You are given an assignment to make 250ml buffer solution having 0.25M weak acid and 0.2M of its conjugate base. You have stock solutions 1.5M acid and 0.5M of the conjugate base. Clearly explain how you are going to make this buffer solution.

Can anyone please help me!!!

This problem is nothing but two dilutions since you have stock solutions. You want 250 mL x 0.25M acid or 250 x 0.25 = 62.5 millimoles acid.

You want 250 of 0.2M base = 250 mL x 0.2M = 50 mmols base.

Since M = mmoles/mL we can rearrange that to mL = mmoles/M.
mL acid stock soln = 62.5/1.5M = 41.67 mL acid stock solution.

mL base = mmoles/M = 50/0.20 = 100 mL.

So place 41.67 mL of the acid stock soln in a 250 mL volumetric flask, add 100 mL of the base stock solution, add water to the mark on the flask and shake thoroughly.