Question

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

Answer 1

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.