Problem 96 - Chemistry The Molecular Science

Question

Ascorbic acid (vitamin \(\left.\mathrm{C}, \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}\right)\) is a diprotic acid \(\left(K_{\mathrm{a}_{1}}=7.9 \times 10^{-5}, K_{\mathrm{a}_{2}}=1.6 \times 10^{-12}\right) .\) Calculate the \(\mathrm{pH}\) of a solution that contains \(5.0 \mathrm{mg}\) acid per mL water. (Assume that only the first ionization is important in determining pH.)

Step-by-Step Solution

Answer

The pH of the solution is approximately 2.82.

Step 1: Convert Mass to Moles

First, convert the mass of ascorbic acid in the solution to moles. We know the molar mass of ascorbic acid (C_{6}H_{8}O_{6} ) is approximately 176.12 g/mol. The solution contains 5.0 mg/mL, so in 1 mL of water, the mass is 0.005 g. Calculate moles using the formula: moles = mass (g) / molar mass (g/mol).\[\text{Moles of } C_6H_8O_6 = \frac{0.005 \text{ g}}{176.12 \text{ g/mol}} \approx 2.84 \times 10^{-5} \text{ mol}\]

Step 2: Calculate Molarity

Determine the molarity (M ) of the solution, assuming the volume of the solution is 1 L. Since we have already calculated the moles in 1 mL solution above, multiply by 1000 to convert to moles per liter (mol/L or M).\[M = 2.84 \times 10^{-5} \text{ mol/mL} \times 1000 \approx 0.0284 \text{ M}\]

Step 3: Calculate Initial Concentration of H^+

Since ascorbic acid is a diprotic acid, it has two ionization steps. However, we assume only the first ionization matters here, with K_{a_1} = 7.9 \times 10^{-5}. Use the formula for a weak acid to find [H^+]:\[K_a = \frac{[H^+][A^-]}{[HA]} \approx \frac{[H^+]^2}{[HA]}\]. Plugging in values gives:\[7.9 \times 10^{-5} = \frac{x^2}{0.0284}\]

Step 4: Solve for H^+ Concentration

To find [x] = [H^+], solve 7.9 \times 10^{-5} = \frac{x^2}{0.0284} for x.\[x^2 = 7.9 \times 10^{-5} \times 0.0284\]\[x^2 = 2.244 \times 10^{-6}\]\[x = \sqrt{2.244 \times 10^{-6}} \approx 1.5 \times 10^{-3} \text{ M}\]

Step 5: Calculate pH

The pH is calculated using the formula \text{pH} = -\log[H^+]. Using [H^+] = 1.5 \times 10^{-3} M:\[\text{pH} = -\log(1.5 \times 10^{-3}) \]Using the logarithm:\[\text{pH} \approx 2.82\]

Key Concepts

Ascorbic Acid

Ascorbic acid, commonly known as vitamin C, is an essential nutrient for humans. It is a vital component in the growth, development, and repair of all body tissues. Ascorbic acid's chemical formula is \( C_6H_8O_6 \), and it exists in a crystalline form that is highly soluble in water.
It serves as a potent antioxidant, protecting the body against the harmful effects of free radicals. Additionally, it supports immune function and helps synthesize collagen, a crucial part of connective tissue.

• Antioxidant: Neutralizes free radicals.
• Immune Support: Boosts the body's defenses.
• Collagen Production: Essential for tissue repair.

In chemistry, ascorbic acid is an organic compound that acts as a diprotic acid, meaning it can donate two protons \( (H^+) \) per molecule during dissociation. Understanding its properties, such as the ability to affect solutions' pH, is key for various applications ranging from food preservation to skincare products.

Diprotic Acid

Diprotic acids are acids capable of donating two protons per molecule in a solution. Ascorbic acid is an example of a diprotic acid, with two dissociation constants: \( K_{a1} = 7.9 \times 10^{-5} \) and \( K_{a2} = 1.6 \times 10^{-12} \). These constants indicate how easily the acid donates its protons during the first and second ionization processes respectively.
When assessing the acidity of a solution, the first ionization step is often the most significant for weak diprotic acids like ascorbic acid, making \( K_{a1} \) crucial for pH calculations, especially when dealing with dilute solutions.
In the exercise given, only the first dissociation step is considered in determining the pH because it has a much larger constant, which means it occurs more readily compared to the second step.

• First Ionization: \( K_{a1} \), predominant in pH calculation.
• Second Ionization: \( K_{a2} \), often negligible in weak acids.

This understanding helps in making accurate predictions about a solution's behavior when utilising acids in chemical processes.

Molarity Calculation

Determining the molarity of a solution is a fundamental skill in chemistry, key for predicting how substances react in solution. Molarity \( (M) \) represents the number of moles of solute per liter of solution. It's calculated using the formula: \[M = \frac{\text{moles of solute}}{\text{liters of solution}}\]
In the given exercise, the molarity is calculated by first converting the mass of ascorbic acid in grams to moles using its molar mass, and then adjusting this value to per liter.
The exercise showed that when 5.0 mg of ascorbic acid is dissolved in 1 mL of water, this can be scaled to a 1-liter measurement, providing a molarity of approximately 0.0284 M. This step helps in further calculations involving chemical reactions, equilibria, and pH values.

Ionization Constant

Ionization constants \( (K_a) \) are critical in understanding an acid's strength and how it behaves in a solution. For ascorbic acid, given the values \( K_{a1} = 7.9 \times 10^{-5} \) and \( K_{a2} = 1.6 \times 10^{-12} \), these constants help us calculate the concentration of ions in a solution, which directly affects the pH.
The ionization constant is used to describe the equilibrium state of an acid in water, reflecting the extent to which the acid donates its protons. For weak acids like ascorbic acid, it typically leads to a partial dissociation in water.

• \( K_a \) Value: Indicates acid strength.
• Higher \( K_a \): Greater ionization, stronger acid.
• Lower \( K_a \): Lesser ionization, weaker acid.

By using the ionization constant in calculations, one can derive the concentration of hydrogen ions \( [H^+] \) and thus find the pH of a solution accurately, offering insight into the acid's dissociative properties.