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 3.60? (Ka for HF is 6.8 × 10⁻⁴)

Answer 1

This solution is a buffer so you can use the Henderson Hasselbalch equation.

pH = pka + log[base/acid]

200 mL HF x 1L = 0.200L HF

1000mL

M= moles / L

moles = M x L

0.200L HF x 0.200 M HF = 0.04 moles HF (acid)

pka = -log(Ka)

pka = -log(6.8 x10^-4)

pka = 3.17

3.60 = 3.17 + log [base / 0.04 moles]

0.43 = log [base /0.04 moles]

10^0.43 = base / 0.04 moles

2.69 = base / 0.04 moles

base = 2.69 x 0.04 moles

base = 0.108 moles NaOH