Question

what will be the ph of a buffer solution prepared from 0.17 mol sodium acetate, 0.17 mol acetic acid, and just enough water to give 1.03 l of solution? use ionization constants from appendices h

Answer 1

The pH of the buffer solution prepared from 0.17 mol sodium acetate, 0.17 mol acetic acid, and 1.03 L of water is 4.74.

Step-by-step explanation:

To calculate the pH of a buffer solution prepared from 0.17 mol sodium acetate, 0.17 mol acetic acid, and just enough water to give 1.03 L of solution, we can use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

First, we need to find the molar concentrations of acetate (A-) and acetic acid (HA). Since the solution volume is 1.03 L, the molar concentrations will be:

[A-] = 0.17 mol / 1.03 L = 0.165 M
[HA] = 0.17 mol / 1.03 L = 0.165 M

These are equal, which is ideal for a buffer solution. Now, we use the ionization constant (Ka) for acetic acid from Appendix H, which is 1.8 × 10-5. Its pKa is -log(1.8 × 10-5) = 4.74.

Inserting these values into the Henderson-Hasselbalch equation gives:

pH = 4.74 + log(0.165/0.165)
pH = 4.74 + log(1)
pH = 4.74

Therefore, the pH of the buffer solution is 4.74.