Farmers who raise cotton once used arsenic acid, H3AsO4, as a defoliant at harvest time. Arsenic acid is a polyprotic acid with Ka1 = 2.5 Γ 10β4, Ka2 = 5.6 Γ 10β8, and Ka3 = 3 Γ 10β13. Suppose that in your lab, you have a small sample of sodium dihydrogen arsenate, NaH2AsO4. It is a colorless (clear) solid with very low solubility in water at room temperature. When it does dissolve appreciably at an elevated temperature, it forms
NaH2AsO4(aq) β Na+(aq) + H2AsO4β(aq).
On the product side of the equation, aqueous dihydrogen arsenate (H2AsO4β) has a Ka of 5.6 Γ 10β8 which has already been mentioned above (as Ka2). Yet, since H2AsO4β itself can accept a proton (H+), it can act as a weak base as well. Calculate its Kb given Ka of H3AsO4 to be 2.5 Γ 10β4 (Ka1). Is Ka or Kb greater for H2AsO4β?
A. Kb of H2AsO4β = 3.0 Γ 10β2 and so its Kb > Ka
B. Kb of H2AsO4β = 2.5 Γ 10β4 and so its Kb > Ka
C. Kb of H2AsO4β = 4.0 Γ 10β11 and so its Ka > Kb
D. Kb of H2AsO4β = 5.6 Γ 10β8 and so its Ka = Kb
E. Kb of H2AsO4β = 1.8 Γ 10β7 and so its Kb > Ka
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