A 15.0 mL solution of Sr(OH)2 is neutralized with 29.8 mL of 0.350 M HCI. What is the concentration of the original Sr(OH)2 solution?

[Sr(OH)₂] = 0.699 M  in Sr(OH)₂ sol'n  (1 sig. fig)  

Explanation:

Concentrated-1 Sr(OH)₂ = M₁ = unknown  Concentration-2 HCl = M₂ = 0.350M          

Volume of Concentrate = V₁ = 15ml           Volume of Diluted = V₂ = 29.8ml

Using the Dilution Equation

Molarity  x  Volume of Concnt'd Soln = Molarity x Volume of Diluted Soln

From definition of molarity (M) = moles solute / volume of solution in Liters

=> moles solute = Molarity x Volume in liters*

*Note: if volume units are the same on both sides of the dilution equation then one may retain the volume in ml.

M₁ x V₁ = M₂ x V₂      

=> 15ml ·  [Sr(OH)₂] = 30ml X 0.350M HCl soln  

=>  [Sr(OH)₂] = 29.8ml X 0.350M / 15ml = 0.695333... (calculator answer)

=> [Sr(OH)₂] = 0.7 M  in Sr(OH)₂ sol'n  (1 sig. fig)  

work check:

M₁ x V₁ = M₂ x V₂

0.6953M x 15ml = 0.35M x 29.8ml

10.43 moles = 10.43 moles                    QED

i belive it is water