Respuesta :

superman1987
first, we have to get moles of HF & F- 

moles HF = molarity * volume 
 
                = 0.3 * 1 L = 0.3 moles
moles F- = molarity * volume
 
               = 0.7 m * 1L = 0.7 moles 

 by using ICE table :

               HF(aq) + NaOH(aq) → NaF(aq) + H2O(l)

initial        0.3              0                    0.7

change  -  0.08             0.08                  + 0.08

Equ      (0.3 - 0.08)                            (0.7+0.08)

∴ [HF] = 0.3 - 0.08 = 0.22 m

∴[F-] = 0.7 + 0.08 = 0.78 m 

when we have Ka for HF = 3.5 x 10 ^-4 

∴Pka = - ㏒ Ka
 
         = - ㏒ (3.5x10^-4) 

         = 3.46

by using the PH formula :

PH = Pka + ㏒[F-]/[HF]

by substitution:

∴PH = 3.46 + ㏒ 0.78 / 0.22

∴PH = 4

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