A 15.0 mL solution of Sr(OH)₂ is neutralized with 34.0 mL of 0.350 M HCl. What is the concentration of the original Sr(OH)₂ solution?

Answer 1

The neutralization reaction is: Sr(OH)₂ + 2 HCl -----> SrCl₂ + 2 H₂O

mol = 0.034 L x 0.35 mol/L = 0.0119 mol HCl

0.0119 mol HCl x 1 mol Sr(OH)₂/ 2 mol HCl = 0.00595 mol Sr(OH)₂

M = 0.00595 mol / 0.015 L = 0.397 M Sr(OH)₂