When a diprotic acid, \(\mathrm{H}_{2} \mathrm{~A}\), is titrated with \(\mathrm{NaOH}\), the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape: a. Notice that the plot has essentially two titration curves. If the first equivalence point occurs at \(100.0 \mathrm{~mL} \mathrm{NaOH}\) added, what volume of \(\mathrm{NaOH}\) added corresponds to the second equivalence point? b. For the following volumes of \(\mathrm{NaOH}\) added, list the major species present after the \(\mathrm{OH}^{-}\) reacts completely. i. \(0 \mathrm{~mL}\) NaOH added ii. between 0 and \(100.0 \mathrm{~mL}\) NaOH added iii. \(100.0 \mathrm{~mL} \mathrm{NaOH}\) added iv. between \(100.0\) and \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added v. \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added vi. after \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added c. If the \(\mathrm{pH}\) at \(50.0 \mathrm{~mL} \mathrm{NaOH}\) added is \(4.0\) and the \(\mathrm{pH}\) at \(150.0 \mathrm{~mL} \mathrm{NaOH}\) added is \(8.0\), determine the values \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) for the diprotic acid.

To find the volume of NaOH added at the second equivalence point, we need to consider that it takes twice the amount of NaOH to neutralize both protons in the diprotic acid. Since it takes \(100.0 \mathrm{~mL}\) of \(\mathrm{NaOH}\) to reach the first equivalence point, we need another \(100.0 \mathrm{~mL}\) to reach the second equivalence point. Therefore, the second equivalence point occurs at \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added.

i. At \(0 \mathrm{~mL}\) NaOH added: The major species present will be the diprotic acid, \(\mathrm{H}_{2} \mathrm{A}\). ii. Between 0 and \(100.0 \mathrm{~mL}\) NaOH added: The major species will be the diprotic acid \(\mathrm{H}_{2} \mathrm{A}\) and its first conjugate base \(\mathrm{HA}^-\), as some protons have been removed. iii. At \(100.0 \mathrm{~mL} \mathrm{NaOH}\) added: The major species present will be the first conjugate base, \(\mathrm{HA}^-\), since all protons have been removed from \(\mathrm{H}_{2}A\). iv. Between \(100.0\) and \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added: The major species will be the first conjugate base \(\mathrm{HA}^-\) and the second conjugate base \(\mathrm{A}^{2-}\), as protons are being removed from \(\mathrm{HA}^-\). v. At \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added: The major species present will be the second conjugate base, \(\mathrm{A}^{2-}\), since all protons have been removed from \(\mathrm{HA}^-\). vi. After \(200.0 \mathrm{~mL} \mathrm{NaOH}\) added: The major species present will still be the second conjugate base, \(\mathrm{A}^{2-}\), but now the \(\mathrm{OH}^-\) concentration will start to increase as further addition of NaOH will not react with a proton anymore.

At \(50.0 \mathrm{~mL} \mathrm{NaOH}\) added, halfway between the first equivalence point and the starting point, the pH is \(4.0\). At this point, the concentrations of \(\mathrm{H_{2}A}\) and \(\mathrm{HA}^-\) are equal, so \(\mathrm{pH}=\mathrm{p}K_{\mathrm{a}_{1}}\). Thus, \(K_{\mathrm{a}_{1}}=10^{-4}\). At \(150.0 \mathrm{~mL} \mathrm{NaOH}\) added, halfway between the first and second equivalence points, the pH is \(8.0\). At this point, the concentrations of \(\mathrm{HA}^-\) and \(\mathrm{A}^{2-}\) are equal, so \(\mathrm{pH}=\mathrm{p}K_{\mathrm{a}_{2}}\). Thus, \(K_{\mathrm{a}_{2}}=10^{-8}\).