When hydrazine is dissolved in water, it acts like a base. N2H4(a q)+H2O(l) ⇌N2H5^+(a q)+OH

step 1 A high rizine, they got two steps to produce hydroxide ion, but we see the KB2 very, very small.

When hydrazine is dissolved in water, it acts like a base. ...

When hydrazine is dissolved in water, it acts like a base. $$ \begin{array}{c} \mathrm{N}_{2} \mathrm{H}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{5}{ }^{+}(a q)+\mathrm{OH}^{-}(a q) \\ K_{\mathrm{b}_{1}}=8.5 \times 10^{-7} \\ \mathrm{~N}_{2} \mathrm{H}_{5}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{6}^{2+}(a q)+\mathrm{OH}^{-}(a q) \\ K_{\mathrm{b}_{2}}=8.9 \times 10^{-16} \end{array} $$ a. Calculate the $K_{\mathrm{b}}$ for the overall reaction of hydrazine forming $\mathrm{N}_{2} \mathrm{H}_{6}^{2+}$ b. Calculate $K_{a_{1}}$, for $N_{2} H_{5}^{+}$. c. Calculate the concentration of hydrazine and both cations in a solution buffered at a pH of 8.5 for a solution that was made by dissolving 0.012 mol of hydrazine in 1 L of water.

When hydrazine is dissolved in water, it acts like a base.
$$
\begin{array}{c}
\mathrm{N}_{2} \mathrm{H}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{5}{ }^{+}(a q)+\mathrm{OH}^{-}(a q) \\
K_{\mathrm{b}_{1}}=8.5 \times 10^{-7} \\
\mathrm{~N}_{2} \mathrm{H}_{5}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{6}^{2+}(a q)+\mathrm{OH}^{-}(a q) \\
K_{\mathrm{b}_{2}}=8.9 \times 10^{-16}
\end{array}
$$
a. Calculate the $K_{\mathrm{b}}$ for the overall reaction of hydrazine forming $\mathrm{N}_{2} \mathrm{H}_{6}^{2+}$
b. Calculate $K_{a_{1}}$, for $N_{2} H_{5}^{+}$.
c. Calculate the concentration of hydrazine and both cations in a solution buffered at a pH of 8.5 for a solution that was made by dissolving 0.012 mol of hydrazine in 1 L of water.

Key Concepts

Acid?Base Equilibrium

This concept involves the reversible reactions between acids and bases in solution. In such reactions, acids donate protons (H? ions) while bases accept them. Understanding acid–base equilibrium is crucial because it helps predict how a substance behaves in water, whether it acts as an acid, base, or remains neutral, and how its equilibrium is established in solution.

Base Dissociation Constant (Kb)

The base dissociation constant, Kb, quantifies the tendency of a weak base to accept a proton and form its conjugate acid. It is derived from the equilibrium expression for the reaction of the base with water. Lower values of Kb indicate a weaker base, and comparing Kb values allows one to understand the relative strengths of different bases in solution.

Overall Equilibrium Constant from Sequential Reactions

When multiple reactions occur in sequence, the overall equilibrium constant for the net reaction can be obtained by multiplying the equilibrium constants of the individual steps. This principle is crucial for complex reaction systems where intermediate species are formed, as it allows one to combine multiple equilibria into a single expression for the overall reaction.

Relationship Between Ka and Kb

For a conjugate acid-base pair, the acid dissociation constant (Ka) and the base dissociation constant (Kb) are related through the ionization constant of water (Kw), as expressed by the relationship Ka × Kb = Kw. This fundamental principle helps in calculating one constant if the other is known, and is particularly useful when analyzing the behavior of weak bases and their conjugate acids in aqueous solutions.

Buffer Solutions

Buffer solutions are systems that resist changes in pH upon the addition of small amounts of acids or bases. They typically consist of a weak acid and its conjugate base (or a weak base and its conjugate acid) in appreciable amounts. Understanding buffers is key to maintaining a desired pH in a chemical or biological system, and involves balancing the concentrations of the buffering species in accordance with the acid–base equilibrium principles.

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