Problem 75 - Chemistry and Chemical Reactivity

Question

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) can interact with water in two steps. $$\begin{aligned} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) & \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \\ K_{\mathrm{b} 1} &=8.5 \times 10^{-7} \\ \mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) & \rightleftarrows \mathrm{N}_{2} \mathrm{H}_{6}^{2+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \\ K_{\mathrm{b} 2} &=8.9 \times 10^{-16} \end{aligned}$$ (a) What is the concentration of \(\mathrm{OH}^{-}, \mathrm{N}_{2} \mathrm{H}_{5}^{+}\) and \(\mathrm{N}_{2} \mathrm{H}_{6}^{2+}\) in a \(0.010 \mathrm{M}\) aqueous solution of hydrazine? (b) What is the \(\mathrm{pH}\) of the 0.010 M solution of hydrazine?

Step-by-Step Solution

Answer

[OH^-] = 2.9 × 10⁻⁵ M, [N₂H₅⁺] ≈ 2.9 × 10⁻⁵ M, [N₂H₆²⁺] ≈ 3.1 × 10⁻¹¹ M; pH ≈ 9.46.

Step 1: Write the Equilibrium Expressions

For the first reaction \( \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq}) + \mathrm{OH}^{-}(\mathrm{aq}) \), the equilibrium expression is:\[K_{\mathrm{b} 1} = \frac{[\mathrm{N}_{2} \mathrm{H}_{5}^{+}][\mathrm{OH}^{-}]}{[\mathrm{N}_{2} \mathrm{H}_{4}]}\]For the second reaction, \( \mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq}) + \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{6}^{2+}(\mathrm{aq}) + \mathrm{OH}^{-}(\mathrm{aq}) \), the equilibrium expression is:\[K_{\mathrm{b} 2} = \frac{[\mathrm{N}_{2} \mathrm{H}_{6}^{2+}][\mathrm{OH}^{-}]}{[\mathrm{N}_{2} \mathrm{H}_{5}^{+}]}\]

Step 2: Set Up the Initial Conditions

Start with a solution that has an initial concentration of \(0.010 \, \text{M}\) for \(\mathrm{N}_{2} \mathrm{H}_{4}\). Before any dissociation, both \([\mathrm{N}_{2} \mathrm{H}_{5}^{+}]\) and \([\mathrm{OH}^{-}]\) are 0. Consider the initial change to be \(x\) for the first reaction and \(y\) for the second reaction.

Step 3: Establish the Change in Concentrations

Let \(x\) be the concentration of \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}\) and \(\mathrm{OH}^{-}\) formed from the first reaction. Therefore, \([ ext{N}_2 ext{H}_4] = 0.010 - x\).For the second equilibrium, let \(y\) be the concentration of \([\mathrm{N}_{2} \mathrm{H}_{6}^{2+}]\) formed. Therefore, \([ ext{OH}^-]_{total} = x + y\) and \([ ext{N}_2 ext{H}_5^+] = x - y\).

Step 4: Solve the First Equilibrium Equation

Substitute values into the expression for the first reaction: \[8.5 \times 10^{-7} = \frac{x^2}{0.010 - x}\]Assuming \(x \ll 0.010\), simplify to find: \[x^2 \approx 8.5 \times 10^{-9}\]Thus, \(x \approx 2.9 \times 10^{-5}\, \text{M}\).

Step 5: Solve the Second Equilibrium Equation

Substitute into the expression of the second reaction:\[8.9 \times 10^{-16} = \frac{y(x + y)}{x - y}\]Assume \(y \ll x\); therefore, \[8.9 \times 10^{-16} \approx \frac{y \cdot x}{x}\]Given \(x \approx 2.9 \times 10^{-5}\, \text{M}\), solve for \(y\):\[y \approx 8.9 \times 10^{-16} \times \frac{1}{x} = 3.1 \times 10^{-11} \, \text{M}\]Now we know \([\text{N}_2\text{H}_6^{2+}] = y\).

Step 6: Calculate Total [OH^-] Concentration

Since \([\mathrm{OH}^-]_{total} = x + y\), \[[\mathrm{OH}^-] = 2.9 \times 10^{-5} + 3.1 \times 10^{-11} \approx 2.9 \times 10^{-5} \, \text{M}\]Given the small value of \(y\) compared to \(x\), \([\text{OH}^-] \approx 2.9 \times 10^{-5}\, \text{M}\).

