Chemistry HW help

Question

What quantity in moles of NaOH need to be added to 200.0 mL of a 0.200 M solution of HF to make a buffer with a pH of 4.00?

(Ka for HF is 6.8 × 10-4)

Answer 1

For this problem, we can use the Henderson Hasselbalch equation:

pH = pKa + log [conj.base] /[acid]

pKa = -log Ka = -log 6.8x10^-4 = 3.17

4.00 = 3.17 + log [CB] / [A]

log [CB] / [A] =0.83

[CB] / [A] = 6.8

This tells us that we want the ratio of NaF / HF to be 6.8 : 1

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

HF + NaOH ==> NaF + H2O

0.04.......x...............0...............

-x..........-x.............+x.............

0.04-x....................x..............

x / 0.04-x = 6.8

x = 0.035 mols NaOH

To check this, let us use the HH equation and solve for pH. It should be 4.00 (or close due to rounding)

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

pH = 3.17 + log (0.035) / (0.005)

pH = 3.17 + log 7

pH = 317 + 0.85

PH = 4.01