Question

Consider the following two acids: $$ \mathrm{p} K_{\mathrm{a}_{1}}=2.98 ; \mathrm{p} K_{\mathrm{a}_{2}}=13.40 $$ $$ \begin{array}{l} \mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H} \\ \text { Adipic acid } \quad \mathrm{p} K_{\mathrm{a}_{1}}=4.41 ; \mathrm{p} K_{\mathrm{a}_{2}}=5.28 \end{array} $$ In two separate experiments the pH was measured during the titration of \(5.00\) mmol of each acid with \(0.200 M \mathrm{NaOH}\). 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 \(\mathrm{NaOH}\), and in the other experiment the stoichiometric point was at \(50.00 \mathrm{~mL} \mathrm{NaOH}\). Explain these results. (See Exercise \(113 .\) )

Short answer

The stoichiometric points in the two experiments can be explained by analyzing the titration curves of the two given acids. Acid 1 has a large difference between its pKa1 and pKa2 values, so only one proton is released during titration, leading to a stoichiometric point at 25.00 mL of 0.200 M NaOH. In contrast, Adipic acid (Acid 2) has similar pKa1 and pKa2 values, causing both protons to be released during titration with NaOH, resulting in a stoichiometric point at 50.00 mL of 0.200 M NaOH.

Step-by-step solution

Calculate the number of moles of NaOH at the stoichiometric point for each experiment.

In order to analyze the problem, we need to know the number of moles of NaOH used at each stoichiometric point. For experiment 1: \(V_{NaOH1} = 25.00\) mL. For experiment 2: \(V_{NaOH2} = 50.00\) mL. The concentration of NaOH used is 0.200 M. So, to find the number of moles of NaOH, we will use the equation: Moles of NaOH = (Volume in L) × (Concentration in mol/L)

Find the moles of NaOH at each stoichiometric point.

Now, let's calculate the number of moles of NaOH for each experiment: For experiment 1: Moles of NaOH = (25.00 mL / 1000) × 0.200 M = 0.00500 mol For experiment 2: Moles of NaOH = (50.00 mL / 1000) × 0.200 M = 0.0100 mol

Determine the equivalence point for each acid.

Now that we have the moles of NaOH at each stoichiometric point, we can analyze the results. For acid 1: Since the pKa2 is significantly higher than pKa1 (13.40 > 2.98), the second deprotonation will barely happen, and we can consider that only one proton is released from each acid molecule. In this case, the number of moles of NaOH should be equal to the number of moles of acid 1 at the stoichiometric point. So, for the 5.00 mmol of acid 1, we expect to use 5.00 mmol of NaOH. For acid 2 (Adipic acid): Since both pKa values are close (4.41 and 5.28), we expect that the titration curve has two inflection points, meaning that the stoichiometric point should be reached when 2 moles of NaOH are added per mole of acid 2. As we have 5.00 mmol of adipic acid, the stoichiometric point should occur when 10.0 mmol of NaOH are added.

Match the acids to their respective experiments.

Finally, we can match each acid to the corresponding experiment based on the number of moles of NaOH used: Experiment 1 (0.00500 mol NaOH): This matches the expected stoichiometric point for acid 1 (5.00 mmol of NaOH), so we have: Acid 1 ⟶ Experiment 1 (25.00 mL of 0.200 M NaOH) Experiment 2 (0.0100 mol NaOH): This matches the expected stoichiometric point for acid 2 (10.0 mmol of NaOH), so we have: Acid 2 (Adipic acid) ⟶ Experiment 2 (50.00 mL of 0.200 M NaOH) These results occur because acid 1 has a very high value of pKa2 compared to its pKa1, so only one proton is released during the titration with NaOH. On the other hand, adipic acid has similar pKa1 and pKa2 values, so both protons are released when titrating with NaOH, leading to a higher amount of NaOH needed to reach the stoichiometric point.

Key concepts

Stoichiometric Point

In the captivating world of acid-base titrations, the stoichiometric point is a moment of perfect harmony - when the number of acid moles equals the number of base moles, leading to complete neutralization.

Imagine you're at a dance, and for every dancer, there's a perfect partner. When the music stops, each one has found their match; this is akin to the stoichiometric point in a titration. Consider the acid as one group of dancers and the base as another. When they come together in just the right numbers, the dance floor is balanced.

No more acid to react with the base, no more base to neutralize the acid. It's important to remember that reaching the stoichiometric point doesn't necessarily mean the pH is neutral. This unique point depends on the nature of the involved substances and the overall context of the titration process.

pKa Values

Delving into the minute details, pKa values are the VIP guests in the grand event of an acid-base reaction. They allow us to predict the strength of an acid - the smaller the pKa, the stronger the acid, and vice versa.

A low pKa value (<1) is like a celebrity with high influence (a strong acid), drawing protons away with ease. As the pKa value rises (up to about 15), the acid's ability to part with its proton mirrors a celebrity's waning clout (a weaker acid).

In our exercise scenario, we meet two different 'celebrities'. Acid 1 is a strong acid with a low initial pKa value and a second, very high pKa, indicating the release of a second proton is highly unlikely under normal conditions. Adipic acid, however, showcased two close pKa values, suggesting that both its protons have a decent chance of reacting during titration, showcasing a delicate balance between its acid strength and weakness.

Equivalence Point

The equivalence point is like the grand finale of the acid-base titration show. It's marked by equal quantities of acid and base that have been reacted, resulting in the complete neutralization of the acid.

Now let's break it down further. For monoprotic acids (capable of donating one proton), one mole of acid reacts with one mole of base; and for diprotic acids (capable of donating two protons), it becomes a double act - one mole of acid requires two moles of base for full neutralization.

The pKa values play a director's role in determining if both protons of a diprotic acid will keenly participate or if one proton will shy away from the reaction. In our illustrative example, the different volumes of NaOH used in the titrations signify the different 'scripts' for our two acids, revealing the underlying chemical 'drama' of protons exchanging partners in this molecular dance.