How much 1 M NaOH will need to be added to 50 ml of 3 M Acetic Acid so that the final pH is 9.1 ? Please show work?

You can easily see that answer of 20.2 mL is wrong. Use the Henderson-Hasselbalch equation.

20.2 mL NaOH converts 0.0202 mole of HOac to AcO^-1 (20.2 mM / 70.2 mL = 0.288 M) and the remaining HOAc is 0.1298 mole (0.1500 mole - 0.0202 =0.1298 mole and 0.1298 mole / 0.0702 L = 1.849 M)

pH = pKa + log(0.288 / 1.849) = 4.76 - 0.81 = 3.95

The addition of 150 mL of 1 M NaOH will yield a solution of 0.15 mole of AcO^-1 in 200 mL

AcO^-1 + H2O <--> HOAc + OH^-1

Kb = [HOAc][OH-] / [AcO-] = 5.7x10^-10

[OH-] = [HOAc]

[AcO-] = 0.75 M - [OH-] or approximately = 0.75

5.75x10^-10 = [OH]^2 / 0.75

[OH-] = 2.08x10^-5

pOH = 4.68 and pH = 9.32

So the addition of 150 mL gives a pH of 9.32 (9.3). Only slightly less NaOH will result in a pH of 9.1

You should be able to solve the Henderson-Hasselbalch equation for the required molarity of AcO^-1 and then calculate the required amount of NaOH to yield that molarity.

pH = 9.1 = pKa + log([AcO-]\ / [HOAc])

we get a weak base with

pOH = 0.5(pKb - lg(A^-))

4.9 = 0.5(9.26 - lg(A^-))

A^- = 0.288 = V*1/(V+50)

thus

V = 20.2 mL NaOH