no of moles of HF = molarity * volume in L
                          Â
= 0.382*1 = 0.382moles
no of moles of NaFÂ Â = molarity * volume in L
                             Â
= 0.509*1 =- 0.509moles
PHÂ Â = Pka + log[NaF]/[HF]
     = 3.17+ log0.509/0.382
     = 3.17+ 0.1246 = 3.2946
by the addition of 0.191 moles of NaOH
no of moles of HF = 0.382-0.191 = 0.191moles
no of moles of NaF = 0.509+0.191 = 0.7 moles
PHÂ Â = Pka + log[NaF]/[HF]
     = 3.17 + log0.7/0.191
     = 3.17+ 0.564  =
3.734
Raise the pH slightly
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