BUFFER QUESTION..PLZZZ HELP!!!!

What is the pH of a solution prepared by mixing 24,6g sodium acetate and 50mL of 2.5 M acetic acid in a 500mL flask and filling it to volume?
If possible please list steps... Thank You

First, you want the moles of sodium acetate, which i will abbreviate as A-. Acetic acid corresponds to HA.

(24.6 g A-) * (1 mol A- / 82.03 g A-) = 0.2998 mol A- (I'm keeping an extra sig fig so I won't have any rounding errors).

Now you want the moles of HA:

(2.5 mol/L HA) * 0.050L HA = 0.125 mol HA

Now you can just plug in those values into the Henderson-Hasselbach equation:

pH = pKa + log ([A-]/[HA])

pH = 4.756 + log (0.2998/0.125)

pH = 5.1359

pH = 5.14

first find the moles of sodium acetate

MM of sodium acetate=82.04 g/mol

now divide your grams of sodium acetate used by the MM

(24.60g)*(one mole/82.04g)=0.29985 mol

now find the moles of acetic acid

0.050L*2.5M=0.125 mol

now divide each of them by the new volume of 500mL

sodium acetate=0.29985 mol/0.500L=0.5997 M
acetic acid=0.250M

now use the Henderson Hasselbalch equation to fnd pH Ka of acetic acid = 1.8E-5 pKa=4.7447

pH=pKa+log([base]/[acid])
pH=4.7447+log(0.5997M/0.250M)
pH=5.12

you could also just use moles right off the back then your equation would look like this
pH=4.7447+log(0.29985mol/0.125mol)
pH=5.12

either way works but using moles is a lot easier and faster