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 2.90? (Ka for HF is 6.8 × 10⁻⁴)

Answer 1

Henderson Hasselbalclh equation:

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

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

2.90 = 3.17 + log [NaF] / [HF]

2.90 = 3.17 + log [NaF] / 0.200

-0.27 = log [NaF] / 0.200

0.537 = [NaF] / 0.200

[NaF] = 0.107 M = 0.107 mol / L

moles NaF needed for 200 ml (0.200 L) is ...

0.107 mol / L x 0.200 L = 0.0214 moles NaF needed (assuming no change in volume upon addition)