PHÂ Â = PKa + log[NaAc]/[HAc]
4.9Â Â Â = 4.76 + log[NaAc]/[HAc]
log[NaAc]/[HAc]Â Â = 4.9-4.76
log[NaAc]/[HAc]Â Â Â = 0.14
[NaAc]/[HAc]Â Â Â Â Â =
100.14
[NaAc]/[HAc]Â Â Â = 1.38
[NaAc]Â Â Â Â Â Â Â Â Â =
1.38[HAc]
no of moles of acetic acid (HAc) = molarity * volume in L
                                                Â
= 0.3*0.5 = 0.15moles
total no of moles in buffer   = 0.15moles
[NaAc]+[HAc]Â Â = 0.15
1.38[HAc]+[HAc] = 0.15
2.38[HAc]Â Â Â Â Â Â Â Â =
0.15
 Â
[HAc]Â Â Â Â Â Â Â Â Â =
0.15/2.38
[HAc]Â Â Â Â Â Â = 0.063moles
[NaAC]Â Â Â = 0.15-0.063 = 0.087moles
NaOH react with HAc to gives NaAC 0.087 moles of NaOH is
needed
no of moles of NaOH = 0.087 moles
molarity of NaOH = no of moles/volume in L
 Â
3Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
= 0.087/volume in L
volume in LÂ Â Â Â Â Â = 0.087/3 =
0.029L = 29ml
