A 15.0 mL solution of Sr(OH)₂ is neutralized with 50.9 mL of 0.350 M HCl. What is the concentration of the original Sr(OH)₂ solution?

Answer 1

To determine the concentration of the original Sr(OH)₂ solution, we can use the concept of stoichiometry and the volume and concentration of the HCl solution.

The balanced equation for the neutralization reaction between Sr(OH)₂ and HCl is:

Sr(OH)₂ + 2HCl → SrCl₂ + 2H₂O

First, let's calculate the number of moles of HCl used in the reaction:

Moles of HCl = Volume of HCl solution (in liters) * Concentration of HCl solution

Volume of HCl solution = 50.9 mL = 50.9/1000 = 0.0509 L (converted to liters)

Moles of HCl = 0.0509 L * 0.350 mol/L = 0.0178 mol

From the balanced equation, we can see that the stoichiometric ratio between Sr(OH)₂ and HCl is 1:2. This means that 1 mole of Sr(OH)₂ reacts with 2 moles of HCl.

Since 2 moles of HCl react with 1 mole of Sr(OH)₂, the number of moles of Sr(OH)₂ can be calculated as half the number of moles of HCl:

Moles of Sr(OH)₂ = 0.0178 mol / 2 = 0.0089 mol

Now, let's calculate the concentration of the original Sr(OH)₂ solution:

Concentration of Sr(OH)₂ solution = Moles of Sr(OH)₂ / Volume of Sr(OH)₂ solution (in liters)

Volume of Sr(OH)₂ solution = 15.0 mL = 15.0/1000 = 0.015 L (converted to liters)

Concentration of Sr(OH)₂ solution = 0.0089 mol / 0.015 L = 0.593 M

Therefore, the concentration of the original Sr(OH)₂ solution is approximately 0.593 M.