Question

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. \(K_{\mathrm{a} 1}=0.0012\), 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 \(\left[\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}\right]\) and \(\left[\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}\right] .\)

Short answer

The pH of the 2.9 M solution of phthalic acid is approximately 1.49. The estimated concentrations for both conjugate bases, HC₈H₄O₄⁻ and C₈H₄O₄²⁻, are approximately 0.102 M and 1.26 x 10⁻⁵ M, respectively.

Step-by-step solution

Determine the ionization steps

Given that phthalic acid (\(\mathrm{H}_{2}\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}\)) is a diprotic acid, it ionizes in two steps: 1. For \(K_{a1}\): \(\mathrm{H}_{2}\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4} \rightleftharpoons \mathrm{H}^{+} + \mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}\) 2. For \(K_{a2}\): \(\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-} \rightleftharpoons \mathrm{H}^{+} + \mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{2-}\)

Calculate the number of moles for the ionization steps

Ignore the second ionization step initially, because the first ionization will have more impact on the concentration of \(\mathrm{H}^{+}\) ions in solution. We set up the equilibrium expression for the first ionization step: \(K_{a1}=(5.6 \times 10^{-13})=\dfrac{[ \mathrm{H}^{+}] [\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}]}{[\mathrm{H}_{2}\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}]}\) \([\mathrm{H}^{+}]\) = x, \([\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}]\) = x, \([\mathrm{H}_{2}\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}]\)= 2.9 - x

Solve for the concentration of \(\mathrm{H}^+\) ions

Plug in the variables into the equilibrium expression and solve for x: \(0.0012=\dfrac{x \cdot x}{2.9 - x}\) Since \(K_{a1}\) is very small, we can assume that \(x \ll 2.9\). Therefore, we can simplify the equation and solve for x: \(0.0012=\dfrac{x^{2}}{2.9}\) \(x = \sqrt{0.0012 * 2.9} \approx 0.032\) We assume that x is the concentration of \(\mathrm{H}^+\) in the solution, which is approximately \(0.032\).

Calculate the pH of the solution

We use the pH formula to calculate the pH of the solution: \(\mathrm{pH} = -\log\left[\mathrm{H}^{+}\right]\) \(\mathrm{pH} = -\log\left(0.032\right) \approx 1.49\) Hence, the pH of the 2.9 M solution of phthalic acid is approximately 1.49.

Estimate the concentration of both conjugate bases

Now that we have the concentration of \(\mathrm{H}^{+}\) ions, we can use the ionization equilibrium expressions to estimate the concentrations of \(\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}\) and \(\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{2-}\): 1. \(\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}: K_{a1}=\dfrac{[\mathrm{H}^{+}] [\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}]}{[\mathrm{H}_{2}\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}]}\) Rearrange the equation: \([\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}]=\dfrac{K_{a1} [\mathrm{H}_{2}\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}]}{[\mathrm{H}^{+}]}\) Plug in the values from the question: \([\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}]=\dfrac{0.0012 \times (2.9 - 0.032)}{0.032} \approx 0.102\) 2. \(\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{2-}: K_{a2}=\dfrac{[\mathrm{H}^{+}] [\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{2-}]}{[\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}]}\) Rearrange the equation: \([\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{2-}]=\dfrac{K_{a2} [\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}]}{[\mathrm{H}^{+}]}\) Plug in the found values for the conjugate base: \([\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{2-}] =\dfrac{3.9 \times 10^{-6} \times 0.102}{0.032} \approx 1.26 \times 10^{-5}\) So, the estimated concentrations for \(\mathrm{HC}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{-}\) and \(\mathrm{C}_{8}\mathrm{H}_{4}\mathrm{O}_{4}^{2-}\) are approximately \(0.102 \mathrm{M}\) and \(1.26 \times 10^{-5} \mathrm{M}\), respectively.