Problem 55 - Chemistry: Principles and Reactions

Question

For an aqueous solution of acetic acid to be called "distilled white vinegar" it must contain \(5.0 \%\) acetic acid by mass. A solution with a density of \(1.05 \mathrm{~g} / \mathrm{mL}\) has a \(\mathrm{pH}\) of \(2.95 .\) Can the solution be called "distilled white vinegar"?

Step-by-Step Solution

Answer

Answer: To determine if the solution can be considered "distilled white vinegar," calculate the mass percentage of acetic acid in the solution using the given density and pH. If the calculated mass percentage is equal to or greater than 5.0%, the solution is considered distilled white vinegar.

Step 1: Find the concentration of hydronium ions in the solution (H3O+)

To find the concentration of acetic acid, we first need to know the concentration of hydronium ions (H3O+) in the solution. This can be determined using the formula for pH: pH = -log10([H3O+]). Given: pH = 2.95 Let [H3O+] be the concentration of hydronium ions in the solution. We have the equation: 2.95 = -log10([H3O+]) Solving for [H3O+], we get: [H3O+] = 10^(-2.95)

Step 2: Calculate the concentration of acetic acid using the ionization constant Ka

Acetic acid (HC2H3O2) ionizes in water to produce hydronium ions (H3O+) and acetate ions (C2H3O2-): HC2H3O2 (aq) + H2O (l) <-> H3O+ (aq) + C2H3O2- (aq) The ionization constant (Ka) for acetic acid can be obtained from a standard table of Ka values: Ka = 1.8 x 10^(-5) Using Ka and the concentration of H3O+ ions, we can determine the concentration of acetic acid ([HC2H3O2]): Let [HC2H3O2] be the concentration of acetic acid in the solution Let [C2H3O2-] be the concentration of acetate ions in the solution Since the concentrations of H3O+ and C2H3O2- are equal, [C2H3O2-] = [H3O+] Hence, Ka = ([H3O+] * [C2H3O2-]) / [HC2H3O2]

Step 3: Solve for the concentration of acetic acid

Replacing the known values into the Ka equation: 1.8 x 10^(-5) = ( [H3O+] * [H3O+] ) / [HC2H3O2] [H3O+] = 10^(-2.95) HC2H3O2] = x 1.8 x 10^(-5) = (10^(-2.95) * 10^(-2.95)) / x Solve for x: x = (10^(-2.95) * 10^(-2.95)) / (1.8 x 10^(-5))

Step 4: Determine the mass percentage of acetic acid in the solution

Now that we have the concentration of acetic acid (x mol/L), We want to find out if this solution meets the 5.0% by mass requirement for distilled white vinegar. First, find the mass of acetic acid in 1L of solution: Mass of acetic acid = concentration of acetic acid * molar mass of acetic acid Mass of acetic acid = x * 60.05 g/mol Then, find the mass of 1L of solution using the given density (1.05 g/mL): mass of 1L solution = density * volume mass of 1L solution = 1.05 g/mL * 1000 mL = 1050 g Finally, calculate the mass percentage of acetic acid as follows: % acetic acid = (mass of acetic acid / mass of 1L solution) * 100 % acetic acid = (x * 60.05 / 1050) * 100

Step 5: Compare the calculated mass percentage with the requirement

If the calculated mass percentage of acetic acid in the solution is equal to or greater than 5.0%, then the solution can be considered distilled white vinegar. Compare the calculated value to the required value: % acetic acid ≥ 5.0% Based on the result, we can conclude whether the solution is distilled white vinegar or not.

Key Concepts

pH Calculation

The pH scale is a measure of how acidic or basic a solution is. It ranges from 0 to 14. A solution with a pH less than 7 is acidic, while one with a pH greater than 7 is basic. A pH of exactly 7 is considered neutral. To calculate the pH of a solution, you can use the formula: \[pH = -\log_{10}([H^+])\]Here,

• \([H^+]\) represents the concentration of hydronium ions.
• The formula shows the relationship between hydronium ion concentration and pH.

In this particular exercise, the pH of the acetic acid solution is given as 2.95. Using the formula, the concentration of hydronium ions \[H_3O^+\], in the solution is determined by solving \[2.95 = -\log_{10}([H_3O^+])\]This calculation involves applying the inverse of the logarithm: \[H_3O^+] = 10^{-2.95}\], which provides the concentration of hydrogen ions in molarity. This value is crucial for further calculations such as determining the concentration of acetic acid.

Ionization Constant (Ka)

Acetic acid (HC_2H_3O_2) is a weak acid, meaning that it does not completely dissociate in water. Instead, it establishes an equilibrium between undissociated acid and its ionized form:\[HC_2H_3O_2 (aq) + H_2O (l) \rightleftharpoons H_3O^+ (aq) + C_2H_3O_2^- (aq)\]The ionization constant, denoted as \(K_a\), describes the extent to which an acid dissociates in solution.

• Higher \(K_a\) values mean greater ionization, indicating a stronger acid.
• The link between pH and ionization is important because it helps us figure out how much of the acid ionizes to form hydronium ions.

For acetic acid, the \(K_a\) value is a well-known constant: \(1.8 \times 10^{-5}\). Once we know the \(K_a\) or the hydronium ion concentration, we can find how much of the undissociated acetic acid is in the solution using the expression: \[K_a = \frac{[H_3O^+] \cdot [C_2H_3O_2^-]}{[HC_2H_3O_2]}\]In this reflection, the concentrations of \([H_3O^+]\) and \([C_2H_3O_2^-]\) are equivalent because each molecule of acetic acid produces one hydronium ion and one acetate ion.

Mass Percentage in Solutions

The mass percentage of a component in a solution is a way of expressing its concentration. It denotes how much of a substance (in mass) is present per 100 units of total solution mass. To find the mass percentage of acetic acid in our solution, follow these steps:

• First, calculate the mass of acetic acid in a given volume of solution. This involves using the concentration obtained from the K_a calculation and the molar mass of acetic acid, which is 60.05 \(g/mol\).
• Next, determine the mass of the entire solution using its density. Given a density of \(1.05 \, g/mL\), a 1-liter solution has a mass of 1050 grams.
• Finally, apply the mass percentage formula: \[\% \text{acetic acid} = \left( \frac{\text{Mass of acetic acid}}{\text{Total mass of solution}} \right) \times 100 \]

This process tells us whether the solution meets the 5% mass requirement characteristic of distilled white vinegar. If the calculated value is equal to or exceeds this requirement, then the solution is considered to be distilled white vinegar.