PH = PKa +log[salt]/[acid]
[CH3COOH] = mole/volume  = 1/2
[CH3COONa[Â Â = moles/2
PH = PKa + log[CH3COONa]/[CH3COOH]
4Â Â Â = 4.75+ logmole/1
4-4.75Â Â =logmoles/1
-0.75Â Â =logmole/1
moles  = 10-0.75 = 0.1778 moles of
CH3COONa
mass of CH3COONa = no of moles* molar mass
                               Â
= 0.1778*82 = 14.58gm of CH3COONa
By the addition of NaOH
no of moles of CH3COONa =0.1778+x
no of moles of CH3COOHÂ Â Â = 1-x
PH = PKa + log[CH3COONa]/[CH3COOH]
5Â Â Â = 4.75+ log0.1778+x/1-x
5-4.75 = log0.1778+x/1-x
0.25Â Â = log0.1778+x/1-x
0.1778+x/1-x = 100.25
0.1778+x  = 1.778(1-x)
0.1778+x-1.778+1.778x =0
-1.6002 = -2.778x
x = 0.576 moles
mass of NaOH = 0.576*40 = 23.04gm
