A buffered solution is made by adding 75.0g sodium acetate to 500.0 mL of a .64 M solution of acetic acid. What is the pH of the final solution? (Ka for acetic acid, CH3COOH, is 1.8 x 10^-5)
Use the Henderson-Hasselbalch equation. The base is sodium acetate and the acid is acetic acid.
To determine the pH of the final solution, we need to calculate the concentration of the hydronium ion (H₃O⁺). The solution consists of acetic acid (CH₃COOH) and its conjugate base, sodium acetate (CH₃COONa).
To start, let's calculate the number of moles of acetic acid in the 500.0 mL of 0.64 M acetic acid solution:
Moles of acetic acid = concentration × volume
Moles of acetic acid = 0.64 mol/L × 0.500 L
Moles of acetic acid = 0.32 mol
Now, let's calculate the number of moles of sodium acetate added:
Moles of sodium acetate = mass / molar mass
Moles of sodium acetate = 75.0 g / (82.03 g/mol + 22.99 g/mol + 16.00 g/mol)
Moles of sodium acetate = 0.75 mol
Next, let's calculate the total volume of the final solution:
Total volume = volume of acetic acid solution + volume of sodium acetate solution
Total volume = 500.0 mL + 0 mL
Total volume = 500.0 mL
Now, we can calculate the concentration of the acetic acid and its conjugate base in the final solution:
Concentration of acetic acid = moles of acetic acid / total volume
Concentration of acetic acid = 0.32 mol / 0.500 L
Concentration of acetic acid = 0.64 M
Concentration of sodium acetate = moles of sodium acetate / total volume
Concentration of sodium acetate = 0.75 mol / 0.500 L
Concentration of sodium acetate = 1.5 M
Since we have a buffered solution with a weak acid and its conjugate base, we can use the Henderson-Hasselbalch equation to calculate the pH:
pH = pKa + log(concentration of conjugate base / concentration of acid)
First, let's calculate the pKa using the given Ka value for acetic acid:
pKa = -log(Ka)
pKa = -log(1.8 x 10^-5)
pKa = 4.74
Now, we can substitute the values into the Henderson-Hasselbalch equation:
pH = 4.74 + log(1.5 M / 0.64 M)
pH = 4.74 + log(2.34)
pH ≈ 4.74 + 0.37
pH ≈ 5.11
Therefore, the pH of the final solution is approximately 5.11.