3)
a) pH of acidic buffer = pka +
log(ch3coona+NaOH/ch3cooh-NaOH)
pka of ch3cooh = -logKa = -log(1.78*10^-5) =
4.75
No of mole of acetic acid present in buffer = 35*1 = 35
mmole
No of mole of sodium acetate = 35*1 = 35
mmol
No of mole of NaOH added = 1.5*0.5 = 0.75
mmole
pH = 4.75 + log((35+0.75)/(35-0.75))
  = 4.77
pH change = 4.77-4.75 = 0.02
b)
Concentration of NaOH solution = n/v
                             Â
= 1.5*0.5/(35+1.5)
                             Â
= 0.0205 M
pOH = - log[OH-]
   = -log0.0205
   = 1.7
pH = 14-pOH = 14-1.7 = 12.3
initial pH of water = 7
pH change = 12.3 - 7 = 5.3
