Questions: pKa = 4.76, 3.00 M NaOH, and water. How many grams of acetic acid will be needed to prepare the 500 mL buffer? Note that the given concentration of acetate refers to the concentration of all acetate species in solution. mass: g

Solution Steps

To prepare a buffer solution, we need to use the Henderson-Hasselbalch equation, which is given by:

\[ \mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

where:

• \([\text{A}^-]\) is the concentration of the acetate ion,
• \([\text{HA}]\) is the concentration of acetic acid,
• \(\mathrm{p}K_a\) is the negative logarithm of the acid dissociation constant.

Given that \(\mathrm{p}K_a = 4.76\), we need to determine the amount of acetic acid required to achieve a specific buffer pH. However, the problem does not specify the desired pH, so we will assume the buffer is to be prepared at the pH equal to the \(\mathrm{p}K_a\) of acetic acid, which is 4.76. This simplifies the equation to:

\[ 4.76 = 4.76 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

This implies:

\[ \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = 0 \]

Thus, \([\text{A}^-] = [\text{HA}]\).

Since \([\text{A}^-] = [\text{HA}]\), the concentration of acetic acid \([\text{HA}]\) will be equal to the concentration of acetate ions. The problem states that the concentration of all acetate species is given, but does not specify the concentration. Assuming a typical buffer concentration, let's say the concentration of acetate is \(0.1 \, \text{M}\).

The volume of the buffer solution is \(500 \, \text{mL} = 0.5 \, \text{L}\). The moles of acetic acid needed are:

\[ \text{moles of HA} = 0.1 \, \text{M} \times 0.5 \, \text{L} = 0.05 \, \text{moles} \]

The molar mass of acetic acid (\(\text{CH}_3\text{COOH}\)) is approximately \(60.05 \, \text{g/mol}\). Therefore, the mass of acetic acid required is:

\[ \text{mass} = 0.05 \, \text{moles} \times 60.05 \, \text{g/mol} = 3.0025 \, \text{g} \]

Final Answer