Step 7: Calculate [N_2H_5^+] Concentration

Calculate \([\mathrm{N}_{2}\mathrm{H}_{5}^{+}]\) after knowing \(y\):\[[\mathrm{N}_{2}\mathrm{H}_{5}^{+}] = x - y = 2.9 \times 10^{-5} - 3.1 \times 10^{-11} \approx 2.9 \times 10^{-5} \, \text{M}\]

Step 8: Calculate the pH

Use the relation \([\mathrm{OH}^-] = 2.9 \times 10^{-5}\) M to find pOH:\[\text{pOH} = -\log([\mathrm{OH}^-]) = -\log(2.9 \times 10^{-5}) \approx 4.54\]Find the pH by subtracting from 14:\[\text{pH} = 14 - \text{pOH} \approx 9.46\]

Step 9: Final Check

Make sure the results are within valid ranges for equilibrium systems. The minor species need to align with assumptions used in simplifications. Double-check the calculations made for accuracy.

Key Concepts

Equilibrium Constant

The equilibrium constant is crucial in understanding chemical reactions at equilibrium. It represents the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced equation.

For an equilibrium like \[A + B \rightleftharpoons C + D\], the equilibrium constant, \(K\), is expressed as:\[K = \frac{[C][D]}{[A][B]}\]

This value remains constant at a given temperature. It allows us to predict the extent of a reaction. A large \(K\) indicates a reaction that heavily favors product formation, while a small \(K\) suggests reactants are favored. For the hydrazine reactions given:

• \(K_{b1} = 8.5 \times 10^{-7}\)
• \(K_{b2} = 8.9 \times 10^{-16}\)

These values help determine the relative amounts of each species at equilibrium.

Acid-Base Reactions

Acid-base reactions involve the transfer of protons (H\(^+\)) between molecules. In these reactions, acids donate protons, while bases accept them. Water often acts as a participant, providing or accepting protons, which makes it amphoteric.

Considering hydrazine, itself a base, its reaction with water can be analyzed using equilibrium constants \(K_b\). This constant helps indicate the strength of a base—the larger \(K_b\), the stronger the base. Hydrazine reacts in two steps:

• In the first step, it partially ionizes to produce \(\mathrm{N}_{2}\mathrm{H}_{5}^{+}\) and \(\mathrm{OH}^-\).
• In the second step, \(\mathrm{N}_{2}\mathrm{H}_{5}^{+}\) can further ionize, albeit less favourably, to form \(\mathrm{N}_{2}\mathrm{H}_{6}^{2+}\).

These reactions reflect both acid-base principles and the weaker basic nature of hydrazine.

pH Calculation

Understanding pH is essential when dealing with acid-base equilibria. pH is a scale that indicates the acidity or basicity of a solution. It is calculated as the negative logarithm of the hydrogen ion concentration: \[\text{pH} = -\log [\text{H}^+]\]

For hydroxide ions \(\mathrm{OH}^-\), we use \(\text{pOH}\) and the relationship:\[\text{pOH} = -\log [\mathrm{OH}^-] \,\text{and} \,\text{pH} = 14 - \text{pOH}\]

The exercise showed how to calculate pOH from the total \([\mathrm{OH}^-]\). With a value of \(2.9 \times 10^{-5} \, \text{M}\), the pOH was 4.54, leading to a pH of 9.46. This result indicates that the solution is basic, which aligns with hydrazine's nature as a base.This straightforward method provides a clear picture of a solution's acidity.

Hydrazine Chemical Reactions

Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), is known for its basic properties and reactivity with water in acid-base equilibria. It undergoes two key reactions involving water:

• The first reaction transforms hydrazine into \(\mathrm{N}_{2}\mathrm{H}_{5}^{+}\) and \(\mathrm{OH}^-\).
• The second converts \(\mathrm{N}_{2}\mathrm{H}_{5}^{+}\) into \(\mathrm{N}_{2}\mathrm{H}_{6}^{2+}\) and \(\mathrm{OH}^-\).

Both reactions illustrate the stepwise nature of hydrazine's interaction with water. With significant declines in \(K_b\) from the first to the second equilibrium, we see the decreasing readiness of hydrazine-derived species to further ionize. This behavioral study of hydrazine provides insights into its applications, particularly in fuels and chemical synthesis.

By understanding the reactions, we can predict the formation of various ionic species in a given hydrazine solution, crucial for applications involving this compound.