Problem 61 - Chemistry: Principles and Reactions

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

Step-by-Step Solution

Answer

Answer: After following the steps to solve for pH, \([\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}]\), and \([\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}]\), the pH of the solution will be approximately 1.18, with \([\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}]\) and \([\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}]\) concentrations being approximately 0.0010 M and 2.6 x 10⁻⁷ M, respectively.

Step 1: Calculate the hydrogen ion concentration

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.

Step 2: Convert the hydrogen ion concentration into pH

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.

Step 3: Calculate the concentrations of \(\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}\) and \(\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}\)

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.

Key Concepts

Diprotic Acid

Phthalic acid, with the chemical formula \(\mathrm{H}_{2} \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}\), is an example of a diprotic acid. This means it can donate two protons (\(\mathrm{H}^{+}\) ions) during its ionization process. Such acids undergo ionization in stages, each losing one proton at a time. Here's how:

• The first stage is the release of the first hydrogen ion:

\[\mathrm{H}_{2} \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4} \rightleftharpoons \mathrm{H}^{+} + \mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-} \] Key points to remember:

• In the first ionization, the acid donates one proton, forming a compound that can further donate another proton.
• Diprotic acids like phthalic acid have two ionization constants, \(K_{a1}\) and \(K_{a2}\), with \(K_{a1}\) being larger than \(K_{a2}\), indicating that the first proton is usually more easily removed than the second.

This quality makes diprotic acids a bit more complex to work with compared to monoprotic acids, as you must consider each step of ionization separately.

Equilibrium Expression

Equilibrium expressions are essential in understanding how chemical reactions behave under different conditions. For an acid, like phthalic acid, at equilibrium, we use the equilibrium expression to describe the extent of ionization.For the ionization of phthalic acid, we can write the equilibrium expressions as follows:### First Ionization Stage:The first equilibrium expression for the ionization of phthalic acid is:\[K_{a1} = \frac{[\mathrm{H}^{+}][\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}]}{[\mathrm{H}_{2} \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}]} \] This set of brackets \([]\) indicates the concentration of each component in moles per liter (M). Thus, \([\mathrm{H}^{+}]\), \([\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}]\), and \([\mathrm{H}_{2} \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}]\) are the concentrations at equilibrium.### Second Ionization Stage:For the second stage:\[K_{a2} = \frac{[\mathrm{H}^{+}][\mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}]}{[\mathrm{HC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{-}]} \] Understanding these expressions helps to determine how much of each species will be present in solution at equilibrium.

Ionization Constant

The ionization constant of an acid, known as \(K_a\), is a measurement of its strength and its tendency to donate a proton to form hydrogen ions in solution. For diprotic acids like phthalic acid, we have two ionization constants:

• \(K_{a1}\) for the first ionization step
• \(K_{a2}\) for the second ionization step

Why is this important?

• \(K_{a1}\) and \(K_{a2}\) values allow us to understand how strongly the acid acts in each step of ionization.
• In phthalic acid, \(K_{a1} = 0.0012\) and \(K_{a2} = 3.9 \times 10^{-6}\). As we can see, \(K_{a1}\) is considerably larger than \(K_{a2}\), suggesting the first ionization occurs more readily.
• These values indicate that the first step significantly affects the pH due to higher ionization, while the second step has a much smaller impact.

Understanding \(K_a\) values allows for precise calculations of pH and the concentrations of species at equilibrium.

Weak Acid Assumption

When dealing with weak acids in equilibrium, scientists often use what is known as the "weak acid assumption." This assumption simplifies complex calculations.How does it work?

• Weak acids only partially ionize in solution.
• This partial ionization allows simplifying assumptions, namely that the concentration of the undissociated acid essentially remains constant.
• For equations like \(K_{a1} = \frac{x^2}{2.9}\), the weak acid assumption allows us to neglect the change \(x\) in the denominator, i.e., \(2.9 - x \approx 2.9\).

This assumption is valid because the actual dissociation of a weak acid in a concentrated solution is often small. This simplifies the quadratic equation to a more straightforward linear equation, making calculations manageable even with minimal computational tools.Remember that this assumption is checked by ensuring \(\frac{x}{[\text{acid}]}\) is small, typically less than 5%, to maintain accuracy.