!! VERY LONG ANSWER !!
Since you're dealing with a buffer solution, you can use the Henderson-Hasselbalch equation to solve for the pH of the solution.
Now, since you mistyped the volume of the solution, I'll assume it to be 0.100 L - you can use whatever value you had.
So, according to the Henderson-Hasselbalch equation,
#pH_"sol" = pK_a + log(([F^(-)])/([HF]))#, #color(blue)((1))#
You also need the acid dissociation constant for #HF#, which is listed as being #K_a = 6.7 * 10^(-4)#.
The #pK_a# will be
#pK_a = -log(K_a) = -log(6.7 * 10^(-4)) = 3.17#
Plug your values into equation #color(blue)((1))# to get the pH
#pH_"sol" = 3.17 + log("0.50 M"/"0.25 M") = 3.17 + log(2) = color(green)(3.47)#
Now you start adding stuff to your buffer. When you add nitric acid, #HNO_3#, which is a strong acid, it will react with the conjugate base of your weak acid, #F^(-)#.
As a result, you'll have less #NaF# in the solution, which implies that more #HF# will be produced.
#HNO_(3(aq)) + NaF_((aq)) -> HF_((aq)) + NaNO_(3(aq))#
Now, you need to calculate how many moles of #NaF# and of #HF# you initially had in the buffer. To do this, use the given volume and molarity
#C = n/V => n = C * V#
#n_(NaF) = "0.50 M" * "0.100 L" = "0.050 moles NaF"#
#n_(HF) = "0.25 M" * "0.100 L" = "0.025 moles HF"#
Set up your ICE table
#" "HNO_(3(aq)) + NaF_((aq)) -> HF_((aq)) + NaNO_(3(aq))#
I.......0.002...............0.05..............0.025
C.....(-0.002)...........(-0.002).......(+0.002)
E.......0......................0.048............0.027
All the moles of #HNO_3# will be consumed; at the same time, 0.002 moles of #HF# will be produced and the number of moles of #NaF# will decrease by 0.002.
Calculate the new molarities of the species present in the buffer
#[HF] = "0.027 moles"/"0.100 L" = "0.27 M"#
#[F^(-)] = "0.048 moles"/"0.100 L" = "0.48 M"#
Use equation #color(blue)((1))# to determine the new pH
#pH_"sol 2" = 3.17 + log("0.48 M"/"0.27 M") = color(green)(3.42)#
Adding a strong acid to your buffer reduced the pH slightly
Now you add potassium hydroxide, #KOH# - a strong base. It will react with the weak acid to produce #F^(-)#.
#KOH_((aq)) + HF_((aq)) -> KF_((aq)) + H_2O_((l))#
SIDE NOTE You can ignore the #K^(+)# cation altogether, the important thing in this reaction is #F^(-)#.
Once again, set up the ICE table
#" "OH_((aq))^(-) + HF_((aq)) -> F_((aq))^(-) + H_2O_((l))#
I......0.004...........0.027............0.048
C...(-0.004).......(-0.004).........(+0.004)
E.........0...............0.023..............0.052
#[HF] = "0.023 moles"/"0.100 L" = "0.23 M"#
#[F^(-)] = "0.052 moles"/"0.100 L" = "0.52 M"#
Finally, use #color(blue)((1))# to solve for the new pH
#pH_"sol 3" = 3.17 + ("0.52 M"/"0.23 M") = color(green)(3.52)#
Adding a strong base increased the pH slightly.
