Calculate the amount of NaoH required to prepare a acetic acid buffer solution of ph 5 ? Given initial concentration of CH3COOH=3M and ka of CH3COOH =1.8× 10^-5

Acetic auid react with NaoH
CH3 C00 H + NaoH =
CH3 COONA + H20

Lets take,'x' mele NAOH required to prepare
a buffer Solution of pH - 5
So, at equlibrium, [CH3 cooNa] = xM
[ CH3 cooH] = (3- x)M

Now Handerson equation becomes-
pH=pka+log [{CH3COONa}÷{CH3COOH}]

Given, PH= 5,
PKa = - logg Ka = - log(1.8 x10^- 5)
= 4.74

by using handerson equation we get
x=1.94 mole
=1.94×40
=77g
77g NaoH required