Question

The number of moles of OH\(^-\) ions present in 100 mL of 0.2 M Al(OH)\(_3\) solution is (assume 100% dissociation)

• A. \( 2 \times 10^{-2} \)
• B. \( 3 \times 10^{-2} \)
• C. \( 6 \times 10^{-2} \)
• D. \( 8 \times 10^{-2} \)

Hint:

For a compound that dissociates in water, the number of moles of ions produced is a multiple of the dissociation coefficient. In this case, Al(OH)\(_3\) produces 3 OH\(^-\) ions per formula unit.

Answer 1

The Correct Option is C

- Al(OH)\(_3\) dissociates in water as:
\[ \text{Al(OH)}_3 \rightarrow \text{Al}^{3+} + 3 \text{OH}^- \] For every 1 mole of Al(OH)\(_3\), 3 moles of OH\(^-\) are produced.
- Moles of Al(OH)\(_3\) in 100 mL (0.1 L) of 0.2 M solution:
\[ \text{Moles of Al(OH)}_3 = 0.2 \times 0.1 = 0.02 \, \text{mol} \] - Moles of OH\(^-\) ions: \[ \text{Moles of OH}^- = 0.02 \times 3 = 0.06 \, \text{mol} \] Therefore, the number of moles of OH\(^-\) ions is \( 6 \times 10^{-2} \), making option (3) the correct answer.