hey think I have managed to figure it out - but please double check...
Q) What Volume of NaOH is required to prepare 200ml of a 0.2M buffer with a pH of 4 and pKa of acid = 3.5?
Step 1: work out ratio of acid and base in buffer using Henderson Hasselbach Eqn
[H+] = Ka (a/b)
H+ = antilog pH = 1x10-4
Ka = antilog pKa = 3.16 x 10-4
putting that into above equation you get a/b = 0.3162
Step 2: Work out concentrations of acid and base in buffer
the total concentration of buffer is 0.2M therefore a+b= 0.2M
so you can set up simultaneous equations to work out a and b
a=0.2 - b and a= 0.3162b
solving these gives you a = 0.0480 M and b = 0.1519 M
Step 3: Working out volume of base required
You know that the total volume of buffer is 200ml
therefore the amount of b is in 200ml
i.e. 0.1519 molar in 200ml
so what volume of 200ml would have 0.2M in?
(0.1519/0.2) X 200ml = 151.9 ml
therefore the volume of NaOH reqd to make this buffer is 151.9ml