What volume of 0.307 m naoh must be added to 200.0ml of 0.425m acetic acid (ka = 1.75 x 10-5 ) to produce a buffer of ph = 4.250?

Answer 1

I assume you mean M and not m. m stands for molality. M is for molarity. I will use HAc for
millimols HAc = mL x M = 200 x 0.425 = 85
...............HAc + NaOH ==> NaAc + H2O
I...............85.........0................0............0
add ......................x.....................................
C.............-x.........-x.................x.............x
E............85-x........x.................x.............x

Plug the E line into the Henderson-Hasselbalch equation and solve for x =- millimols NaOH to be added. Then M = millimols/mL. You know M and millimols, solve for mL.