Question

Calculate the ph of a buffer solution made from 0.30 m hydrofluoric acid and 0.70 m sodium fluoride after the addition of 0.08 moles of naoh to 1 l of this solution

Answer 1

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