What volumes of 0.49 M HF and 0.49 M NaF must be mixed to prepare 1.00 L of a solution buffered at pH = 2.77? HF ____L NaF ____L?

HF(aq) + H₂O(l) ⇌ F⁻(aq) + H₃O⁺(aq) …… Ka = 7.4 × 10⁻⁴

Henderson-Hasselbalch equation:
pH = pKa + log([F⁻]/[HF])
2.77 = -log(7.4 × 10⁻⁴) + log([F⁻]/[HF])
log([F⁻]/[HF]) = -0.36
[F⁻]/[HF] = 10⁻⁰˙³⁶

Let V L is the volume of HF.
Then, volume of NaF = (1.00 - V) L

As HF and F⁻ are equi-molar, [F⁻]/[HF] = (Volume of NaF)/(Volume of HF)
(1.00 - V) / V = 10⁻⁰˙³⁶
1.00 - V = 10⁻⁰˙³⁶V
(1 + 10⁻⁰˙³⁶)V = 1
V = 1/(1 + 10⁻⁰˙³⁶)
V = 0.70

Volume of HF = V L = 0.70 L
Volume of NaF = (1.00 - V) L = 0.30 L