Final answer:
Approximately 3.934 mL of 0.381 M hydroiodic acid is required to neutralize 12.0 mL of 0.499 M strontium hydroxide, using the stoichiometry of the neutralization reaction.
Step-by-step explanation:
The question involves a titration problem in which hydroiodic acid (HI) is used to neutralize strontium hydroxide (Sr(OH)2). To find the volume of HI required, we need to establish the stoichiometry of the neutralization reaction. The balanced equation is 2HI (aq) + Sr(OH)2 (aq) → SrI2 (aq) + 2H2O (l), indicating a 2:1 molar ratio of HI to Sr(OH)2. Using the formula M1V1 = M2V2 (where M and V refer to molarity and volume), the molarity of Sr(OH)2 is 0.499 M and the volume is 12.0 mL. To neutralize this, the volume of HI needed (V1) is calculated as:
V1 = (M2V2 / M1) × stoichiometric ratio
V1 = (0.499 M × 12.0 mL / 0.381 M) × (1/2)
V1 = (5.988 mL / 0.381 M) × 1/2
V1 = 7.868 mL × (1/2)
V1 = 3.934 mL
Therefore, approximately 3.934 mL of 0.381 M hydroiodic acid is required to neutralize 12.0 mL of 0.499 M strontium hydroxide.