A solution of Sr(OH) is prepared at . Assuming complete dissociation, what is in the solution?
- 6.0×10^-8
- 6.0×10^-4
- 1.0×10^-14
- 1.2×10^-3 (correct answer)
- 3.0×10^-4
Explanation: This question tests pH and pOH of strong acids and bases. Sr(OH)₂ is a strong base that completely dissociates as Sr(OH)₂ → Sr²⁺ + 2OH⁻, producing 2 moles of OH⁻ per mole of Sr(OH)₂. With [Sr(OH)₂] = 6.0×10⁻⁴ M, we calculate [OH⁻] = 2 × 6.0×10⁻⁴ = 1.2×10⁻³ M. A common mistake is forgetting the stoichiometric coefficient and using [OH⁻] = 6.0×10⁻⁴ M (choice A). For bases containing multiple hydroxide ions, multiply the base concentration by the number of OH⁻ ions in the formula.