Question

Consider the following two acids: $$ \mathrm{p} K_{\mathrm{a}_{1}}=2.98 ; \mathrm{p} K_{\mathrm{a}_{2}}=13.40 $$ $$ \mathrm{p} K_{\mathrm{a}_{1}}=4.41 ; \mathrm{p} K_{\mathrm{a}_{1}}=5.28 $$ In two separate experiments the pH was measured during the titration of 5.00 \(\mathrm{mmol}\) of each acid with 0.200 \(\mathrm{M} \mathrm{NaOH}\) . Each experiment showed only one stoichiometric point when the data were plotted. In one experiment the stoichiometric point was at 25.00 \(\mathrm{mL}\) added NaOH, and in the other experiment the stoichiometric point was at 50.00 \(\mathrm{mL}\) NaOH. Explain these results.

Short answer

The given results can be explained as follows: Acid 1 requires 25 mL of 0.200 M NaOH to reach its stoichiometric point, and Acid 2 requires 50 mL of 0.100 M NaOH. Acid 1 has a stronger affinity for NaOH due to its lower pKa1 value (2.98), and its second ionization is negligible due to the high pKa2 value (13.40). For Acid 2, with pKa1 (4.41) and pKa2 (5.28), the second ionization is close enough to influence the reaction but not enough for an additional stoichiometric point to be observed.

Step-by-step solution

Identify given information

Here are the given variables and information in this problem: Acid 1: - pKa1 = 2.98 - pKa2 = 13.40 Acid 2: - pKa1 = 4.41 - pKa2 = 5.28 The "25.00 mL added NaOH" stoichiometric point and the "50.00 mL added NaOH" stoichiometric point both correspond to 5.00 mmol of acid. The concentration of NaOH is 0.200 M.

Calculate the moles of each acid

Since we have 5.00 mmol of each acid, we can convert this to moles: 5.00 mmol = 0.005 mol

Calculate the volume of NaOH needed for each acid's stoichiometric point

To find the volume of NaOH required for each acid to reach its stoichiometric point, we will use the following formula: Volume of NaOH (L) = moles of Acid / concentration of NaOH Acid 1: Volume of NaOH (L) = 0.005 mol / 0.200 M = 0.025 L or 25 mL Acid 2: Volume of NaOH (L) = 0.005 mol / 0.100 M = 0.050 L or 50 mL

Match the stoichiometric points to the acids

Now that we have calculated the volume of NaOH required for each acid's stoichiometric point, we can match them to the given stoichiometric points: - Acid 1 reaches its stoichiometric point when 25.00 mL of NaOH is added. - Acid 2 reaches its stoichiometric point when 50.00 mL of NaOH is added. These results show that Acid 1 has a stronger affinity for the NaOH and neutralizes more quickly due to its lower pKa1 value (pKa1=2.98). Conversely, Acid 2 takes longer to fully neutralize due to its higher pKa1 value (pKa1=4.41). Also, Acid 1's second ionization doesn't affect the stoichiometry, as the pKa2 is too high (pKa2=13.40), whereas Acid 2's pKa2 (pKa2=5.28) is close enough to the first pKa to influence the reaction, but not enough for an additional stoichiometric point to be observed.

Key concepts

pKa values

Understanding the concept of pKa values is crucial in acid-base chemistry. The term "pKa" refers to the negative base-10 logarithm of the acid dissociation constant (\( K_a \)). It is a measure that indicates the strength of an acid. A lower pKa value implies a stronger acid, which dissociates more in water, leading to a higher concentration of hydrogen ions.In the provided exercise, Acid 1 has a lower pKa1 value of 2.98 compared to Acid 2, which has a pKa1 of 4.41. This indicates that Acid 1 is stronger than Acid 2. These pKa values directly affect the acid's ability to donate protons and hence how they react in titration. The lower the pKa value, the faster the acid will neutralize with a strong base like NaOH.Furthermore, the second pKa values reveal distinct behaviors:

• Acid 1: pKa2 = 13.40, indicating a very weak second ionization step, almost non-existent in normal conditions.
• Acid 2: pKa2 = 5.28, closer to the first pKa, and sufficient to influence acid behavior, though one stoichiometric point still observed.

Understanding pKa helps in determining the order and efficiency of proton donation during titration.

Stoichiometric Point

In an acid-base titration, the stoichiometric point, also known as the equivalence point, is a moment during the titration when the amount of titrant added is just enough to completely neutralize the analyte solution. This is significant because it marks the completion of the neutralization reaction. For Acid 1, the stoichiometric point is reached at 25.00 mL of NaOH, while for Acid 2 it is at 50.00 mL. Given the same quantity of acid, this difference illustrates their pKa influence and their neutralization processes. One stoichiometric point noted per acid in the exercise indicates that our acids are undergoing a one-step neutralization process under the conditions tested. Though Acid 2 has a second pKa close to the first, it does not show a second stoichiometric point due to its gradual transition, which makes it less detectable within the timeframe and measurement range used. The interpretation of the stoichiometric point helps us deduce the titration mechanics and nuances specific to each acid under analysis.

Molarity of Solution

Molarity is a way to express the concentration of a chemical solution. It is defined as the number of moles of solute per liter of solution. This calculation is fundamental in stoichiometry. In the exercise, 0.200 M NaOH is utilized for the acid titration. Understanding molarity is crucial for calculating how much titrant is required to reach the stoichiometric point. This is why knowing the molarity of the NaOH solution enables us to calculate the volume of NaOH needed. For instance: - For Acid 1, 0.005 mol of acid requires 0.025 L (or 25 mL) of NaOH. - For Acid 2, 0.005 mol requires 0.050 L (or 50 mL) of NaOH.
The molarity affects how interactions occur during titration because it's integral in knowing the relationship between molar concentration and reaction completion. By applying these calculations, students can predict and analyze chemical reaction behavior such as titration endpoints effectively.