no of moles of CH3COOH = molarity * volume in L
                                         Â
= 0.2*0.01 = 0.002moles
no of moles of CH3COONa = molarity * volume in L
                                         Â
= 0.2*0.01 = 0.002 moles
no of moles of HCl   = molarity * volume in L
                                            Â
= 0.1*0.001 = 0.0001moles
no of moles of CH3COOH after addition of 0.0001 moles of
HCl  = 0.002+0.0001 = 0.0021 moles
no of moles of CH3COONa after addition of 0.0001 moles of
HCl  = 0.002-0.0001  = 0.0019 moles
PKa  = -logKa
          Â
= -log1.8*10^-5
           Â
=4.74
PHÂ Â = Pka + log[CH3COONa]/[CH3COOH]
         = 4.74 +
log0.0019/0.0021
         = 4.74
-0.04346
          =
4.6965 >>>>answer
