CH3COOH+H2O(l)⇌CH3COO−(aq)+H3O+(aq)t=00.1M00t=t0.1−xxx
Ka=x20.1−x
As it is a weak acid then (0.1−x)≈0.1
[H3O+]=x=√Ka×0.1=10−3
∴ pH = 3
On addition of CH3COONa, pH will change. So we don't have add any CH3COONa, to form a solution of pH = 3.
So answer is 0.