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 × 10–4, Ka2 = 5.6 × 10–8, and Ka3 = 3 × 10–13. 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) → Na+(aq) + H2AsO4–(aq).

On the product side of the equation, aqueous dihydrogen arsenate (H2AsO4–) has a Ka of 5.6 × 10–8 which has already been mentioned above (as Ka2). Yet, since H2AsO4– 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 × 10–4 (Ka1). Is Ka or Kb greater for H2AsO4–?

A. Kb of H2AsO4– = 3.0 × 10–2 and so its Kb > Ka

B. Kb of H2AsO4– = 2.5 × 10–4 and so its Kb > Ka

C. Kb of H2AsO4– = 4.0 × 10–11 and so its Ka > Kb

D. Kb of H2AsO4– = 5.6 × 10–8 and so its Ka = Kb

E. Kb of H2AsO4– = 1.8 × 10–7 and so its Kb > Ka