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

To form a pH 5.00 buffer solution, approximately 7.4 g of sodium acetate must be added to 500.0 mL of 0.200 M acetic acid. This is calculated using the Henderson-Hasselbalch equation and the given information.

Step-by-step solution

Write the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentrations of weak acid and its conjugate base: pH = \( pK_a + \log \frac{[A^-]}{[HA]} \) In this case, [A^-] stands for the concentration of acetate ions and [HA] stands for the concentration of acetic acid. We will be given the \( pK_a \) of acetic acid and the final pH of the buffer solution.

Find the \( pK_a \) value for acetic acid

You may find the \( pK_a \) value for acetic acid in a textbook or other reference material. Alternatively, you can use the relationship between \( K_a \) and \( pK_a \), where: \( pK_a= -\log K_a \) The \( K_a \) value for acetic acid is \( 1.8 \times 10^{-5} \). Thus, the \( pK_a \) of acetic acid is: \( pK_a = - \log (1.8 \times 10^{-5}) = 4.74 \)

Calculate the concentration of acetate ions using the Henderson-Hasselbalch equation

Now that we have the \( pK_a \) value, we can use the Henderson-Hasselbalch equation to find the concentration of acetate ions in the buffer solution: 5.00 = 4.74 + \( \log \frac{[A^-]}{0.2 M} \) First, we need to isolate the ratio of concentrations by subtracting the \( pK_a \) value: \( 0.26 = \log \frac{[A^-]}{0.2 M} \) Next, we can remove the logarithm by taking the antilog of both sides: \( 10^{0.26} = \frac{[A^-]}{0.2 M} \) Finally, we can solve for [A^-]: \( [A^-] = (0.2 M) \times 10^{0.26} ≈ 0.18 M \) The concentration of acetate ions required in the buffer solution is approximately 0.18 M.

Calculate the mass of sodium acetate needed

Now that we have the concentration of acetate ions, we can calculate the mass of sodium acetate that needs to be added to the buffer solution. First, we need to find the number of moles of sodium acetate to be added, which is the product of the volume of the solution and the required concentration of acetate ions: Moles of sodium acetate = \( (0.18 M) \times (0.5 L) = 0.09 \textrm{ moles} \) Next, we can find the molar mass of sodium acetate (NaC\(_2\)H\(_3\)O\(_2\)). The molar mass of each element is as follows: Na: 22.99 g/mol C: 12.01 g/mol H: 1.01 g/mol O: 16.00 g/mol Molar mass of sodium acetate = 22.99 + (2 × 12.01) + (4 × 1.01) + (2 × 16.00) = 82.03 g/mol Finally, we can calculate the mass of sodium acetate: Mass = moles × molar mass = \( (0.09 \textrm{ moles}) \times (82.03 \, \frac{\text{g}}{\text{mol}}) \approx 7.4 \, \text{g} \) So, approximately 7.4 g 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

Understanding the Henderson-Hasselbalch equation is pivotal when preparing buffer solutions. This equation offers a simple relationship between the pH of a buffer solution and the ratio of the concentrations of a weak acid and its conjugate base:\[\begin{equation}\text{pH} = pK_a + \log \frac{[A^-]}{[HA]}\end{equation}\]In this context, \(pK_a\) is a constant that reflects the strength of the weak acid, \( [A^-] \) is the molar concentration of the conjugate base, and \( [HA] \) is the molar concentration of the weak acid. To use the equation effectively, you'll first isolate the variables by performing algebraic manipulations. The ease of solving the equation makes it a handy tool for quickly determining the necessary additives to achieve a desired pH in buffer preparations.

pKa and pH Relationship

The relationship between \( pK_a \) and pH is integral to understanding acid-base chemistry. Here's a simple breakdown of what each term signifies:

• pKa: It's a quantitative measure of the strength of an acid in solution. A lower \( pK_a \) signifies a stronger acid, which dissociates more in water.
• pH: Represents the acidity or basicity of a solution. A pH less than 7 indicates acidity, while a value greater than 7 indicates basicity.

Their relationship is laid out in the Henderson-Hasselbalch equation, which predicts how the pH of a solution changes as the relative amounts of acid and base change. This predictive capability is especially important when creating a buffer solution with specific pH requirements, such as the \( pH = 5 \) buffer solution in the example.

Molar Mass Calculation

Molar mass calculation is a fundamental concept in chemistry that involves determining the mass of one mole of a substance. The molar mass is usually given in grams per mole (g/mol) and is calculated by summing the atomic masses of all the atoms in a molecule. For instance, sodium acetate \( \text{NaC}_2\text{H}_3\text{O}_2 \) has a molar mass that is the sum of the molar masses of sodium (Na), carbon (C), hydrogen (H), and oxygen (O) atoms. Knowing the molar mass allows chemists to convert between the mass of a substance and the amount in moles, facilitating precise measurements for reactions and solutions such as the required mass of sodium acetate in the buffer solution.