Problem 86 - Chemical Principles

Question

\(12.86\) A species that can accept two protons is classified as dibasic. The dibasic molecule 1,2 -ethanediamine, \(\mathrm{H}_{2} \mathrm{NC}_{2} \mathrm{H}_{4} \mathrm{NH}_{2}\), which has \(\mathrm{p} K_{\mathrm{b} 1}=3.19\) and \(\mathrm{p} K_{\mathrm{b} 2}=6.44\), is titrated with \(\mathrm{HCl}(\mathrm{aq})\). (a) What is the \(\mathrm{pH}\) when \(\left[\mathrm{H}_{2} \mathrm{NC}_{2} \mathrm{H}_{4} \mathrm{NH}_{2}\right]=\left[\mathrm{H}_{2} \mathrm{NC}_{2} \mathrm{H}_{4} \mathrm{NH}_{3}{\underline{\phantom{xx}}}^{+}\right]\)? (b) What is the \(\mathrm{pH}\) when \(\left[\mathrm{H}_{2} \mathrm{NC}_{2} \mathrm{H}_{4} \mathrm{NH}_{3}{\underline{\phantom{xx}}}^{+}\right]=\left[{ }^{+} \mathrm{H}_{3} \mathrm{NC}_{2} \mathrm{H}_{4} \mathrm{NH}_{3}{\underline{\phantom{xx}}}^{+}\right]\)? (c) Which is the predominant species at \(\mathrm{pH}=4.8\) ?

Step-by-Step Solution

Answer

The pH at point (a) is 10.81, at point (b) is 7.56. For part (c), the predominant species at pH 4.8 is \(\mathrm{H}_2\mathrm{NC}_2\mathrm{H}_4\mathrm{NH}_3^+\) since pH is closer to \(\text{p}K_{\text{a}2}\) than \(\text{p}K_{\text{a}1}\).

Step 1: Introduction to pH Calculation

To find the pH when \[\mathrm{H}_2\mathrm{NC}_2\mathrm{H}_4\mathrm{NH}_2\] equals \[\mathrm{H}_2\mathrm{NC}_2\mathrm{H}_4\mathrm{NH}_3^+\], we recognize this point as the first equivalence point in the titration. The Henderson-Hasselbalch equation can be used to calculate the pH at this point: \[\text{pH} = \text{p}K_{\text{a}} + \log\left(\frac{\text{[base]}}{\text{[acid]}}\right)\]. Since the concentrations of the base and the conjugate acid are equal, the log term will be zero.

Step 2: Calculating pH for Part (a)

Using the pH equation for the first equivalence point, the \(\text{p}K_{\text{a}1}\) is needed and can be obtained from the given \(\text{p}K_{\text{b}1}\) using the relation \(\text{p}K_{\text{a}} = 14 - \text{p}K_{\text{b}}\). Calculate \(\text{p}K_{\text{a}1}\) and substitute in the Henderson-Hasselbalch equation to find pH.

Step 3: Determining \(\text{p}K_{\text{a}1}\) for First Equivalence Point

Calculate \(\text{p}K_{\text{a}1}\) from the given \(\text{p}K_{\text{b}1}\): \[\text{p}K_{\text{a}1} = 14 - 3.19 = 10.81\].

Step 4: Calculating pH for Part (b)

For the second equivalence point, use the given \(\text{p}K_{\text{b}2}\) to find \(\text{p}K_{\text{a}2}\) and then apply the Henderson-Hasselbalch equation as per step 1. The concentrations of the secondary base and its conjugate acid are equal.

Step 5: Determining \(\text{p}K_{\text{a}2}\) for Second Equivalence Point

Calculate \(\text{p}K_{\text{a}2}\) from the given \(\text{p}K_{\text{b}2}\): \[\text{p}K_{\text{a}2} = 14 - 6.44 = 7.56\].

Step 6: Identifying the Predominant Species for Part (c)

At \(\text{pH} = 4.8\), compare the given pH with \(\text{p}K_{\text{a}1}\) and \(\text{p}K_{\text{a}2}\). The predominant species is the one for which the corresponding pKa is closest to the given pH without going over it.

Key Concepts

Henderson Hasselbalch Equation

The Henderson Hasselbalch equation is a fundamental expression in acid-base chemistry. It describes the pH of a solution in terms of the pKa (acid dissociation constant) and the ratio of the concentrations of the conjugate base and its corresponding acid. Presented in the form:
\[\begin{equation}\text{pH} = \text{p}K_{\text{a}} + \log\left(\frac{\text{[base]}}{\text{[acid]}}\right)\end{equation}\]
An important application of this equation is during titrations. It serves as a guide for predicting the pH when certain amounts of acid or base are added to the system. Especially at the halfway point to the equivalence point, where the concentrations of the acid and its conjugate base are equal, the logarithmic term becomes zero, and thus the pH equals the pKa of the acid. This relationship simplifies calculations and aids in understanding buffer systems in solutions.

Equivalence Point

The equivalence point in a titration is the moment where the amount of titrant added is stoichiometrically equivalent to the amount of substance present in the sample. For an acid-base titration, it signifies the point where the moles of acid equal the moles of base. This is a pivotal moment in titration, as the pH can change rapidly around this point.
At the equivalence point of a weak acid with a strong base (or vice versa), the solution is not neutral (pH 7), but rather is slightly basic or acidic depending on the nature of the weak component. In the case of dibasic molecules with two stages of proton acceptance or donation, two equivalence points exist. These points are crucial for determining the end-point of a titration and are typically indicated by a sharp change in pH, observed with a pH meter or indicated by a change in color of an appropriate indicator.

Dibasic Molecule

A dibasic molecule, such as the textbook exercise's 1,2-ethanediamine, is capable of accepting two protons (H+). This characteristic makes it a polyprotic base, in this particular scenario. Polyprotic acids or bases have more than one ionizable hydrogen or accepting site, respectively, each with a distinct pKa or pKb value.
During titration, such molecules undergo stepwise dissociation or protonation, evident in the existence of multiple buffering regions and equivalence points. In a dibasic molecule's titration curve, we can observe two distinct stages corresponding to the acceptance of the first and second protons. The identification of these two stages is essential when interpreting titration data for dibasic or polyprotic substances and is critical for understanding their buffering capacity at different pH levels.

pKa and pKb Relationship

The relationship between pKa (acid dissociation constant) and pKb (base dissociation constant) is inverse and complementary within the aqueous solution context. They are related by the simple equation:
\[\begin{equation}\text{p}K_{\text{w}} = \text{p}K_{\text{a}} + \text{p}K_{\text{b}}\end{equation}\]
where pKw is the ionic product of water, which is a constant at a given temperature, typically 14 at room temperature. This relationship indicates that as the strength of an acid increases (lower pKa), the strength of its conjugate base decreases (higher pKb), and vice versa.
In practical terms for titrations, this equation allows us to convert from pKb to pKa when dealing with weak bases. By knowing one, we can always find the other and predict the behavior of the conjugate acid-base pair in a solution. This calculation is particularly useful when we deal with substances like 1,2-ethanediamine, where pKb values are given, and we need to find the corresponding pKa values to use in the Henderson Hasselbalch equation for pH calculations.