Question

Calculate the ph of a buffer consisting of 0.50 m hf and 0.45 m kf (a) before and (b) after addition of 0.40 g of naoh to 1.0 l of the buffer (ka of hf = 6.8×10–4).

Answer 1

part A): According to the reaction equation before adding NaOH:

by using the ICE table:

HF ↔ H+ + F-

initial 0.5 0 0.45

change -X +X +X

Equ (0.5-X) X (0.45+X)

when Ka expression = concentration of products/ concentration of reactants

= [H+][F-] / [HF]
when Ka = 6.8 x 10^-4

and we assumed that [H+] = X

and [F-] = (0.45+X)

and [HF] = (0.5-X)

so, by substitution:

6.8 x 10^-4 = X * (0.45+X) / (0.5-X) by solving for X

∴ X = 7.5 x 10^-4

when [H+] = X = 7.5 x 10^-4

∴ PH = -㏒[H+]

= -㏒ 7.5 x 10^-4

= 3.12

part B): according to the reaction equation after adding NaOH:

first, we need to have the moles of NaOH which is added:

the initial moles of NaOH = mass/molar mass of NaOH

= 0.4 g / 39.997 g/mol

= 0.01 moles

then according to the reaction equation after adding 0.01 M of NaOH:

and by using the ICE table:

HF(aq) + NaOH(aq) ↔ H2O(l) + NaF(aq)

initial 0.5 0.01 0.45

change - 0.01 -0.01 +0.01

Equ 0.49 0 0.46

when Ka = 6.8 x 10^-4 so we can get Pka:

Pka = -㏒Ka

= -㏒ 6.8 x 10^-4
= 3.17

by using H-H equation we can get the PH:

when PH = Pka + ㏒[Salt/acid]

PH = Pka +㏒ [NaF] /[ HF]

∴ PH = 3.17 + ㏒(0.46/0.49)

= 3.14