pH of 0.100 M HF

• x2 + 7.0 x 10-4 x - 7.0 x 10-5 = 0

• -7.0 x 10-4 + [(7.0 x 10-4)2 - (4)(1)(-7.0 x 10-5)]1/2

  x = ———————————————————
  
  (2)(1)

• x = 0.00802

• [H3O+] = [F-] = 8.02 x 10-3 M

• [HF] = 0.100 - 0.00802 = 0.0920 M

• [H3O+][F-]/[HF] = (8.02 x 10-3)2/0.0920 = 6.99 x 10- 4 which compares well to Ka = 7.0 x 10-4, so the calculation is correct

• pH = - log (8.02 x 10-3) = 2.096 (or 2.10)

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