Question

Determine the pH of an HF solution of each of the following concentrations. 0.270 M, 4.90×10−2 M, 2.30×10−2 M

Answer 1

HF is a weak acid

so

HF <-> H+ + F-

Ka = [H+][F-]/[HF]

for HF = 7.2*10^-4

so

[H+] = [F-] = x in equilibrium

[HF] = M-x

substituting inf Ka

Ka = x*x/(M-x)

7.2*10^-4 = x^2 / (M-x)

x^2 + Ka*x -MKa = 0

solve for each

M = 0.27

x^2 + Ka*x -0.27*7.2*10^-4 = 0

x = H+ = 0.0135

ph = -log(0.0135) = 1.8696

x^2 + Ka*x -4.9*10^-2*7.2*10^-4 = 0

x = H+ = 0.00559

pH = -log(0.00559) = 2.2525

x^2 + Ka*x - 2.3*10^-2*7.2*10^-4 = 0

x = H+ = 0.0037

pH = -log(0.0037) = 2.4317