How many grams of sodium acetate, NaC2H3O2, would have to be added to 5.00 L of 0.65 M acetic acid (pKa 4.74) to make the solution a buffer for pH 4.00?
To answer this question, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the ratio of the concentration of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
In this case, acetic acid (HC2H3O2) is the acid (HA), and sodium acetate (NaC2H3O2) is its conjugate base (A-).
We have the following information:
- pH = 4.00
- pKa of acetic acid = 4.74
To make the solution a buffer at pH 4.00, we need to find the concentration ratio ([A-]/[HA]) that satisfies the Henderson-Hasselbalch equation.
Step 1: Calculate the concentration of acetic acid (HA):
Using the given Molarity (0.65 M) and volume (5.00 L), we can calculate the number of moles of acetic acid:
number of moles = Molarity × volume = 0.65 M × 5.00 L = 3.25 moles
Step 2: Convert the pH to [A-]/[HA] ratio:
Using the Henderson-Hasselbalch equation, rearrange it to solve for [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
Substituting the given values:
[A-]/[HA] = 10^(4.00 - 4.74) = 0.478
Step 3: Calculate the concentration of sodium acetate (A-):
We know that the concentration ratio [A-]/[HA] is equal to the concentration of A-. Let's denote the concentration of sodium acetate as C(A-):
C(A-) = [A-] = 0.478
Step 4: Calculate the mass of sodium acetate (NaC2H3O2):
The molar mass of sodium acetate is the sum of the molar masses of its individual elements:
Molar mass of NaC2H3O2 = (22.99 g/mol) + (2 × 12.01 g/mol) + (3 × 1.01 g/mol) + (2 × 16.00 g/mol) = 82.03 g/mol
Now, we can calculate the mass of sodium acetate needed to obtain the desired concentration using the formula:
mass = moles × molar mass
mass = 0.478 × 82.03 g/mol ≈ 39.18 grams
Therefore, approximately 39.18 grams of sodium acetate (NaC2H3O2) would need to be added to 5.00 L of 0.65 M acetic acid to make the solution a buffer for pH 4.00.
Use the Henderson-Hasselbalch equation.
Post your work if you get stuck.