Question

Calculate the mass of sodium acetate that must be added to 500.0 \(\mathrm{mL}\) of 0.200\(M\) acetic acid to form a \(\mathrm{pH}=5.00\) buffer solution.

Short answer

Approximately 11.98 grams of sodium acetate must be added to 500.0 mL of 0.200 M acetic acid to form a pH=5.00 buffer solution.

Step-by-step solution

Calculate the pKa of acetic acid

First, we need to find the pKa of acetic acid. Acetic acid is a weak acid with a known dissociation constant, \(K_a = 1.8 \times 10^{-5}\). We can find its pKa by taking the negative logarithm of its \(K_a\) value: \( pK_a = -\log{K_a} = -\log{(1.8 \times 10^{-5})} \) Calculating this yields: \( pK_a \approx 4.74 \)

Determine the required concentration of sodium acetate

Now that we have the pKa, we can use the Henderson-Hasselbalch equation to determine the concentration of the conjugate base, sodium acetate: \(pH = pK_a + \log{\frac{[A^-]}{[HA]}} \) We are given the desired pH (5.00) and the initial concentration of acetic acid (\(0.200 M\)). The concentration of the conjugate base (\([A^-]\)) is unknown. Plugging the given values into the equation: \(5.00 = 4.74 + \log{\frac{[A^-]}{0.200}} \) Now we will solve for \([A^-]\).

Solve for the unknown concentration of sodium acetate

To isolate \([A^-]\), we first need to subtract the pKa from both sides of the equation: \( 5.00 - 4.74 = \log{\frac{[A^-]}{0.200}} \) \( 0.26 = \log{\frac{[A^-]}{0.200}} \) Next, we will exponentiate both sides of the equation with base 10 to remove the logarithm: \( 10^{0.26} = \frac{[A^-]}{0.200} \) Now, we will multiply both sides of the equation by 0.200 to solve for \([A^-]\): \( [A^-] = 10^{0.26} \times 0.200 \) Calculating this yields: \( [A^-] \approx 0.292 M \)

Calculate the mass of sodium acetate needed

We have the required concentration of sodium acetate (0.292 M), and we know the volume of the solution (500.0 mL). Now we can calculate the moles of sodium acetate needed, then convert it to mass using its molar mass: First, convert the volume to liters: \( Volume = 500.0 \, mL \times \frac{1 \, L}{1000 \, mL} = 0.500 \, L \) Next, calculate the moles of sodium acetate needed for the buffer solution: \( Moles \, of \, sodium \, acetate = [A^-] \times Volume = 0.292 \, M \times 0.500 \, L \) Calculating this yields: \( Moles \, of \, sodium \, acetate \approx 0.146 \, mol \) Finally, convert moles of sodium acetate to mass using its molar mass (\(Molar \, mass \, of \, sodium \, acetate = 82.03 \, g/mol\)): \[Mass \, of \, sodium \, acetate = Moles \times Molar \, mass = 0.146 \, mol \times 82.03 \, g/mol \] Calculating this yields: \( Mass \, of \, sodium \, acetate \approx 11.98 \, g \) Thus, approximately 11.98 grams of sodium acetate must be added to 500.0 mL of 0.200 M acetic acid to form a pH=5.00 buffer solution.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation provides a straightforward way to calculate the pH of buffer solutions, which involve a weak acid and its conjugate base. This equation is expressed as: \[ pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) \] Where:

• \([A^-]\) is the concentration of the conjugate base.
• \([HA]\) is the concentration of the weak acid.
• \(pK_a\) is the negative logarithm of the acid dissociation constant \(K_a\).

To use this equation effectively, we need to know the concentrations of the acid and its conjugate base, as well as the \(pK_a\) of the weak acid in question. This relationship helps us understand how the components of a buffer solution interact to resist changes in pH when an acid or base is added. The idea is by adjusting the ratio of the weak acid \([HA]\) to its conjugate base \([A^-]\), we can target a specific pH value for the buffer solution. This makes it an essential tool in both the laboratory and in various industrial applications.

molar mass

Understanding molar mass is crucial when converting moles of a substance into grams, particularly in chemical calculations such as buffer solutions. Molar mass, often expressed in grams per mole (g/mol), is the mass of one mole of a given substance. For instance, the molar mass of sodium acetate, as mentioned in the exercise, is 82.03 g/mol. To find the mass of a particular number of moles, you multiply the number of moles by the molar mass:\[ \text{mass} = \text{moles} \times \text{molar mass} \]For example, if you have 0.146 moles of sodium acetate, the mass would be:\[ \text{mass} = 0.146 \, \text{mol} \times 82.03 \, \text{g/mol} \approx 11.98 \, \text{g} \] This process of calculating the mass from the moles is straightforward but vital. It allows chemists to measure out precise quantities of a compound needed for chemical reactions and solutions, ensuring the correct mixture of substances to achieve the desired pH balance or other chemical properties.

weak acid

A weak acid is an acid that only partially dissociates in solution. Unlike strong acids, which dissociate completely and release a large number of hydronium ions \(H_3O^+\), weak acids form an equilibrium between the undissociated acid \([HA]\) and the ions \([A^-]\) and \([H^+]\) in solution. The degree of ionization of a weak acid is represented by its acid dissociation constant \(K_a\), which is a measure of the strength of the acid. For acetic acid, the \(K_a\) is relatively low (\(1.8 \times 10^{-5}\)), which means it doesn't release a lot of protons \([H^+]\) in solution and thus maintains an equilibrium. Being a weak acid makes acetic acid ideal for forming buffer solutions. In a buffer, this property ensures that the solution can resist drastic pH changes even when small amounts of an acid or a base are introduced.In constructing a buffer using a weak acid like acetic acid and its conjugate base (such as sodium acetate), the equilibrium can be adjusted to maintain a desired pH within a certain range. This feature is critical in many biochemical processes where pH levels need to remain constant for optimal function.