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).

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

The answer is 323 g.