Chemistry Acid and Base

Question

What quantity in moles of HCl need to be added to 200.0 mL of a 0.200 M solution of NaF to make a buffer with a pH of 3.00? (Ka for HF is 6.8 × 10⁻⁴)

Answer 1

You can use the Henderson Hasselbalch equation for this problem.

pH = pKa + log [salt]/[acid]

pH = 3.00

pKa = -log Ka = - log 6.8x10-4 = 3.18

log [NaF]/[HF] = ?

3.00 = 3.18 + log [NaF]/[HF]

log [NaF]/[HF] = -0.18

[NaF]/[HF] = 0.66

To find the values of NaF and HF, we proceed as follows:

Initial mols NaF = 200.0 ml x 1 L/1000 ml x 0.200 mol/L = 0.04 mols NaF

F- + H+ ==> HF

0.04 - x / x = 0.66

x = 0.024 mols HCl

Check:

NaF + HCl ===> HF + NaCl

0.04......0.024.........0.................Initial

-0.024.....+0.024....+0.024...............Change

0.016..........0..........0.024.............Equilibrium

From Henderson Hasselbalch equation:

Solve for pH as a check:

pH = 3.18 + log (0.016/0.024)

pH = 3.18 + log 0.667

pH = 3.18 -0.18

pH = 3.00 √