Problem 39 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Ascorbic acid (vitamin \(\mathrm{C}\) ) is a diprotic acid, \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}\). What is the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution? The acid ionization constants are \(K_{\mathrm{al}}=9.0 \times 10^{-5}\) and \(K_{\mathrm{a} 2}=1.6 \times 10^{-12} \cdot(\log 2=0.3, \log 3=0.48)\) (a) \(3.52\) (b) \(2.52\) (c) \(1.52\) (d) \(2.48\)

Step-by-Step Solution

Answer

The pH of the 0.10 M ascorbic acid solution is approximately 3.52.

Step 1: Identify the Relevant Equations

Since ascorbic acid is a diprotic acid, it can release two protons successively. The ionization can be represented with two steps, and each step has its own equilibrium constant: for the first ionization, we use \(K_{a1}\) and for the second ionization \(K_{a2}\). We'll first deal with the first ionization as it is the dominant source of \(H^+\) ions. The equation representing the first ionization is: \(H_2C_6H_6O_6 \leftrightarrow H^+ + HC_6H_6O_6^-\). To calculate the pH of the solution, we will need to find the concentration of \(H^+\) ions using \(K_{a1}\) and the initial concentration of ascorbic acid.

Step 2: Set Up the Equilibrium Expression for the First Ionization

Using the acid dissociation constant \(K_{a1}\), we can write the expression: \(K_{a1} = \frac{[H^+][HC_6H_6O_6^-]}{[H_2C_6H_6O_6]}\). Assuming that the change in concentration for \(H^+\) and \(HC_6H_6O_6^-\) is \(x\) and \(x\) is much smaller compared to the initial concentration, the equilibrium concentrations become: \(\frac{x^2}{0.10 - x} \approx \frac{x^2}{0.10}\).

Step 3: Solve for \(x\)

Substituting \(K_{a1}\) into the equilibrium expression gives: \((9.0 \times 10^{-5} = \frac{x^2}{0.10}\), solving for \(x\) gives \(x = \sqrt{9.0 \times 10^{-5} \times 0.10} = \sqrt{9.0 \times 10^{-6}} = 3.0 \times 10^{-3}\). This \(x\) represents the concentration of \(H^+\) ions in the solution.

Step 4: Calculate the \(pH\)

The \(pH\) is calculated using the formula \(pH = -\log[H^+]\). Substituting the value of \(x\), we get \(pH = -\log(3.0 \times 10^{-3})\).

Step 5: Approximate the \(pH\) Using Logarithm Properties

Using logarithm rules, \(\log(xy) = \log(x) + \log(y)\) and approximate logarithm values given for \(\log(2) = 0.3\) and \(\log(3) = 0.48\), we get \(\log(3.0 \times 10^{-3}) = \log(3) + \log(10^{-3}) = 0.48 - 3\). So, the approximate \(pH\) of the ascorbic acid solution is \((3.52\).

Step 6: Select the Correct Answer

The calculation yields a \(pH\) of approximately \(3.52\), therefore, the correct answer is (a) \(3.52\).

Key Concepts

Understanding Ascorbic Acid

Ascorbic acid, also known as vitamin C, plays an essential role in human nutrition. Its chemical formula is \( \mathrm{H}_{2}\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6} \), illustrating that it's a molecule with two acidic protons, hence it's referred to as a diprotic acid. In the context of pH calculation, ascorbic acid can release these protons sequentially into an aqueous solution, each through its own ionization process.

These ionizations occur in stages. The first stage involves the loss of one proton to form the ascorbate ion (\( HC_6H_6O_6^- \)), and the second stage involves this ion losing another proton. In a typical solution of ascorbic acid, such as the one under study, the first ionization is more significant because the first dissociation constant (\( K_{a1} \)) is relatively larger than that of the second (\( K_{a2} \)). Consequently, when calculating the pH of a moderately concentrated solution of ascorbic acid, the contribution from the second ionization is often so small that it can be neglected in the pH calculation.

Understanding this two-step ionization process is key to accurately determining the pH of ascorbic acid solutions. The nature of ascorbic acid also stresses the importance of pKa values, which influence the strength of the acid and its behavior in solution.

Acid Dissociation Constant Clarified

The acid dissociation constant, often denoted as \( K_a \), quantifies the strength of an acid in solution. It serves as an equilibrium constant for the dissociation reaction of the acid, where the acid (\( HA \)) donates a proton (\( H^+ \)) to water, producing the conjugate base (\( A^- \)) and a hydronium ion (\( H_3O^+ \)), or as expressed in the context of ascorbic acid, a hydrogen ion (\( H^+ \)) and an ascorbate ion (\( HC_6H_6O_6^- \)).

The value of \( K_a \) informs us about the extent of ionization at equilibrium. A larger \( K_a \) suggests a stronger acid, which means a higher concentration of \( H^+ \) ions and, consequently, a lower pH value. Conversely, a smaller \( K_a \) indicates a weaker acid with less ionization. When dealing with a diprotic acid like ascorbic acid, we encounter two \( K_a \) values, one for each ionization step. Predictably, the first dissociation constant \( K_{a1} \) is always higher than the second \( K_{a2} \).

Our exercise revealed that the pH calculation is contingent upon understanding and applying \( K_{a1} \) because it represents the predominant ionization step that most influences the pH. Indeed, manipulating the \( K_a \) values is fundamental to predicting the behavior of acids in solution.

Chemical Equilibrium in Acids

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in no apparent change in the system over time. This dynamic state is essential in the context of acids in solution, such as ascorbic acid, where it defines the balance between the undissociated acid and the ions produced through dissociation.

When solvated in water, ascorbic acid reaches an equilibrium state where the rate at which it donates protons to water matches the rate at which the ascorbate ions and hydrogen ions recombine to form the acid again. This equilibrium can be disturbed by changes in conditions such as concentration, temperature, or the addition of other substances.

Importantly, when calculating the pH of an acid solution, we assume that equilibrium has been achieved. This allows us to use equilibrium constants like \( K_{a1} \) and expressions that reflect the stoichiometry of the dissociation reaction to find the concentrations of all species at equilibrium. The exercise guided us through this process and demonstrated that accurate use of equilibrium concepts is vital for determining the acidity of a solution.