Adipic acid, \(\mathrm{HOOC}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{COOH},\) is among the top 50 manufactured chemicals in the United States (nearly 1 million metric tons annually). Its chief use is in the manufacture of nylon. It is a diprotic acid having \(K_{\mathrm{a}_{1}}=3.9 \times 10^{-5}\) and \(K_{\mathrm{a}_{2}}=3.9 \times 10^{-6} .\) A saturated solution of adipic acid is about \(0.10 \mathrm{M}\) \(\mathrm{HOOC}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{COOH} .\) Calculate the concentration of each ionic species in this solution.

Adipic acid is a diprotic acid, meaning it can lose two protons. Therefore, we need to write two separate ionization equations for the two stages. The first ionization is given by: \(HOOC(CH_{2})_{4}COOH \Rightarrow H^{+} + HOOC(CH_{2})_{4}COO^{-}\) and the second ionization by: \(HOOC(CH_{2})_{4}COO^{-} \Rightarrow H^{+} + -(CH_{2})_{4}COO^{2-}\). Each reaction shows the dissociation of adipic acid into protons and the corresponding anionic forms.

Using the formula for equilibrium constant \(K_a = [H+][A−]/[HA]\), and the approximation that \(x << 0.10 M\) where \(x\) represents the concentration change, we can find the concentrations after the first ionization. Using \(K_{a1}\) we can calculate \(x\), representing the \(H^+\) concentration, which would also be the \(HOOC(CH_{2})_{4}COO^{-}\) concentration due to the 1:1 ratio. We also find the remaining adipic acid concentration, \(HA\), as \([HA] = 0.10 M - x\). Given the small \(K_{a1}\) value, we can make the approximation that \([HA] \approx 0.10M\). By substituting \(K_{a1}\) and these concentrations into the equilibrium constant formula, we can solve for \(x\), the hydrogen ion concentration after the first ionization.

Next, consider the ionization of the \(HOOC(CH_{2})_{4}COO^-\) ion using the second ionization equation. We use the same equilibrium constant formula but now with \(K_{a2}\). The removal of a proton from the \(HOOC(CH_{2})_{4}COO^-\) ion causes the formation of \(-(CH_{2})_{4}COO^{2-}\) ion and an additional \(H^+\). We consider the changes in concentration in the same method as in step 2 but using \(K_{a2}\). From this, we can find the new concentrations of each ion in the solution.

After determining concentrations from both ionizations, we add concentrations of like species to get total concentrations in the solution: total \(H^+\) concentration would be the sum of \(H^+\) from first and second ionization; total \((CH_{2})_{4}COO^{2-}\) concentration would equal concentration after the second ionization; total \(HOOC(CH_{2})_{4}COO^{-}\) concentration would equal its concentration after first ionization minus concentration after the second ionization as some of it further ionizes during second ionization.