[Sis290025]

HF is a weak acid (K_a = 3.52E-4). If 0.2800 mol of gaseous HF are dissolved in 0.2900 litres of water. Determine the formed HF solution's pH.


HF + H2O <-> H3O+ + F-

k_a = [x^2]/[0.947220598-x] = 3.52E-4

0.2800 mol HF*(20.006 g/1 mol HF) = 5.60168 g HF

290 g H2O + 5.60168 g HF = 295.60168 g solution = 0.29560168 L

M = 0.2800 mol HF/0.29560168 L = 0.9472206 M

x = sqrt(3.52E-4*0.947220598) = 0.0182078887 = [H3O+] assuming 0.947220598-x = 0.947220598

pH = -log(0.0182078887) = 1.73974 = 1.74

Thanks.

[Albert]

The result is correct, but I think you chose a rather complicated way of determining the concentration of hydrofluoric acid.

[Alberto_Kravina]

Yeah! It's more simple to use the formula:

pKa = - lg(Ka)
pH = 0.5 (pKa - log c_o)

[Albert]

0.2800 mol HF*(20.006 g/1 mol HF) = 5.60168 g HF

290 g H2O + 5.60168 g HF = 295.60168 g solution = 0.29560168 L

I meant this procedure.

[Sis290025]

Instead of adding the volume of water and HF, is it less erroneous to just
use the 0.2900 L in calculating the molarity?

Which is the correct way?

Thanks again.

[AWK]

The last way is sufficiently correct

[Sis290025]

Using the the 0.2900 L in the molarity formula?

[Albert]

Yes, exactly.