Question

Ethylenediamine \(\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right.\), abbreviated en \()\) is an organic base that can accept two protons: $$ \begin{aligned} \mathrm{en}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{enH}^{+}(a q)+\mathrm{OH}^{-}(a q) & K_{\mathrm{b} 1}=5.2 \times 10^{-4} \\ \mathrm{enH}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{enH}_{2}^{2+}(a q)+\mathrm{OH}^{-}(a q) & K_{\mathrm{b} 2}=3.7 \times 10^{-7} \end{aligned} $$ (a) Consider the titration of \(30.0 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) ethylenediamine with \(0.100 \mathrm{M}\) HCl. Calculate the \(\mathrm{pH}\) after the addition of the following volumes of acid, and construct a qualitative plot of pH versus milliliters of HCl added: (i) \(0.0 \mathrm{~mL}\) (ii) \(15.0 \mathrm{~mL}\) (iii) \(30.0 \mathrm{~mL}\) (iv) \(45.0 \mathrm{~mL}\) (v) \(60.0 \mathrm{~mL}\) (vi) \(75.0 \mathrm{~mL}\) (b) Draw the structure of ethylenediamine, and explain why it can accept two protons. (c) What hybrid orbitals do the \(\mathrm{N}\) atoms use for bonding?

Short answer

(a) pH values are approximately: (i) 11.30, (ii) 10.28, (iii) 8.34, (iv) 7.43, (v) 4.27, (vi) 2.00. (b) Ethylenediamine is H\(_2\)N-CH\(_2\)-CH\(_2\)-NH\(_2\). (c) N atoms use sp\(^3\) hybrid orbitals.

Step-by-step solution

Structure of Ethylenediamine

Ethylenediamine has the chemical structure:H\(_2\)N-CH\(_2\)-CH\(_2\)-NH\(_2\). This allows it to accept two protons due to the presence of two nitrogen atoms, each of which can act as a proton acceptor because of their lone pairs of electrons.

Hybrid Orbitals Used by Nitrogen Atoms

In ethylenediamine, the nitrogen atoms use sp\(^3\) hybrid orbitals for bonding. This is because each nitrogen atom forms three sigma bonds (with hydrogen and carbon) and retains a lone pair, giving it a tetrahedral geometry.

Titration Setup

We are titrating 30.0 mL of 0.100 M ethylenediamine with 0.100 M HCl solution. Since ethylenediamine is a diprotic base, each titration point may involve the neutralization of one or both of the base's protons.

Calculate pH at Given Volumes

Let's calculate the pH at each given volume of HCl added. The pH change reflects the stages of protonation of ethylenediamine:

Part (i) 0.0 mL HCl

Here, no HCl has been added, so ethylenediamine acts as a base. The initial concentration is 0.100 M. Use the first base dissociation constant, K\(_{b1}\):\[K_{b1} = 5.2 \times 10^{-4}\]Set up the equilibrium expression for en:\[ \text{en} + \text{H}_2\text{O} \leftrightarrows \text{enH}^+ + \text{OH}^-\]Use the expression \[ K_{b1} = \frac{[\text{enH}^+][\text{OH}^-]}{[\text{en}]} \]Solve for [OH\(^-\)] and calculate pOH, then pH. Use: \[ [OH^-] = \sqrt{K_{b1} \times [en]} \]Calculate pOH and then pH using: \[ \text{pOH} = -\log[\text{OH}^-] \] \[ \text{pH} = 14 - \text{pOH} \]Substituting the values gives pH \( \approx \) 11.30.

Part (ii) 15.0 mL HCl

At 15.0 mL HCl:- Half-equivalence point for the first protonation.With equal moles of added HCl and remaining en, the solution forms a buffer system and pH = pK\(_a1\) (where pK\(_a1\) is for the deprotonation of enH\(^+\)).Use: \[pK_{a1} = 14 - \log K_{b1} \]Calculate: \[pK_{a1} = 14 + \log(5.2 \times 10^{-4}) \approx 10.28\]Thus, pH = 10.28.

Part (iii) 30.0 mL HCl

At 30.0 mL HCl:- Complete neutralization of the first proton acceptor site.All en is converted to enH\(^+\). Since enH\(^+\) is the major species in the solution, set up equilibrium for enH\(^+\) with K\(_{b2}\):\[K_{b2} = \frac{[\text{enH}_2^{2+}][\text{OH}^-]}{[\text{enH}^+]} \]Solve for [OH\(^-\)], then find pOH and pH:\[ \text{pH} \approx \frac{1}{2}(14 + pK_{a1} + pK_{a2}) = 8.34\] (approximated mediation point formula).

