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A 15.0 mL solution of Sr(OH)₂ is neutralized with 36.6 mL of 0.350 M HCl.

What is the concentration of the original Sr(OH)₂ solution?

Answer :

Final Answer:

The concentration of the original Sr(OH)2 solution is approximately 0.262 M.

Explanation:

The given question involves a neutralization reaction between a solution of strontium hydroxide (Sr(OH)2) and hydrochloric acid (HCl). In a neutralization reaction, an acid and a base react to form a salt and water. The balanced chemical equation for this reaction is:

Sr(OH)2 + 2HCl -> SrCl2 + 2H2O

We're given the volume and concentration of the HCl solution used in the neutralization (36.6 mL of 0.350 M HCl) and the volume of the Sr(OH)2 solution (15.0 mL).

Using the stoichiometry of the balanced equation, we can determine the moles of HCl that reacted, and subsequently the moles of Sr(OH)2 that were neutralized. Then, by dividing the moles of Sr(OH)2 by the initial volume of the solution (15.0 mL), we can find the concentration of the original Sr(OH)2 solution.

Calculations:

  1. Moles of HCl = (0.350 mol/L) * (0.0366 L) = 0.1281 mol
  2. Moles of Sr(OH)2 = (0.1281 mol HCl) * (1 mol Sr(OH)2 / 2 mol HCl) = 0.0641 mol
  3. Concentration of Sr(OH)2 = 0.0641 mol / 0.0150 L = 0.262 M

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