times 10^{-8} + 0.3}$$ $$[\mathrm{X}^{2-}] \approx 8.6 \times 10^{-15} \mathrm{M}$$ Thus, the concentration of \(\mathrm{X}^{2-}\) in the solution is approximately \(8.6 \times 10^{-15} \mathrm{M}\).

Key Concepts

Hydride Formation

Elements in Group 16 of the periodic table, like oxygen, sulfur, and selenium, form compounds with hydrogen, known as hydrides. These hydrides follow a simple formula, which is typically \(\mathrm{H}_{2}\mathrm{X}\). Here, \(\mathrm{X}\) stands for a Group 16 element. They form by a direct combination of hydrogen and the respective element, often in the presence of heat.
These compounds exhibit distinctive properties based on their molecular structure. For instance, water (\(\mathrm{H}_{2}\mathrm{O}\)) is a hydride of oxygen, and hydrogen sulfide (\(\mathrm{H}_{2}\mathrm{S}\)) is a hydride of sulfur.

• Hydrides can display different types of chemical behavior, such as acting as acids or bases.
• This chemical behavior is crucial in studying various reactions, including those involving acid-base equilibria.

The ability of these compounds to participate in equilibrium reactions illustrates their importance in both lab-based and real-world chemistry.

Equilibrium Constants

In chemistry, equilibrium constants like \(K_1\) and \(K_2\) quantify the extent to which a chemical reaction occurs. They help us understand how products and reactants are distributed when a reaction is at equilibrium.
The equilibrium constant depends on the concentrations of the chemical species involved in the reaction. For instance, in our case, \(K_1\) and \(K_2\) represent two different equilibrium reactions:

• The first reaction involves \(\mathrm{H}_{2}\mathrm{X}(aq)\) and the formation of ions \(\mathrm{HX}^{-}(aq)\) and \(\mathrm{H}_{3}\mathrm{O}^{+}(aq)\).
• The second reaction further converts \(\mathrm{HX}^{-}(aq)\) into \(\mathrm{X}^{2-}(aq)\) along with more \(\mathrm{H}_{3}\mathrm{O}^{+}(aq)\).

Understanding equilibrium helps in predicting the direction of a reaction and what conditions are necessary to achieve a particular set of products. In acid-base chemistry, equilibrium constants are vital in calculating the concentration of various ions in solution, as they allow chemists to predict behaviors under different conditions.

Acid-Base Equilibrium

Acid-base equilibrium involves the balance between acids and bases in a solution. It refers to the state reached when the rates of the forward reaction (acid donating protons to base to form products) and the backward reaction (products recomposing to form reactants) are equal.
In our given scenario, as we look at \(\mathrm{H}_{2}\mathrm{X}\), which behaves like an acid, donating protons to water to form \(\mathrm{HX}^{-}(aq)\) and \(\mathrm{H}_{3}\mathrm{O}^{+}(aq)\), this represents an example of acid-base equilibrium.

• The strength of an acid or base is determined by its ability to donate or accept protons, and is quantified by its equilibrium constant.
• Strong acids have high equilibrium constants, meaning they dissociate more completely.
• Conversely, weak acids like \(\mathrm{H}_{2}\mathrm{X}\) typically have lower constants.

These principles are essential for understanding how complex chemical balances can shift depending on concentrations and conditions, helping chemists design reactions with desired endpoint balances.

Quadratic Equations in Chemistry

Sometimes, solving chemical equilibrium problems require the use of quadratic equations. This is often necessary when dealing with reactions that involve more complex relationships between reactants and products.
These equations arise from substituting into the expressions for the equilibrium constants. For example, when determining the concentration of \([\mathrm{HX}^{-}]\), we come up with a quadratic equation:
\[ [\mathrm{HX}^{-}]^2 + 0.3[\mathrm{HX}^{-}] - 8.3\times 10^{-8} \cdot 0.1 = 0 \]

• This equation helps in finding the concentration of a species by solving for the variable using standard quadratic formula methods.
• It's key to remember that these concentrations must be positive, thus only positive solutions are valid.
• Quadratic equations in chemistry allow us to predict concentrations and dynamics of reactions that are not straightforward.

By understanding how to form and solve these equations, students can aptly navigate more intricate chemical scenarios, giving them greater predictive power and understanding of chemical reactions and equilibria.