no of moles of HF = molarity * volume in L
                          Â
= 0.25*1.5Â Â = 0.375 moles
no of moles of NaF = molarity * volume in L
                             Â
= 0.25*1.5Â Â = 0.375 moles
By the addition of NaOH
no of moles of HF = 0.375-0.05Â Â = 0.325 moles
no of moles of NaF = 0.375 + 0.05 = 0.425 moles
   Pka = -logKa
           Â
= -log3.5*10-4Â Â = 3.4559
PHÂ Â = PKa + log[NaF]/[HF]
        = 3.4559 +
log0.425/0.325
        = 3.4559 +
0.1165Â Â = 3.5724
