Please solve the question for me

The acid dissociation constant ($K_a$) for \ce{HF} is given as $3.5 \times 10^{-4}$.

The solution contains a weak acid (\ce{HF}) and its conjugate base (\ce{F-}, from \ce{NaF}). This is a buffer solution.
Given concentrations: $[\ce{HF}] = 0.250\ \text{M}$ and $[\ce{F-}] = 0.250\ \text{M}$.

The pH of the solution is:

The initial volume of the solution is 1.00 L.
Moles of \ce{HF} = $0.250\ \text{M} \times 1.00\ \text{L} = 0.250\ \text{mol}$
Moles of \ce{F-} = $0.250\ \text{M} \times 1.00\ \text{L} = 0.250\ \text{mol}$

When 0.050 mol of \ce{HCl} (a strong acid) is added, it reacts with the conjugate base (\ce{F-}).

Initial moles: $\ce{F-} = 0.250\ \text{mol}$, $\ce{H+} = 0.050\ \text{mol}$, $\ce{HF} = 0.250\ \text{mol}$
Change: $\ce{F-}$ reacts with $\ce{H+}$, so $\ce{F-}$ decreases by $0.050\ \text{mol}$, $\ce{H+}$ decreases by $0.050\ \text{mol}$, and $\ce{HF}$ increases by $0.050\ \text{mol}$.
After reaction:
Moles of \ce{F-} = $0.250\ \text{mol} - 0.050\ \text{mol} = 0.200\ \text{mol}$
Moles of \ce{HF} = $0.250\ \text{mol} + 0.050\ \text{mol} = 0.300\ \text{mol}$

Since the volume remains 1.00 L:
$[\ce{F-}] = \frac{0.200\ \text{mol}}{1.00\ \text{L}} = 0.200\ \text{M}$
$[\ce{HF}] = \frac{0.300\ \text{mol}}{1.00\ \text{L}} = 0.300\ \text{M}$

The $\text{p}K_a$ for \ce{HF} is 3.46 (calculated in part a).

The pH of the solution after adding \ce{HCl} is:

The initial volume of the solution is 1.00 L.
Moles of \ce{HF} = $0.250\ \text{M} \times 1.00\ \text{L} = 0.250\ \text{mol}$
Moles of \ce{F-} = $0.250\ \text{M} \times 1.00\ \text{L} = 0.250\ \text{mol}$

When 0.050 mol of \ce{NaOH} (a strong base) is added, it reacts with the weak acid (\ce{HF}).

Initial moles: $\ce{HF} = 0.250\ \text{mol}$, $\ce{OH-} = 0.050\ \text{mol}$, $\ce{F-} = 0.250\