Phthalic acid, \(\mathrm{H}_{2} \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}\), is a diprotic acid. It is used to make phenolphthalein indicator. \(K_{\mathrm{al}}=0.0012\), and \(K_{\mathrm{a} 2}=3.9 \times 10^{-6} .\) Calculate the \(\mathrm{pH}\) of a \(2.9 \mathrm{M}\) solution of phthalic acid. Estimate \(\left[\mathrm{HC}_{3} \mathrm{H}_{4} \mathrm{O}_{4}^{-}\right]\) and \(\left[\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}\right]\)

We'll begin by writing the equilibrium expressions for the ionization of phthalic acid. Since it is a diprotic acid, it ionizes in two steps: \(\mathrm{H}_{2} \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}\) with \(K_{a1}=0.0012\) \(\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}\) with \(K_{a2}=3.9 \times 10^{-6}\) Ignoring \(K_{a2}\) for a moment, we can focus on the \(K_{a1}\) expression: \(K_{a1} =\frac{[\mathrm{H}^{+}][\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}]}{[\mathrm{H}_{2} \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}]}\) Let \(x\) be the \(\left[\mathrm{H}^{+}\right]\). Since it is a weak acid, we assume [H+] is equal to \(x\). Also, the initial concentration of phthalic acid is 2.9 M, and the amount dissociated will be as well as \(x\), so: \(0.0012 = \frac{x * x}{2.9-x}\) This equation is a quadratic equation, but given the weak nature of the acid, we can simplify it assuming that \(x\) is much smaller than \(2.9\). Therefore, the equation becomes: \(0.0012 = \frac{x^2}{2.9}\) Now, solve for x.

Once we have the hydrogen ion concentration, we can calculate the pH of the solution. The formula for pH is: \(\mathrm{pH} = -\log_{10}[\mathrm{H}^{+}]\) Calculate the pH from the value obtained for \(x\) in step 1.

Now that we have the \(\left[\mathrm{H}^{+}\right]\), we can calculate the concentrations of the ionized forms of phthalic acid using the equilibrium constant expressions from step 1: \([\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}] = \frac{[\mathrm{H}^{+}][\mathrm{H}_{2} \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}]}{K_{a1}}\) \([\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}] = \frac{[\mathrm{H}^{+}][\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}]}{K_{a2}}\) Find the values of \(\left[\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}\right]\) and \(\left[\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}\right]\) using the equilibrium constants and \(\left[\mathrm{H}^{+}\right]\) value from step 1.