217k views
0 votes
What is the pH of a solution prepared by adding 0.025 mol of NaOH to 400 mL of 0.10 M C₆H₅COOH? (Ka = 6.5 × 10^-5 for C₆H₅COOH)

a) 1.64
b) 2.23
c) 4.18
d) 8.91

User Holli
by
7.9k points

1 Answer

3 votes

Final answer:

The pH of the solution is calculated using the Henderson-Hasselbalch equation after determining the moles of unreacted C6H5COOH and the moles of the formed C6H5COO- post-reaction with NaOH.

Step-by-step explanation:

To determine the pH of the solution after adding 0.025 mol of NaOH to 400 mL of 0.10 M C6H5COOH, we must first consider the reaction of NaOH with C6H5COOH. This will form sodium benzoate and water. Since NaOH is a strong base, it will react in a 1:1 ratio with the weak acid C6H5COOH.

The initial amount of C6H5COOH is 0.10 M * 0.4 L = 0.040 mol. After reacting with 0.025 mol of NaOH, we will have 0.040 mol - 0.025 mol = 0.015 mol of C6H5COOH left, and 0.025 mol of conjugate base C6H5COO-.

Now, we can apply the Henderson-Hasselbalch equation to find the pH:

pH = pKa + log([C6H5COO-]/[C6H5COOH])
pKa = -log(Ka) = -log(6.5 × 10-5)

By calculating the pKa and plugging in the concentrations, we can solve for the pH of the solution.

User Heddy
by
7.3k points

Related questions

asked Feb 26, 2024 144k views
Logan Guo asked Feb 26, 2024
by Logan Guo
8.1k points
1 answer
1 vote
144k views
1 answer
3 votes
40.7k views