Given:
pKa of HF = 3.45
Concentration of HF = 0.1 M
Concentration of KF = 0.300 M

To find: pH of the buffer solution

Formula:
pH = pKa + log ([A-]/[HA])
where,
pH = the pH of the buffer solution
pKa = -log(Ka)
[A-] = concentration of the conjugate base
[HA] = concentration of the acid

Calculation:
Ka for HF = 10^(-pKa) = 6.31 × 10^(-4)
[H+] = 10^(-pH)
[HF] = 0.1 M
[KF] = 0.300 M
[OH-] = Kw/[H+] = 1.0 × 10^(-14)/[H+]

The balanced chemical equation for HF dissociation is given below:
HF + H2O ↔ H3O+ + F-

Using the equation, we can write the equilibrium expression as follows:
Ka = [H3O+][F-]/[HF]

Let's assume x is the concentration of H3O+ and F- ions formed at equilibrium.

[H3O+] = x
[F-] = x
[HF] = 0.1 - x
Ka = 6.31 × 10^(-4)

Substituting the values in the equilibrium expression:

6.31 × 10^(-4) = x^2/(0.1 - x)

Since x is very small compared to 0.1, we can assume that (0.1 - x) is equal to 0.1.

6.31 × 10^(-4) = x^2/0.1
x^2 = 6.31 × 10^(-5)
x = 0.00795

[H3O+] = 0.00795 M
pH = -log[H3O+]
pH = 2.10

Now, we need to check whether our assumption (x is very small compared to 0.1) is valid or not. We can do this by calculating the percent ionization of HF.

% ionization = (concentration of H3O+ ions formed/initial concentration of HF) × 100
% ionization = (0.00795/0.1) × 100
% ionization = 7.95%

Since the percent ionization is less than 10%, our assumption is valid.

Next, we need to calculate the concentration of F- ions.

[F-] = 0.300 + 0.00795
[F-] = 0.30795 M

Now, we can calculate the pH of the buffer solution using the Henderson-Hasselbalch equation.

pH = pKa + log ([A-]/[HA])
pH = 3.45 + log (0.30795/0.1)
pH = 3.93

Therefore, the pH of the buffer solution is 3.93.