In: Chemistry
Calculate the pH of a 25 mg/L solution of hydrofluroic acid.
HF = H+ + F- Ka=10^-3.2
25 mg / L = 25 x 10^-3 g / L
molarity = 25 x 10^-3 / 20 mol / L
= 0.00125 M
HF <------------------> H+ + F-
0.00125 0 0 -------------> initial
-x +x +x ------------> change
0.00125-x x x -------------> equilibrium
Ka = [H+][F-]/[HF]
10^-32 = x^2 / 0.00125 - x
x^2 + 6.31 x 10^-4 x - 7.89 x 10^-7 = 0
x = 6.27 x 10^-4
x = [H+] = 6.27 x 10^-4 M
pH = -log [H+
pH = -log (6.27 x 10^-4 )
pH = 3.20