If 25 mL of 0.5 M acetic acid is mixed with 25 mL of 0.5 M sodium acetate, what is the resulting pH of the buffer solution?
Practice Questions
1 question
Q1
If 25 mL of 0.5 M acetic acid is mixed with 25 mL of 0.5 M sodium acetate, what is the resulting pH of the buffer solution?
4.76
5.00
5.76
6.00
Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). pKa of acetic acid is approximately 4.76. Since [A-] = [HA], pH = 4.76.
Questions & Step-by-step Solutions
1 item
Q
Q: If 25 mL of 0.5 M acetic acid is mixed with 25 mL of 0.5 M sodium acetate, what is the resulting pH of the buffer solution?
Solution: Using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). pKa of acetic acid is approximately 4.76. Since [A-] = [HA], pH = 4.76.
Steps: 9
Step 1: Identify the components of the buffer solution. We have acetic acid (HA) and sodium acetate (A-).
Step 2: Determine the concentrations of acetic acid and sodium acetate. Both are 0.5 M.
Step 3: Calculate the total volume of the solution after mixing. 25 mL + 25 mL = 50 mL.
Step 4: Calculate the concentrations of acetic acid and sodium acetate in the mixed solution. Since the volumes are equal, the concentrations remain 0.5 M for both.
Step 5: Use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]).
Step 6: Find the pKa of acetic acid, which is approximately 4.76.
Step 7: Since the concentrations of A- and HA are equal ([A-] = [HA]), the log term becomes log(1) = 0.
Step 8: Substitute the values into the equation: pH = 4.76 + 0.
