describe the preparation of 2litres of 0.2M acetate buffer of pH 5.5,starting from 2.5M solution of acetic acid and 2.5M solution of KOH{4.77}

Answer 1

I assume the 4.77 has nothing to do with the problem or perhaps that was meant to be pKa for acetic acid. I've always used 4.74 for pKa acetic acid.
You want the solution to be 0.2M so if you start with 0.2M HAc it will react with the KOH to produce acetate. Whatever you lose of the HAc will be formed into Ac^- and the total will be 0.2M which is what you want. I will work in mols and not M. You want 2L of 0.2M which is 0.4 mol.

........HAc + KOH ==> KAc + H2O
I.......0.4...0........0......0
add...........x..................
C........-x..-x........x
E.....0.4-x...0........x

5.5 = 4.77 + log (x/0.4-x)
Solve for x = mols KOH to be added.
Convert to L of 2.5M to get that number of mols.
To start, calculate L of 2.5M acetic acid to give 0.4 mol.

I obtained approx 160 mL acetic acid and approx 135 mL KOH and those are approximate. You should confirm all of this. Also, I like to plug in the numbers at the end and see if the final pH is in fact 5.5 AND that the molarity of the final solution is 0.2M.