Calculate the pH and the molarities of all species (H₃AsO₄, H₂AsO₃^–, HAsO₄^2–, AsO₄^3–, H₃O⁺, and OH^–) in a 9.50 % (by mass) solution of arsenic acid with a density of 1.1033 g/mL. The equilibrium constants are:
K_a1 = 6.46×10^–3K_a2 = 1.14×10^–7
K_a3 = 3.16×10^–12

• A. [H₃AsO₄] = 0.672 M
  [H₂AsO₃^–] = 0.0659 M
  [HAsO₄^2–] = 1.14×10^–7 M
  [AsO₄^3–] = 5.47×10^–18 M
  [H₃O⁺] = 0.0659 M
  [OH^–] = 2.52×10^–13 M
  pH = 2.181
• B. [H₃AsO₄] = 0.672 M
  [H₂AsO₃^–] = 0.0659 M
  [HAsO₄^2–] = 1.14×10^–7 M
  [AsO₄^3–] = 5.47×10^–18 M
  [H₃O⁺] = 0.0659 M
  [OH^–] = 1.52×10^–13 M
  pH = 1.181
• C. [H₃AsO₄] = 0.714 M
  [H₂AsO₃^–] = 0.0812 M
  [HAsO₄^2–] = 1.14×10^–7 M
  [AsO₄^3–] = 5.47×10^–18 M
  [H₃O⁺] = 0.0659 M
  [OH^–] = 2.52×10^–13 M
  pH = 2.181
• D. [H₃AsO₄] = 0.714 M
  [H₂AsO₃^–] = 0.0812 M
  [HAsO₄^2–] = 1.14×10^–7 M
  [AsO₄^3–] = 6.18×10^–18 M
  [H₃O⁺] = 0.0812 M
  [OH^–] = 2.52×10^–13 M
  pH = 3.181

Answer: (B)