Part (iv) 45.0 mL HCl

At 45.0 mL of HCl:- First equivalence point of second protonation.The solution becomes a mix of enH\(^+\) and enH\(_2\)\(^{2+}\). The concentration relationship gives a pH between 7 and the respective pK\(_a2\) (pH at this point is typically close to 7 because of buffering action). Here, pH adjusts more towards 5.4 (pK\(_a2\) calculations):Use: \[pK_{a2} = 14 + \log (3.7 \times 10^{-7}) \approx 7.43\].

Part (v) 60.0 mL HCl

For 60.0 mL HCl:- Full protonation of both amine sites or second equivalence. Complete formation of enH\(_2\)\(^{2+}\), a cationic species.The solution contains predominantly enH\(_2\)\(^{2+}\), contributing to a mildly acidic solution:Calculated or estimated pH \( \approx \) 4.27 using calculated equilibrium position of hydrolysis reactions.

Part (vi) 75.0 mL HCl

At 75.0 mL HCl:- A strong excess of HCl over protonation sites.The solution is strongly acidic with high HCl concentration dictating:- The pH dictated by excess HCl concentration entirely, calculated directly:Since moles added gives a shifted balance to free HCl calculation, \[ [H^+] = [HCl] = \frac{\text{(excess moles)}}{\text{(total solution volume)}} \], dominating pH towards low value,- Giving a pH\( \approx 2.00 \).

Key concepts

Protonation Equilibria

Ethylenediamine, often abbreviated as en, is capable of accepting two protons, as indicated by its basic nature. This characteristic is integral to understanding its behavior in protonation equilibria during titration. The equilibria are defined by two sequential reactions, each with a distinct equilibrium constant, symbolized as \( K_{b1} \) and \( K_{b2} \).

In the first equilibrium, ethylenediamine takes up a proton from water, forming \( \text{enH}^+ \) and releasing \( \text{OH}^- \). The strength of this reaction is portrayed by \( K_{b1} = 5.2 \times 10^{-4} \). This initial stage highlights the molecule's initial proton accommodating capacity.

In the second equilibrium, \( \text{enH}^+ \) accepts another proton to become \( \text{enH}_2^{2+} \), again paired with \( \text{OH}^- \). The equilibrium constant here is \( K_{b2} = 3.7 \times 10^{-7} \), which is smaller, indicating less affinity compared to the first proton uptake. Understanding these equilibria is essential for interpreting the concentration dynamics of reactants and products during the titration process.

pH Calculation

Performing pH calculations during a titration involving ethylenediamine and HCl hinges upon the stage of protonation and the volume of HCl added. The pH reflects the balance between unprotonated and protonated forms at particular titration points.

- **Starting Point (0.0 mL HCl):** Ethylenediamine is measured in its basic state. Calculating its pH involves determining \( \text{OH}^- \) concentration via \( K_{b1} \), following with pH from pOH conversion:
\[ [\text{OH}^-] = \sqrt{K_{b1} \times [\text{en}]} \]
- **Half Equivalence (15.0 mL):** The system acts as a buffer. Here, \( \text{pH} = \text{pK}_{a1} \), where \( \text{pK}_{a1} = 14 + \log(K_{b1}) \), stabilizing the pH around 10.28.

- **First Equivalence (30.0 mL):** pH is a function of \( K_{b2} \); use a median approximation for simplicity, resulting in pH \( \approx 8.34 \).

- **Second Equivalence (60.0 mL):** As fully protonated, pH is ruled by excess \( HCl \), calculated from its concentration, typically leading to a lower pH around 4.27.

Each calculated pH represents specific protonation completion, pivotal for understanding the transition from base to a fully protonated species.

Nitrogen Hybridization

In ethylenediamine, the nitrogen atoms are significant contributors to its dual role as a proton acceptor. These nitrogen atoms are involved in \( sp^3 \) hybridization.

Hybridization refers to the process where atomic orbitals mix to form new hybrid orbitals, which are essential for bonding. In \( sp^3 \) hybridization:

• The nitrogen atom combines one 2s and three 2p orbitals to create four equivalent \( sp^3 \) hybrid orbitals.
• Due to tetrahedral arrangement, this leads to three sigma bonds (two with hydrogen and one with carbon in each nitrogen’s local environment) and one lone pair of electrons.

This lone pair on each nitrogen allows them to act as a Lewis base, efficiently accepting protons and facilitating the protonation equilibria during titration. The \( sp^3 \) hybridization ensures a geometry that permits maximum separation of these electron pairs, further rationalizing its protonation capacity.