Question

Identify P and Q in the following reaction: $$ \text{P} + \text{Q} \rightarrow [\text{B(OH)}_4]^– + \text{H}_3\text{O}^+ $$

• A. \( \text{P} = \text{H}_3\text{BO}_3,\ \text{Q} = 3\text{H}_2\text{O} \)
• B. \( \text{P} = \text{H}_3\text{BO}_3,\ \text{Q} = 2\text{H}_2\text{O} \)
• C. \( \text{P} = \text{HBO}_2,\ \text{Q} = 2\text{H}_2\text{O} \)
• D. \( \text{P} = 2\text{H}_3\text{BO}_3,\ \text{Q} = \text{H}_2\text{O} \)

Hint:

Boric acid is not proton-donating; it acts as a Lewis acid by accepting OH⁻ from water, forming \( \text{B(OH)}_4^- \).

Answer 1

The Correct Option is B

Boric acid (\( \text{H}_3\text{BO}_3 \)) is a weak monobasic acid that behaves as a Lewis acid by accepting hydroxide ions. In aqueous solution: \[ \text{H}_3\text{BO}_3 + 2\text{H}_2\text{O} \rightleftharpoons [\text{B(OH)}_4]^– + \text{H}_3\text{O}^+ \] This explains why Q is 2 water molecules and P is \( \text{H}_3\text{BO}_3 \).