Question

Calculation of \(\mathrm{p} K_{\mathrm{a}}\) An unknown compound, \(\mathrm{X}\), is thought to have a carboxyl group with a \(\mathrm{p} K_{\mathrm{n}}\) of \(2.0\) and another ionizable group with a \(\mathrm{p} K_{\mathrm{n}}\) between 5 and 8 . When \(75 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{NaOH}\) is added to \(100 \mathrm{~mL}\) of a \(0.1 \mathrm{~m}\) solution of \(\mathrm{X}\) at \(\mathrm{pH} 2.0\), the \(\mathrm{pH}\) increases to \(6.72\). Calculate the \(\mathrm{p} K_{\mathrm{a}}\) of the second ionizable group of \(X\).

Short answer

The pKa of the second ionizable group is 6.72.

Step-by-step solution

Identify Initial Information

Given that the initial concentration of the solution X is 0.1 M, and its initial volume is 100 mL, we multiply these to get the amount of moles of X. Since NaOH is added to this solution, the information provided about NaOH is 0.1 M volume is 75 mL.

Calculate Moles of X and NaOH

Calculate the moles of X: \[ \text{moles of } X = 100 \, \text{mL} \times 0.1 \, \text{M} = 0.01 \, \text{mol} \]Calculate the moles of NaOH: \[ \text{moles of } \text{NaOH} = 75 \, \text{mL} \times 0.1 \, \text{M} = 0.0075 \, \text{mol} \]

Analyze Titration Point

The pH increases from 2.0 to 6.72, indicating that the second ionizable group has been half-neutralized by NaOH. This means 0.0075 mol of NaOH has reacted with the same amount of the ionizable group.

Determine Concentration of Second Group

After the reaction, 0.0075 mol of X with the carboxyl group has reacted. The remaining 0.0025 mol represents the second ionizable group after 0.0075 mol of OH- neutralized the same amount of X molecules. The solution's pH reflects the second ionizable group reaching the midpoint of its titration.

Calculate pKa of Second Ionizable Group

At this half-neutralization point (where \([HA] = [A^-]\)), the pH equals the pKa. Thus, \(\text{pKa}_{2} = \text{pH} = 6.72\). This matches with Henderson-Hasselbalch principle that when the pH is equal to pKa, the molecule is exactly half-neutralized. The pKa of the second ionizable group is 6.72.

Key concepts

Titration

Titration is a useful analytical technique used to determine the concentration of an unknown solution. In our exercise, titration helps us figure out the \(( ext{p}K_a)\) of the second ionizable group of compound X. The process involves gradually adding a known solution, like NaOH, to react with a substance whose concentration needs to be measured. Key components of titration are:

• The \(\text{analyte}\)—the unknown solution, in this case, compound X.
• The \(\text{titrant}\)—a solution with known concentration, here being 0.1 M NaOH.
• The \(\text{equivalence point}\)—where the amount of titrant added neutralizes the solution completely.
• The \(\text{endpoint}\)—indicated by a pH change, pinpointing completion of the titration.

In the scenario presented, by observing the pH at which the second ionizable group of X is half-neutralized, we can deduce when enough NaOH has been added. This pH change is crucial as it is directly used to calculate the \(\text{p}K_a\) of the second ionizable group.

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is central to calculating the \(\text{p}K_a\) from the titration data. It is expressed as: \[\text{pH} = \text{p}K_a + \log\left(\frac{[A^-]}{[HA]}\right)\]In this context, \(\text{pH}\) is the current pH of the solution, \(\text{p}K_a\) is the dissociation constant we aim to find, \(\left[ A^- \right]\) represents the concentration of the deprotonated form of the molecule, and \(\left[ HA \right]\) the protonated form.When the solution reaches half-neutralization, it signifies that the concentrations of \( [A^-] \) and \( [HA] \) are equal:\[\log\left(\frac{[A^-]}{[HA]}\right) = \log(1) = 0\]In this scenario, the equation simplifies to \(\text{pH} = \text{p}K_a\). The measured pH at this point directly gives the \(\text{p}K_a\) value. Thus, for our compound X, the pH measurement of 6.72 during the half-neutralization of its second ionizable group equates to its \(\text{p}K_a\). This demonstrates how helpful and straightforward this equation can be in lab calculations.

Ionizable groups

Ionizable groups in a molecule can significantly impact its chemical properties and reactivity. Each ionizable group has a specific \(\text{p}K_a\) value indicating the pH at which half of the molecules in a solution are deprotonated (or ionized).In compound X, we dealt with two ionizable groups:

• The \(\text{carboxyl group}\), which had a \(\text{p}K_a\) of 2.0.
• The \(\text{second ionizable group}\), which had an unknown \(\text{p}K_a\) until the titration revealed its value of 6.72.

Each ionizable group's \(\text{p}K_a\) helps predict how the molecule behaves in different pH conditions. At pH lower than the \(\text{p}K_a\), the group will usually be in its protonated form. At pH higher than the \(\text{p}K_a\), it will be primarily deprotonated.Understanding these ionizable groups in chemical analysis is key to evaluating not only the molecule's acidic or basic nature, but also its role in biological systems such as enzymatic reactions and cell signaling.