no of moles of NaC7H5O2Â Â = W/G.M.Wt
                                     Â
= 132.8/144Â Â = 0.923moles
no of moles of HC7H5O2Â Â = molarity * volume in L
                                    Â
= 2.15*0.3Â Â = 0.645moles
PKa = -logKa
        =
-log(6.3*10^-5)
        = 4.2
no of moles of NaC7H5O2 after adding 0.750 mol of H^+ =
0.923-0.75Â Â = 0.173 moles
no of moles of HC7H5O2 after adding 0.750mol of H^+ =
0.645+0.75Â Â Â = 1.395moles
PHÂ Â = Pka + log[NaC7H5O2]/[HC7H5O2]
     = 4.2 + log0.173/1.395
     = 4.2 -0.9065
      =
3.2935>>>>answer
