What volumes of 0.55 M HF and 0.55 M LiF must be mixed to prepare 1.00 L of a solution buffered at pH = 3.55?

Question

What volumes of 0.55 M HF and 0.55 M LiF must be mixed to prepare 1.00 L of a solution buffered at pH = 3.55?

(a) HF ______ L

(b) LiF ______ L

Answer 1

Taking the pKa for HF to be 3.1, and using the Henderson Hasselbalch equation...

pH = pKa + log [LiF]/[HF]

3.55 = 3.1 + log [LiF][HF]

log [LiF]/[HF] = 0.45

[LiF]/[HF] = 2.8

To make 1 liter and have a 2.8 : 1 mole ratio, we would do the following;

x = liters of LiF

x / 1 - x = 2.8

2.8 - 2.8x = x

3.8x = 2.8

x = 0.737 = 737 mls of 0.55 M LiF

Add to this 263 mls of 0.55 M HF

Final moles will be ....

0.737 L x 0.55 mol/L LiF = 0.41 moles LiF

0.263 L x 0.55 mol/L HF = 0.14 moles HF

Mole ratio = 0.41/0.14 = 2.9 (close to 2.8 desired ratio, but error due to rounding)