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
