Consider HA - , the intermediate form of a diprotic acid. K a for this species i

Question

Consider HA-, the intermediate form of a diprotic acid. Ka for this species is 10-4 and Kb is 10-8. Nonetheless, the Ka and Kb reactions proceed to nearly the same extent when NaHA is dissolved in water. Explain. (Select all that apply.)

Consider HA-, the intermediate form of a diprotic acid. Ka for this species is 10-4 and Kb is 10-8. Nonetheless, the Ka and Kb reactions proceed to nearly the same extent when NaHA is dissolved in water. Explain. The Kb reaction, with a much greater equilibrium constant than Ka, releases H+: HA- equilibrium reaction arrow H+ + A2- Kb The Ka reaction, with a much greater equilibrium constant than Kb, releases H+: HA- equilibrium reaction arrowH+ + A2- Ka The net result is that the Kb reaction is driven almost as far towards completion as the Ka reaction. The net result is that the Ka reaction is driven almost as far towards completion as the Kb reaction. Each mole of H+ reacts with one mole of OH - from the Kb reaction: HA- + H2O equilibrium reaction arrow H2A + OH -.Each mole of H+ reacts with one mole of OH - from the Ka reaction: HA- + H2O equilibrium reaction arrow H2A + OH -.

Explanation / Answer

The Kb reaction has a smaller equilibrium constant, thus it will not be favored over the Ka reaction. Ka is the acid dissociation constant, and will result in release of an H+ ion. The second to last choice is also correct, as H+ ions will react with the Kb reactions product, OH-. Thus, the Kb constant, though it has a smaller equilibrium constant, will proceed almost as far as the Ka reaction.

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