Problem 130 - General Chemistry

Question

Malic acid is a weak diprotic organic acid with \(K_{a 1}=4.0 \times 10^{-4}\) and \(K_{a 2}=9.0 \times 10^{-6}\) a. Letting the symbol \(\mathrm{H}_{2}\) A represent malic acid, write the chemical equations that represent \(K_{a 1}\) and \(K_{a 2}\). Write the chemical equation that represents \(K_{a 1} \times K_{a 2}\) b. Qualitatively describe the relative concentrations of \(\mathrm{H}_{2} \mathrm{~A}, \mathrm{HA}^{-}, \mathrm{A}^{2-},\) and \(\mathrm{H}_{3} \mathrm{O}^{+}\) in a solution that is about one molar in malic acid. c. Calculate the \(\mathrm{pH}\) of a \(0.0175 \mathrm{M}\) malic acid solution and the equilibrium concentration of \(\left[\mathrm{H}_{2} \mathrm{~A}\right]\) d. What is the \(\mathrm{A}^{2-}\) concentrationin in solutions \(\mathrm{D}\) and \(\mathrm{c}\) ?

Step-by-Step Solution

Answer

The chemical equations are defined, pH of 0.0175 M solution is 2.58, and \([A^{2-}]\) is \(9.0 \times 10^{-6}\) M.

Step 1: Define Chemical Equilibrium Expressions

For part (a), we need to write the chemical reactions for each dissociation step of malic acid, denoted as \( \text{H}_2\text{A} \). The reaction for \( K_{a1} \) is \( \text{H}_2\text{A} \leftrightharpoons \text{H}^+ + \text{HA}^- \). The equilibrium expression is \( K_{a1} = \frac{[\text{H}^+][\text{HA}^-]}{[\text{H}_2\text{A}]} = 4.0 \times 10^{-4} \). For \( K_{a2} \), the reaction is \( \text{HA}^- \leftrightharpoons \text{H}^+ + \text{A}^{2-} \) with the expression \( K_{a2} = \frac{[\text{H}^+][\text{A}^{2-}]}{[\text{HA}^-]} = 9.0 \times 10^{-6} \). The combined equilibrium (\( K_{a1} \times K_{a2} \)) is \( \text{H}_2\text{A} \leftrightharpoons 2\text{H}^+ + \text{A}^{2-} \) with the expression \( K_{a1} \times K_{a2} = \frac{[\text{H}^+]^2[\text{A}^{2-}]}{[\text{H}_2\text{A}]} \).

Step 2: Describe Relative Concentrations

For part (b), we describe the concentrations in a solution that is about one molar malic acid. Initially, \( [\text{H}_2\text{A}] \) is high. After partial dissociation, \( [\text{HA}^-] \) is less than \( [\text{H}_2\text{A}] \), and \( [\text{A}^{2-}] \) is even smaller, given the significantly smaller \( K_{a2} \) versus \( K_{a1} \). The \( [\text{H}_3\text{O}^+] \) concentration depends on the first dissociation, so \( [\text{H}_3\text{O}^+] \approx \left[K_{a1}\right]. \) Thus, we expect: \( [\text{H}_2\text{A}] > [\text{HA}^-] > [\text{A}^{2-}] \).

Step 3: Calculate pH and \([\text{H}_2\text{A}]\)

For part (c), use the initial concentration \( [\text{H}_2\text{A}]_0 = 0.0175 \text{ M} \). Apply the first dissociation expression. \[ \begin{align*} \text{H}_2\text{A} &\leftrightharpoons \text{H}^+ + \text{HA}^- \ K_{a1} &\approx \frac{x^2}{0.0175-x} \approx \frac{x^2}{0.0175} \x &= [\text{H}^+] = [\text{HA}^-] = \sqrt{K_{a1} \times 0.0175} \ &\approx \sqrt{4.0 \times 10^{-4} \times 0.0175} = \sqrt{7.0 \times 10^{-6}} \ &\approx 2.645 \times 10^{-3} \end{align*} \] The pH is then \( \text{pH} = -\log(2.645 \times 10^{-3}) \approx 2.58 \), and \( [\text{H}_2\text{A}]\approx 0.0175 - 2.645 \times 10^{-3} = 0.0149 \text{ M} \).

