A 15.0 mL solution of Sr(OH)₂ is neutralized with 22.8 mL of 0.350 M HCl. What is the concentration of the original Sr(OH)₂ solution?

Answer 1

This is a titration problem. You need to use the equation M₁V₁ = M₂V₂.

M₁ = ?

V₁ = 15.0mL

M₂ = 0.350M

V₂ = 22.8mL

You need to solve for M₁. M₁ = M₂V₂/V1 = (0.350M)(22.8mL)/15.0mL = 0.532M Sr(OH)₂. Now check to see if this answer makes sense. It took 22.8mL of your HCl solution to neutralize 15.0mL of your Sr(OH)₂ solution. This would mean that your Sr(OH)₂ solution is MORE concentrated than your HCl solution and that is exactly what we found.