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Answer :
Final answer:
To find the concentration of the original Sr(OH)2 solution, we first determine the moles of HCl used in the reaction, then use the stoichiometry of the neutralization reaction to find the moles of Sr(OH)2, and finally calculate its molarity. The original concentration of the Sr(OH)2 solution is 0.449 M.
Explanation:
To calculate the concentration of the original Sr(OH)2 solution, let's use the titration data provided:
- Volume of HCl used: 38.5 mL
- Molarity of HCl: 0.350 M
- Volume of Sr(OH)2 solution: 15.0 mL
First, we need to determine the moles of HCl used in the titration:
Moles of HCl = Molarity of HCl × Volume of HCl in liters
= 0.350 mol/L × 0.0385 L
= 0.013475 mol
Next, we use the chemical reaction for the neutralization of Strontium hydroxide by hydrochloric acid, which is:
Sr(OH)2 + 2 HCl → SrCl2 + 2 H2O
This equation shows us that 1 mole of Sr(OH)2 reacts with 2 moles of HCl. Therefore, we can calculate the moles of Sr(OH)2 that were present:
Moles of Sr(OH)2 = Moles of HCl ÷ 2
= 0.013475 mol ÷ 2
= 0.0067375 mol
Finally, to find the molarity of the original Sr(OH)2 solution, divide the moles of Sr(OH)2 by the volume of Sr(OH)2 solution in liters:
Molarity of Sr(OH)2 = Moles of Sr(OH)2 ÷ Volume of Sr(OH)2 in liters
= 0.0067375 mol ÷ 0.015 L
= 0.44917 M
The original concentration of the Sr(OH)2 solution is therefore 0.449 M.
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