# Chemist needs a buffer with pH of 4.5.

how many milliliters of pure CH3COOH (density;1.049g/ml) must be added to 500ml of 0.1M NaOH solution to obtain such a buffer?

1. 👍
2. 👎
3. 👁
4. ℹ️
5. 🚩
1. The Henderson-Hasselbalch equation correctly uses concentrations; however, since the volumes is the same (because it's the same solution), the volumes cancel and one may use mols. I like to use millimoles.
millimols NaOH = mL x M = 500 mL x 0.1 M = 50
Let HAc = CH3COOH. You will need 50 mmols HAc to neutralize the NaOH you have initially. Then HAc must be added to this to form the buffer. You start, therefore, with 50 mmols Ac^- as base. We add acid to this.
pH = pKaHAc + log (base)/(acid)
4.50 = 4.76 + log [50/(50+x)] where x = mmols HAc that must be added. to the 5o mmols Ac^- That works out to be approx 40 mL. Again, you should get a better number than the estimate. To get the buffer you want, you must add 50 mmols HAc to neutralize the NaOH initially PLUS approx 90 mmols to form the buffer or a total of about 140 mmols HAc. Convert that to grams HAc, then use density to convert to mL HAc.
Post your work if you get stuck.

1. 👍
2. 👎
3. ℹ️
4. 🚩