pH of acidic buffer = pka +
log(ch3cooK/CH33COOH)
pH = 4.9
pka of CH3COOH = 4.76
no of mol of CH3COOH = M*V
                    Â
= 0.881*0.875
                    Â
= 0.771 mol
No of mol of CH3COOK = KOH = x mol
5.52 = 4.76+log(x/(0.771-x))
x = 0.657
No of mol of CH3COOK = KOH = 0.657 mol
volume of KOH required = n/M
                     Â
= 0.657/2.64
                     Â
= 0.25 l
                     Â
= 250 ml