Step 4: Determine \([\text{A}^{2-}]\) Concentration

For part (d), the \( \text{A}^{2-} \) concentration in 0.0175 M solution (as calculated) is determined using the second dissociation equilibrium: \[ K_{a2} = \frac{x \cdot 2.645 \times 10^{-3}}{2.645 \times 10^{-3}} \] Substituting the known values, \[ \begin{align*} x &\approx K_{a2} = 9.0 \times 10^{-6} \end{align*} \] Therefore, \( [\text{A}^{2-}] \approx 9.0 \times 10^{-6} \text{ M} \)

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a crucial concept in understanding how reactions reach a state of balance. In the context of malic acid, a weak diprotic acid, the chemical equilibrium involves the dissociation of the acid into its constituent ions. Malic acid can dissociate in two steps:

• The first dissociation (\( K_{a1} \)) where one hydrogen ion is released, turning \( ext{H}_2 ext{A} \) into \( ext{HA}^- \).
• The second dissociation (\( K_{a2} \)) where another hydrogen ion is released, converting \( ext{HA}^- \) into \( ext{A}^{2-} \).

The equilibrium expressions for these reactions involve the equilibrium constant \( K_a \), which represents the ratio of the products and reactants when the reaction is at equilibrium. The equilibrium constants, \( K_{a1} \) and \( K_{a2} \), provide a measure of the extent to which malic acid dissociates in solution. Smaller constants indicate weaker dissociation.Understanding these equilibria helps predict how malic acid behaves in aqueous solutions and its role in maintaining the balance of hydrogen ions, which directly impacts its acidity levels.

Diprotic Acid

A diprotic acid is an acid that contains two ionizable hydrogen ions, also known as protons. Malic acid is an example of a diprotic acid. It goes through two stages of dissociation as it releases its protons:

• The first stage involves the hydrogen dissociation turning \( ext{H}_2 ext{A} \) into \( ext{HA}^- \).
• The second stage further dissociates \( ext{HA}^- \) into \( ext{A}^{2-} \).

Each dissociation has its own equilibrium constant (\( K_{a1} \) and \( K_{a2} \)), reflecting how easily each proton is lost. Typically, the first ionization constant (\( K_{a1} \)) is larger because the initial dissociation requires less energy compared to the second. This results in the second equilibrium constant (\( K_{a2} \)) being significantly smaller.The concept of diprotic acids is important in practical applications such as buffer solutions where two dissociation steps provide versatility in maintaining pH levels. Exploring these dissociation stages helps in understanding how diprotic acids like malic acid contribute to acid-base balance in chemical and biological environments.

pH Calculation

The pH of a solution is an important measurement that indicates its acidity or basicity. Calculating the pH of a malic acid solution involves understanding its dissociation and the equilibrium constants. For a 0.0175 M malic acid solution, we primarily consider the first dissociation due to its larger constant \( K_{a1} \). This expression helps us estimate the concentration of hydrogen ions:
\[ \text{H}_2\text{A} \leftrightharpoons \text{H}^+ + \text{HA}^- \quad K_{a1} = \frac{[\text{H}^+][\text{HA}^-]}{[\text{H}_2\text{A}]} \]
The small change in concentration \( x \) for \( \text{H}^+ \) gives us an approximate solution:

• Calculate \( x \) as \( \sqrt{K_{a1} \times 0.0175} \)
• Use the relation \( \text{pH} = -\log(x) \) to find the pH value

For this solution, \( x \approx 2.645 \times 10^{-3} \), so the pH is approximately 2.58.Understanding how to calculate the pH accurately is crucial, especially when assessing the acidity of organic acids like malic acid, which is widely found in fruits and other natural products.

Organic Acids

Organic acids are a class of acids that contain carbon-hydrogen bonds and typically have at least one carboxyl (-COOH) group. Malic acid is a well-known organic acid with the formula \( ext{C}_4 ext{H}_6 ext{O}_5 \). It is naturally found in fruits and contributes to their sour taste. Organic acids like malic acid often serve vital roles in both nature and industry.
Malic acid, being a diprotic acid, exhibits unique properties in its dissociation, producing different ionic forms such as \( ext{HA}^- \) and \( ext{A}^{2-} \). The presence of these ions contributes to its usefulness in forming stable complexes and having an impact on pH levels in solutions.

• Malic acid is commonly used as a food additive to enhance flavor and preserve freshness.
• It plays a part in the Krebs cycle, an essential metabolic pathway in cellular respiration.
• Due to its buffering ability, it's used in skincare products to adjust acidity levels.

Understanding organic acids helps us appreciate their chemical behaviors and applications, from daily life activities to biological processes.