PH = 4.9
PKa = 4.74
volume of CH3COOH = 1L
no of moles of CH3COOH = molarity * volume in L
                             Â
= 1*1 = 1 moles
no of moles of NaOHÂ Â Â = molarity * volume in
L
                                 Â
= 3*v
---------- CH3COOH (aq) + NaOH(aq) --------------->
CH3COONa(aq) + H2O(l)
I--------------- 1 ------------------ 3v
---------------------------------- 0
C------------- -3v --------------- -3v
----------------------------------- 3v
E ---------- 1-3v ---------------- 0
------------------------------------ 3v
         PH = Pka
+ log[CH3COONa]/[CH3COOH]
        4.9 = 4.74 +
log[CH3COONa]/[CH3COOH]
log[CH3COONa]/[CH3COOH] = 4.9-4.74
log[CH3COONa]/[CH3COOH] =0.16
[CH3COONa]/[CH3COOH]Â Â Â Â = 1.4454
3v/(1-3v)Â Â = 1.4454
3v  = 1.4454(1-3v)
v = 0.197L
volume of NaOH = 197ml  >>>>answer
