Question

The hydrogen phthalate ion, \(\mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4}^{-},\) is a weak acid with \(K_{\mathrm{a}}=3.91 \times 10^{-6}\). $$\mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{C}_{8} \mathrm{H}_{4} \mathrm{O}_{4}^{2-}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})$$ What is the pH of a 0.050 M solution of potassium hydrogen phthalate, \(\mathrm{KC}_{8} \mathrm{H}_{5} \mathrm{O}_{4}\) ? Note. To find the pH for a solution of the anion, we must take into account that the ion is amphiprotic. It can be shown that, for most cases of amphiprotic ions, the \(\mathrm{H}_{3} \mathrm{O}^{+}\) concentration is $$\left[\mathbf{H}_{3} \mathbf{O}^{+}\right]=\sqrt{K_{1} \times K_{2}}$$ For phthalic acid, \(\mathrm{C}_{8} \mathrm{H}_{6} \mathrm{O}_{4}, K_{1}\) is \(1.12 \times 10^{-3},\) and \(K_{2}\) is \(3.91 \times 10^{-6}\).

Short answer

The pH of the 0.050 M solution of potassium hydrogen phthalate is approximately 4.18.

Step-by-step solution

Understanding Problem and Equations

Potassium hydrogen phthalate (KHP) solution involves an amphiprotic anion: \( \mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4}^{-} \). The equation provided \( \left[\mathrm{H}_{3}\mathrm{O}^{+}\right]=\sqrt{K_{1} \times K_{2}} \) helps determine \( \mathrm{H}_{3}\mathrm{O}^{+} \) for such ions.

Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\)

Using the provided equation \( \left[\mathrm{H}_{3}\mathrm{O}^{+}\right]=\sqrt{K_{1} \times K_{2}} \), substitute \( K_1 = 1.12 \times 10^{-3} \) and \( K_2 = 3.91 \times 10^{-6} \). Calculate this: \( \sqrt{1.12 \times 10^{-3} \times 3.91 \times 10^{-6}} = \sqrt{4.3792 \times 10^{-9}} \approx 6.62 \times 10^{-5} \).

Calculate pH

With \( \left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \approx 6.62 \times 10^{-5} \), use the formula \( \text{pH} = -\log_{10}\left( \left[\mathrm{H}_{3} \mathrm{O}^{+} \right] \right) \). Thus, \( \text{pH} = -\log_{10}\left( 6.62 \times 10^{-5} \right) \approx 4.18 \).

Verify Assumptions and Simplifications

The method used is suitable for amphiprotic species, assuming contributions from other equilibria are minor. This simplifies the calculation by focusing on the provided \( K_1 \times K_2 \) relation.

Key concepts

Weak Acid

In chemistry, a weak acid is characterized by its partial ionization in water. Unlike strong acids, which fully dissociate, weak acids only partially release their hydrogen ions into the solution. This means not all the acid molecules dissociate to produce hydronium ions (

• A weak acid forms an equilibrium between the undissociated acid and the ions it releases.
• The extent to which a weak acid dissociates is quantified by its acid dissociation constant, known as Ka.
• The smaller the Ka value, the weaker the acid, indicating less ionization.

Potassium hydrogen phthalate, (\(\mathrm{KC}_{8} \mathrm{H}_{5} \mathrm{O}_{4} \)), is an example of a weak acid, specifically due to its amphiprotic nature.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, maintaining constant concentrations of reactants and products in a reaction mixture. For weak acids, this balance includes both the ionized and unionized forms of the acid.

• In the context of weak acids, equilibrium is essential in understanding how much acid is ionized.
• The expression for equilibrium is often denoted by a double arrow (\( \rightleftarrows \)).
• The acid dissociation constant (Ka) is used to quantify the dynamic balance.

This concept is crucial while analyzing the behavior of the hydrogen phthalate ion in water. The given equation in the problem indicates the presence of an equilibrium involving hydrogen phthalate and its ionized forms.

Amphiprotic Ion

An amphiprotic ion has the capacity to either donate or accept a proton, behaving as an acid or a base depending on the surrounding environment. In the case of (\(\mathrm{C}_{8} \mathrm{H}_{5} \mathrm{O}_{4}^{-} \)), it can both donate a proton to form \(\mathrm{H}_{3} \mathrm{O}^{+} \) and accept one.

• This dual nature makes the hydrogen phthalate ion a typical amphiprotic ion, crucial in equilibrium reactions.
• The presence of different equilibrium constants (\(K_1\) and \(K_2\)) explains its behavior in solution.
• The calculation for hydronium ion concentration uses the geometric mean formula \([\mathbf{H}_{3}\mathbf{O}^{+}] = \sqrt{K_{1} \times K_{2}}\),

which effectively predicts the hydrogen ion concentration in such systems.

Hydronium Ion Concentration

The hydronium ion concentration is pivotal in determining the pH of a solution. It reflects the concentration of hydrogen ions, which directly influence acidity.

• For a weak acid solution, hydronium ion concentration can be calculated using its equilibrium constant(s).
• It is often depicted using \([\mathrm{H}_{3}\mathrm{O}^{+}] \).
• To find the pH, which is a measure of acidity, we calculate:\
\[\text{pH} = -\log_{10} \left( [\mathrm{H}_{3}\mathrm{O}^{+}] \right)\]

In the provided exercise, the concentration is established through the square root of the product of two equilibrium constants \( K_1 \) and \( K_2 \). This approach simplifies calculating pH for solutions containing amphiprotic ions.