Final answer:
The concentration of the original Sr(OH)₂ solution is 0.3629 M.
Step-by-step explanation:
To find the concentration of the original Sr(OH)₂ solution, we can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction: Sr(OH)₂ + 2HCl ⟶ SrCl₂ + 2H₂O.
First, we calculate the moles of HCl used: 31.1 mL (0.350 M) = 10.885 mmol. Since the reaction ratio is 1:2 between Sr(OH)₂ and HCl, the moles of Sr(OH)₂ used is half of that: 10.885 mmol / 2 = 5.4425 mmol.
To find the concentration, we divide the moles of Sr(OH)₂ used by the volume of the solution: 5.4425 mmol / 15.0 mL = 0.3629 M.